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+    <title>
+      The Project Gutenberg eBook of Amusements In Mathematics, by Henry Ernest Dudeney.
+    </title>
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+<pre>
+
+Project Gutenberg's Amusements in Mathematics, by Henry Ernest Dudeney
+
+This eBook is for the use of anyone anywhere at no cost and with
+almost no restrictions whatsoever.  You may copy it, give it away or
+re-use it under the terms of the Project Gutenberg License included
+with this eBook or online at www.gutenberg.org
+
+
+Title: Amusements in Mathematics
+
+Author: Henry Ernest Dudeney
+
+Release Date: September 17, 2005 [EBook #16713]
+[Most recently updated: October 28, 2018]
+
+Language: English
+
+Character set encoding: ISO-8859-1
+
+*** START OF THIS PROJECT GUTENBERG EBOOK AMUSEMENTS IN MATHEMATICS ***
+
+
+
+
+Produced by Stephen Schulze, Jonathan Ingram and the Online
+Distributed Proofreading Team at https://www.pgdp.net
+
+
+
+
+
+
+</pre>
+
+
+<div style="background: #EEEEEE; padding: 2em; border: dashed 1px;"><p>Transcribers note: Many of the puzzles in this book assume a
+familiarity with the currency of Great Britain in the early 1900s. As
+this is likely not common knowledge for those outside Britain (and possibly
+many within,) I am including a chart of relative values.</p>
+
+<p>The most common units used were:</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='left'>the Penny,</td><td align='left'>abbreviated: <i>d</i>. (from the Roman penny, denarius)</td></tr>
+<tr><td align='left'>the Shilling,</td><td align='left'>abbreviated: <i>s</i>.</td></tr>
+<tr><td align='left'>the Pound,</td><td align='left'>abbreviated: �</td></tr>
+</table></div>
+
+<p>There was 12 Pennies to a Shilling and 20 Shillings to a Pound, so there
+was 240 Pennies in a Pound.</p>
+
+<p>To further complicate things, there were many coins which were various
+fractional values of Pennies, Shillings or Pounds.</p>
+
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='left'>Farthing</td><td align='right'>�<i>d</i>.</td></tr>
+<tr><td align='left'>Half-penny</td><td align='right'>�<i>d</i>.</td></tr>
+<tr><td align='left'>Penny</td><td align='right'>1<i>d</i>.</td></tr>
+<tr><td align='left'>Three-penny</td><td align='right'>3<i>d</i>.</td></tr>
+<tr><td align='left'>Sixpence (or tanner)</td><td align='right'>6<i>d</i>.</td></tr>
+<tr><td align='left'>Shilling (or bob)</td><td align='right'>1<i>s</i>.</td></tr>
+<tr><td align='left'>Florin or two shilling piece</td><td align='right'>2<i>s</i>.</td></tr>
+<tr><td align='left'>Half-crown (or half-dollar)</td><td align='right'>2<i>s</i>. 6<i>d</i>.</td></tr>
+<tr><td align='left'>Double-florin</td><td align='right'>4<i>s</i>.</td></tr>
+<tr><td align='left'>Crown (or dollar)</td><td align='right'>5<i>s</i>.</td></tr>
+<tr><td align='left'>Half-Sovereign</td><td align='right'>10<i>s</i>.</td></tr>
+<tr><td align='left'>Sovereign (or Pound)</td><td align='right'>�1 or 20<i>s</i>.</td></tr>
+</table></div>
+
+<p>This is by no means a comprehensive list, but it should be adequate to
+solve the puzzles in this book.</p></div>
+
+
+
+
+<h1>AMUSEMENTS IN MATHEMATICS</h1>
+
+<h4><i>by</i></h4>
+
+<h2>HENRY ERNEST DUDENEY</h2>
+
+<table summary=''><tr><td>
+<div class="poem center"><div class="stanza">
+<span class="i0">In Mathematicks he was greater<br /></span>
+<span class="i0">Than Tycho Brahe or Erra Pater:<br /></span>
+<span class="i0">For he, by geometrick scale,<br /></span>
+<span class="i0">Could take the size of pots of ale;<br /></span>
+<span class="i0">Resolve, by sines and tangents, straight,<br /></span>
+<span class="i0">If bread or butter wanted weight;<br /></span>
+<span class="i0">And wisely tell what hour o' th' day<br /></span>
+<span class="i0">The clock does strike by algebra.<br /></span>
+</div><div class="stanza">
+<span class="i8">BUTLER'S <i>Hudibras</i>.<br /></span>
+</div></div>
+</td></tr></table>
+
+
+<h5>1917</h5>
+
+<hr style="width: 65%;" /><p><span class='pagenum'>Pg v<a name="Page_v" id="Page_v"></a></span></p>
+<h2><a name="PREFACE" id="PREFACE"></a>PREFACE</h2>
+
+<p>In issuing this volume of my Mathematical Puzzles, of which some have
+appeared in periodicals and others are given here for the first time,
+I must acknowledge the encouragement that I have received from many
+unknown correspondents, at home and abroad, who have expressed a
+desire to have the problems in a collected form, with some of the
+solutions given at greater length than is possible in magazines and
+newspapers. Though I have included a few old puzzles that have
+interested the world for generations, where I felt that there was
+something new to be said about them, the problems are in the main
+original. It is true that some of these have become widely known
+through the press, and it is possible that the reader may be glad to
+know their source.</p>
+
+<p>On the question of Mathematical Puzzles in general there is, perhaps,
+little more to be said than I have written elsewhere. The history of
+the subject entails nothing short of the actual story of the
+beginnings and development of exact thinking in man. The historian
+must start from the time when man first succeeded in counting his ten
+fingers and in dividing an apple into two approximately equal parts.
+Every puzzle that is worthy of consideration can be referred to
+mathematics and logic. Every man, woman, and child who tries to
+"reason out" the answer to the simplest puzzle is working, though not
+of necessity consciously, on mathematical lines. Even those puzzles
+that we have no way of attacking except by haphazard attempts can be
+brought under a method of what has been called "glorified trial"&mdash;a
+system of shortening our labours by avoiding or eliminating what our
+reason tells us is useless. It is, in fact, not easy to say sometimes
+where the "empirical" begins and where it ends.</p>
+
+<p>When a man says, "I have never solved a puzzle in my life," it is
+difficult to know exactly what he means, for every intelligent
+individual is doing it every day. The unfortunate inmates of our
+lunatic asylums are sent there expressly because they cannot solve
+puzzles&mdash;because they have lost their powers of reason. If there were
+no puzzles to solve, there would be no questions to ask; and if there
+were no questions to be asked, what a world it would be! We should all
+be equally omniscient, and conversation would be useless and idle.</p>
+
+<p>It is possible that some few exceedingly sober-minded mathematicians,
+who are impatient of any terminology in their favourite science but
+the academic, and who object to the elusive <i>x</i> and <i>y</i> appearing
+under any other names, will have wished that various problems had
+been presented in a less popular dress and introduced with a less
+flippant phraseology. I can only refer them to the first word of my
+title and remind them that we are primarily out to be amused&mdash;not, it
+is true, without some hope of picking up morsels of knowledge by the
+way. If the manner is light, I can only say, in the words of
+Touchstone, that it is "an ill-favoured thing, sir, but my own; a poor
+humour of mine, sir."</p>
+
+<p>As for the question of difficulty, some of the puzzles, especially in
+the Arithmetical and Algebraical category, are quite easy. Yet some of
+those examples that look the simplest should not be passed over
+without a little consideration, for now and again it will be found
+that there is some more or less subtle pitfall or trap into which the
+reader may be apt to fall. It is good exercise to cultivate the habit
+of being very wary over the exact wording of a puzzle. It teaches
+exactitude and caution. But some of the problems are very hard nuts
+indeed, and not unworthy of the attention of the advanced
+mathematician. Readers will doubtless select according to their
+individual tastes.</p>
+
+<p>In many cases only the mere answers are given. This leaves the
+beginner something to do on his own behalf in working out the method
+of solution, and saves space that would be wasted from the point of
+view of the advanced student. On the other hand, in particular cases
+where it seemed likely to interest, I have given rather extensive
+solutions and treated problems in a general manner. It will often be
+found that the notes on one problem will serve to elucidate a good
+many others in the book; so that the reader's difficulties will
+sometimes be found cleared up as he advances. Where it is possible to
+say a thing in a manner that may be "understanded of the people"
+generally, I prefer to use this simple phraseology, and so engage the
+attention and interest of a larger public. The mathematician will in
+such cases have no difficulty in expressing the matter under
+consideration in terms of his familiar symbols.</p>
+
+<p>I have taken the greatest care in reading the proofs, and trust that
+any errors that may have crept in are very few. If any such should
+occur, I can only plead, in the words of Horace, that "good Homer
+sometimes nods," or, as the bishop put it, "Not even the youngest
+curate in my diocese is infallible."</p>
+
+<p>I have to express my thanks in particular to the proprietors of <i>The
+Strand Magazine</i>, <i>Cassell's Magazine</i>, <i>The Queen</i>, <i>Tit-Bits</i>, and
+<i>The Weekly Dispatch</i> for their courtesy in allowing me to reprint
+some of the puzzles that have appeared in their pages.</p>
+
+<p class='right'>THE AUTHORS' CLUB<br/>
+<i>March</i> 25, 1917</p>
+
+<hr style="width: 65%;" />
+<h2><a name="CONTENTS" id="CONTENTS"></a>CONTENTS</h2>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='left'><a href="#PREFACE"><b>PREFACE</b></a></td><td align='right'><a href="#Page_v">v</a></td></tr>
+<tr><td align='left'><a href="#ARITHMETICAL_AND_ALGEBRAICAL_PROBLEMS"><b>ARITHMETICAL AND ALGEBRAICAL PROBLEMS.</b></a></td><td align='right'><a href="#Page_1">1</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#MONEY_PUZZLES"><b>Money Puzzles.</b></a></span></td><td align='right'><a href="#Page_1">1</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#AGE_AND_KINSHIP_PUZZLES"><b>Age and Kinship Puzzles.</b></a></span></td><td align='right'><a href="#Page_6">6</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#CLOCK_PUZZLES"><b>Clock Puzzles.</b></a></span></td><td align='right'><a href="#Page_9">9</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#LOCOMOTION_AND_SPEED_PUZZLES"><b>Locomotion and Speed Puzzles.</b></a></span></td><td align='right'><a href="#Page_11">11</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#DIGITAL_PUZZLES"><b>Digital Puzzles.</b></a></span></td><td align='right'><a href="#Page_13">13</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#VARIOUS_ARITHMETICAL_AND_ALGEBRAICAL_PROBLEMS"><b>Various Arithmetical and Algebraical Problems.</b></a></span></td><td align='right'><a href="#Page_17">17</a></td></tr>
+<tr><td align='left'><a href="#GEOMETRICAL_PROBLEMS"><b>GEOMETRICAL PROBLEMS.</b></a></td><td align='right'><a href="#Page_27">27</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#DISSECTION_PUZZLES"><b>Dissection Puzzles.</b></a></span></td><td align='right'><a href="#Page_27">27</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#GREEK_CROSS_PUZZLES"><b>Greek Cross Puzzles.</b></a></span></td><td align='right'><a href="#Page_28">28</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#VARIOUS_DISSECTION_PUZZLES"><b>Various Dissection Puzzles.</b></a></span></td><td align='right'><a href="#Page_35">35</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#PATCHWORK_PUZZLES"><b>Patchwork Puzzles</b></a></span></td><td align='right'><a href="#Page_46">46</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#VARIOUS_GEOMETRICAL_PUZZLES"><b>Various Geometrical Puzzles.</b></a></span></td><td align='right'><a href="#Page_49">49</a></td></tr>
+<tr><td align='left'><a href="#POINTS_AND_LINES_PROBLEMS"><b>POINTS AND LINES PROBLEMS.</b></a></td><td align='right'><a href="#Page_56">56</a></td></tr>
+<tr><td align='left'><a href="#MOVING_COUNTER_PROBLEMS"><b>MOVING COUNTER PROBLEMS.</b></a></td><td align='right'><a href="#Page_58">58</a></td></tr>
+<tr><td align='left'><a href="#UNICURSAL_AND_ROUTE_PROBLEMS"><b>UNICURSAL AND ROUTE PROBLEMS.</b></a></td><td align='right'><a href="#Page_68">68</a></td></tr>
+<tr><td align='left'><a href="#COMBINATION_AND_GROUP_PROBLEMS"><b>COMBINATION AND GROUP PROBLEMS.</b></a></td><td align='right'><a href="#Page_76">76</a></td></tr>
+<tr><td align='left'><a href="#CHESSBOARD_PROBLEMS"><b>CHESSBOARD PROBLEMS.</b></a></td><td align='right'><a href="#Page_85">85</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#THE_CHESSBOARD"><b>The Chessboard.</b></a></span></td><td align='right'><a href="#Page_85">85</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#STATICAL_CHESS_PUZZLES"><b>Statical Chess Puzzles.</b></a></span></td><td align='right'><a href="#Page_88">88</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#THE_GUARDED_CHESSBOARD"><b>The Guarded Chessboard.</b></a></span></td><td align='right'><a href="#Page_95">95</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#DYNAMICAL_CHESS_PUZZLES"><b>Dynamical Chess Puzzles.</b></a></span></td><td align='right'><a href="#Page_96">96</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#VARIOUS_CHESS_PUZZLES"><b>Various Chess Puzzles.</b></a></span></td><td align='right'><a href="#Page_112">112</a></td></tr>
+<tr><td align='left'><a href="#MEASURING_WEIGHING_AND_PACKING_PUZZLES"><b>MEASURING, WEIGHING, AND PACKING PUZZLES.</b></a></td><td align='right'><a href="#Page_109">109</a></td></tr>
+<tr><td align='left'><a href="#CROSSING_RIVER_PROBLEMS"><b>CROSSING RIVER PROBLEMS</b></a></td><td align='right'><a href="#Page_112">112</a></td></tr>
+<tr><td align='left'><a href="#PROBLEMS_CONCERNING_GAMES"><b>PROBLEMS CONCERNING GAMES.</b></a></td><td align='right'><a href="#Page_114">114</a></td></tr>
+<tr><td align='left'><a href="#PUZZLE_GAMES"><b>PUZZLE GAMES.</b></a></td><td align='right'><a href="#Page_117">117</a></td></tr>
+<tr><td align='left'><a href="#MAGIC_SQUARE_PROBLEMS"><b>MAGIC SQUARE PROBLEMS.</b></a></td><td align='right'><a href="#Page_119">119</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#SUBTRACTING_MULTIPLYING_AND_DIVIDING_MAGICS"><b>Subtracting, Multiplying, and Dividing Magics.</b></a></span></td><td align='right'><a href="#Page_124">124</a></td></tr>
+<tr><td align='left'><span style="margin-left: 2em;"><a href="#MAGIC_SQUARES_OF_PRIMES"><b>Magic Squares of Primes.</b></a></span></td><td align='right'><a href="#Page_125">125</a></td></tr>
+<tr><td align='left'><a href="#MAZES_AND_HOW_TO_THREAD_THEM"><b>MAZES AND HOW TO THREAD THEM.</b></a></td><td align='right'><a href="#Page_127">127</a></td></tr>
+<tr><td align='left'><a href="#THE_PARADOX_PARTY"><b>THE PARADOX PARTY.</b></a></td><td align='right'><a href="#Page_137">137</a></td></tr>
+<tr><td align='left'><a href="#UNCLASSIFIED_PROBLEMS"><b>UNCLASSIFIED PROBLEMS.</b></a></td><td align='right'><a href="#Page_142">142</a></td></tr>
+<tr><td align='left'><a href="#SOLUTIONS"><b>SOLUTIONS.</b></a></td><td align='right'><a href="#Page_148">148</a></td></tr>
+<tr><td align='left'><a href="#INDEX"><b>INDEX.</b></a></td><td align='right'><a href="#Page_253">253</a></td></tr>
+</table></div>
+
+<hr style="width: 65%;" /><p><span class='pagenum'>Pg 1<a name="Page_1" id="Page_1"></a></span></p>
+<h1>AMUSEMENTS IN MATHEMATICS.</h1>
+
+
+<hr style="width: 65%;" />
+<h2><a name="ARITHMETICAL_AND_ALGEBRAICAL_PROBLEMS" id="ARITHMETICAL_AND_ALGEBRAICAL_PROBLEMS"></a><a href="#CONTENTS">ARITHMETICAL AND ALGEBRAICAL PROBLEMS.</a></h2>
+
+<p class='center'>
+<span style="margin-left: 7em;">"And what was he?</span><br />
+<span style="margin-left: 0em;">Forsooth, a great arithmetician."</span><br />
+<span style="margin-left: 10em;"><i>Othello</i>, I. i.</span><br />
+</p>
+
+<p>The puzzles in this department are roughly thrown together in classes
+for the convenience of the reader. Some are very easy, others quite
+difficult. But they are not arranged in any order of difficulty&mdash;and
+this is intentional, for it is well that the solver should not be
+warned that a puzzle is just what it seems to be. It may, therefore,
+prove to be quite as simple as it looks, or it may contain some
+pitfall into which, through want of care or over-confidence, we may
+stumble.</p>
+
+<p>Also, the arithmetical and algebraical puzzles are not separated in
+the manner adopted by some authors, who arbitrarily require certain
+problems to be solved by one method or the other. The reader is left
+to make his own choice and determine which puzzles are capable of
+being solved by him on purely arithmetical lines.</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="MONEY_PUZZLES" id="MONEY_PUZZLES"></a><a href="#CONTENTS">MONEY PUZZLES.</a></h2>
+
+<p class='center'>
+<span style="margin-left: 0em;">"Put not your trust in money, but put your money in trust."<br /></span>
+<span style="margin-left: 10em;">OLIVER WENDELL HOLMES.<br /></span>
+</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_1_A_POST-OFFICE_PERPLEXITY" id="X_1_A_POST-OFFICE_PERPLEXITY"></a><a href="#X_1_A_POST-OFFICE_PERPLEXITYa"><b>1.&mdash;A POST-OFFICE PERPLEXITY.</b></a></p>
+
+<p>In every business of life we are occasionally perplexed by some chance
+question that for the moment staggers us. I quite pitied a young lady
+in a branch post-office when a gentleman entered and deposited a crown
+on the counter with this request: "Please give me some twopenny
+stamps, six times as many penny stamps, and make up the rest of the
+money in twopence-halfpenny stamps." For a moment she seemed
+bewildered, then her brain cleared, and with a smile she handed over
+stamps in exact fulfilment of the order. How long would it have taken
+you to think it out?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_2_YOUTHFUL_PRECOCITY" id="X_2_YOUTHFUL_PRECOCITY"></a><a href="#X_2_YOUTHFUL_PRECOCITYa"><b>2.&mdash;YOUTHFUL PRECOCITY.</b></a></p>
+
+<p>The precocity of some youths is surprising. One is disposed to say on
+occasion, "That boy of yours is a genius, and he is certain to do
+great things when he grows up;" but past experience has taught us that
+he invariably becomes quite an ordinary citizen. It is so often the
+case, on the contrary, that the dull boy becomes a great man. You
+never can tell. Nature loves to present to us these queer paradoxes.
+It is well known that those wonderful "lightning calculators," who now
+and again surprise the world by their feats, lose all their mysterious
+powers directly they are taught the elementary rules of arithmetic.</p>
+
+<p>A boy who was demolishing a choice banana was approached by a young
+friend, who, regarding him with envious eyes, asked, "How much did you
+pay for that banana, Fred?" The prompt answer was quite remarkable in
+its way: "The man what I bought it of receives just half as many
+sixpences for sixteen dozen dozen bananas as he gives bananas for a
+fiver."</p>
+
+<p>Now, how long will it take the reader to say correctly just how much
+Fred paid for his rare and refreshing fruit?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_3_AT_A_CATTLE_MARKET" id="X_3_AT_A_CATTLE_MARKET"></a><a href="#X_3_AT_A_CATTLE_MARKETa"><b>3.&mdash;AT A CATTLE MARKET.</b></a></p>
+
+<p>Three countrymen met at a cattle market. "Look here," said Hodge to
+Jakes, "I'll give you six of my pigs for one of your horses, and then
+you'll have twice as many animals here as I've got." "If that's your
+way of doing business," said Durrant to Hodge, "I'll give you fourteen
+of my sheep for a horse, and then you'll have three times as many
+animals as I." "Well, I'll go better than that," said Jakes to
+Durrant; "I'll give you four cows for a horse, <span class='pagenum'>Pg 2<a name="Page_2" id="Page_2"></a></span>and then you'll have
+six times as many animals as I've got here."</p>
+
+<p>No doubt this was a very primitive way of bartering animals, but it is
+an interesting little puzzle to discover just how many animals Jakes,
+Hodge, and Durrant must have taken to the cattle market.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_4_THE_BEANFEAST_PUZZLE" id="X_4_THE_BEANFEAST_PUZZLE"></a><a href="#X_4_THE_BEANFEAST_PUZZLEa"><b>4.&mdash;THE BEANFEAST PUZZLE.</b></a></p>
+
+<p>A number of men went out together on a bean-feast. There were four
+parties invited&mdash;namely, 25 cobblers, 20 tailors, 18 hatters, and 12
+glovers. They spent altogether &pound;6, 13<i>s</i>. It was found that five
+cobblers spent as much as four tailors; that twelve tailors spent as
+much as nine hatters; and that six hatters spent as much as eight
+glovers. The puzzle is to find out how much each of the four parties
+spent.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_5_A_QUEER_COINCIDENCE" id="X_5_A_QUEER_COINCIDENCE"></a><a href="#X_5_A_QUEER_COINCIDENCEa"><b>5.&mdash;A QUEER COINCIDENCE.</b></a></p>
+
+<p>Seven men, whose names were Adams, Baker, Carter, Dobson, Edwards,
+Francis, and Gudgeon, were recently engaged in play. The name of the
+particular game is of no consequence. They had agreed that whenever a
+player won a game he should double the money of each of the other
+players&mdash;that is, he was to give the players just as much money as
+they had already in their pockets. They played seven games, and,
+strange to say, each won a game in turn, in the order in which their
+names are given. But a more curious coincidence is this&mdash;that when
+they had finished play each of the seven men had exactly the same
+amount&mdash;two shillings and eightpence&mdash;in his pocket. The puzzle is to
+find out how much money each man had with him before he sat down to
+play.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_6_A_CHARITABLE_BEQUEST" id="X_6_A_CHARITABLE_BEQUEST"></a><a href="#X_6_A_CHARITABLE_BEQUESTa"><b>6.&mdash;A CHARITABLE BEQUEST.</b></a></p>
+
+<p>A man left instructions to his executors to distribute once a year
+exactly fifty-five shillings among the poor of his parish; but they
+were only to continue the gift so long as they could make it in
+different ways, always giving eighteenpence each to a number of women
+and half a crown each to men. During how many years could the charity
+be administered? Of course, by "different ways" is meant a different
+number of men and women every time.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_7_THE_WIDOWS_LEGACY" id="X_7_THE_WIDOWS_LEGACY"></a><a href="#X_7_THE_WIDOWS_LEGACYa"><b>7.&mdash;THE WIDOW'S LEGACY.</b></a></p>
+
+<p>A gentleman who recently died left the sum of &pound;8,000 to be divided
+among his widow, five sons, and four daughters. He directed that every
+son should receive three times as much as a daughter, and that every
+daughter should have twice as much as their mother. What was the
+widow's share?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_8_INDISCRIMINATE_CHARITY" id="X_8_INDISCRIMINATE_CHARITY"></a><a href="#X_8_INDISCRIMINATE_CHARITYa"><b>8.&mdash;INDISCRIMINATE CHARITY.</b></a></p>
+
+<p>A charitable gentleman, on his way home one night, was appealed to by
+three needy persons in succession for assistance. To the first person
+he gave one penny more than half the money he had in his pocket; to
+the second person he gave twopence more than half the money he then
+had in his pocket; and to the third person he handed over threepence
+more than half of what he had left. On entering his house he had only
+one penny in his pocket. Now, can you say exactly how much money that
+gentleman had on him when he started for home?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_9_THE_TWO_AEROPLANES" id="X_9_THE_TWO_AEROPLANES"></a><a href="#X_9_THE_TWO_AEROPLANESa"><b>9.&mdash;THE TWO AEROPLANES.</b></a></p>
+
+<p>A man recently bought two aeroplanes, but afterwards found that they
+would not answer the purpose for which he wanted them. So he sold them
+for &pound;600 each, making a loss of 20 per cent, on one machine and a
+profit of 20 per cent, on the other. Did he make a profit on the whole
+transaction, or a loss? And how much?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_10_BUYING_PRESENTS" id="X_10_BUYING_PRESENTS"></a><a href="#X_10_BUYING_PRESENTSa"><b>10.&mdash;BUYING PRESENTS.</b></a></p>
+
+<p>"Whom do you think I met in town last week, Brother William?" said
+Uncle Benjamin. "That old skinflint Jorkins. His family had been
+taking him around buying Christmas presents. He said to me, 'Why
+cannot the government abolish Christmas, and make the giving of
+presents punishable by law? I came out this morning with a certain
+amount of money in my pocket, and I find I have spent just half of it.
+In fact, if you will believe me, I take home just as many shillings as
+I had pounds, and half as many pounds as I had shillings. It is
+monstrous!'" Can you say exactly how much money Jorkins had spent on
+those presents?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_11_THE_CYCLISTS_FEAST" id="X_11_THE_CYCLISTS_FEAST"></a><a href="#X_11_THE_CYCLISTS_FEASTa"><b>11.&mdash;THE CYCLISTS' FEAST.</b></a></p>
+
+<div class="poem"><div class="stanza">
+<span class="i0">'Twas last Bank Holiday, so I've been told,<br /></span>
+<span class="i2">Some cyclists rode abroad in glorious weather.<br /></span>
+<span class="i0">Resting at noon within a tavern old,<br /></span>
+<span class="i2">They all agreed to have a feast together.<br /></span>
+<span class="i0">"Put it all in one bill, mine host," they said,<br /></span>
+<span class="i2">"For every man an equal share will pay."<br /></span>
+<span class="i0">The bill was promptly on the table laid,<br /></span>
+<span class="i2">And four pounds was the reckoning that day.<br /></span>
+<span class="i0">But, sad to state, when they prepared to square,<br /></span>
+<span class="i2">'Twas found that two had sneaked outside and fled.<br /></span>
+<span class="i0">So, for two shillings more than his due share<br /></span>
+<span class="i2">Each honest man who had remained was bled.<br /></span>
+<span class="i0">They settled later with those rogues, no doubt.<br /></span>
+<span class="i2">How many were they when they first set out?<br /></span>
+</div></div>
+
+<hr style="width: 30%;" />
+<p><a name="X_12_A_QUEER_THING_IN_MONEY" id="X_12_A_QUEER_THING_IN_MONEY"></a><a href="#X_12_A_QUEER_THING_IN_MONEYa"><b>12.&mdash;A QUEER THING IN MONEY.</b></a></p>
+
+<p>It will be found that &pound;66, 6<i>s</i>. 6<i>d</i>. equals 15,918 pence. Now, the four
+6's added together make 24, and the figures in 15,918 also add to 24.
+It is a curious fact that there is only one other sum of money, in
+pounds, shillings, and pence (all similarly repetitions of one
+figure), of which the digits shall add up the same as the digits of
+the amount in pence. What is the other sum of money?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_13_A_NEW_MONEY_PUZZLE" id="X_13_A_NEW_MONEY_PUZZLE"></a><a href="#X_13_A_NEW_MONEY_PUZZLEa"><b>13.&mdash;A NEW MONEY PUZZLE.</b></a></p>
+
+<p>The largest sum of money that can be written in pounds, shillings,
+pence, and farthings, using each of the nine digits once and only
+once, is <span class='pagenum'>Pg 3<a name="Page_3" id="Page_3"></a></span>&pound;98,765, 4<i>s</i>. 3�<i>d</i>. Now, try to discover the smallest sum
+of money that can be written down under precisely the same conditions.
+There must be some value given for each denomination&mdash;pounds,
+shillings, pence, and farthings&mdash;and the nought may not be used. It
+requires just a little judgment and thought.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_14_SQUARE_MONEY" id="X_14_SQUARE_MONEY"></a><a href="#X_14_SQUARE_MONEYa"><b>14.&mdash;SQUARE MONEY.</b></a></p>
+
+<p>"This is queer," said McCrank to his friend. "Twopence added to
+twopence is fourpence, and twopence multiplied by twopence is also
+fourpence." Of course, he was wrong in thinking you can multiply money
+by money. The multiplier must be regarded as an abstract number. It is
+true that two feet multiplied by two feet will make four square feet.
+Similarly, two pence multiplied by two pence will produce four square
+pence! And it will perplex the reader to say what a "square penny" is.
+But we will assume for the purposes of our puzzle that twopence
+multiplied by twopence is fourpence. Now, what two amounts of money
+will produce the next smallest possible result, the same in both
+cases, when added or multiplied in this manner? The two amounts need
+not be alike, but they must be those that can be paid in current coins
+of the realm.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_15_POCKET_MONEY" id="X_15_POCKET_MONEY"></a><a href="#X_15_POCKET_MONEYa"><b>15.&mdash;POCKET MONEY.</b></a></p>
+
+<p>What is the largest sum of money&mdash;all in current silver coins and no
+four-shilling piece&mdash;that I could have in my pocket without being able
+to give change for a half-sovereign?</p>
+<hr style="width: 30%;" />
+<p><a name="X_16_THE_MILLIONAIRES_PERPLEXITY" id="X_16_THE_MILLIONAIRES_PERPLEXITY"></a><a href="#X_16_THE_MILLIONAIRES_PERPLEXITYa"><b>16.&mdash;THE MILLIONAIRE'S PERPLEXITY.</b></a></p>
+
+<p>Mr. Morgan G. Bloomgarten, the millionaire, known in the States as the
+Clam King, had, for his sins, more money than he knew what to do with.
+It bored him. So he determined to persecute some of his poor but happy
+friends with it. They had never done him any harm, but he resolved to
+inoculate them with the "source of all evil." He therefore proposed to
+distribute a million dollars among them and watch them go rapidly to
+the bad. But he was a man of strange fancies and superstitions, and it
+was an inviolable rule with him never to make a gift that was not
+either one dollar or some power of seven&mdash;such as 7, 49, 343, 2,401,
+which numbers of dollars are produced by simply multiplying sevens
+together. Another rule of his was that he would never give more than
+six persons exactly the same sum. Now, how was he to distribute the
+1,000,000 dollars? You may distribute the money among as many people
+as you like, under the conditions given.</p>
+<hr style="width: 30%;" />
+<p><a name="X_17_THE_PUZZLING_MONEY_BOXES" id="X_17_THE_PUZZLING_MONEY_BOXES"></a><a href="#X_17_THE_PUZZLING_MONEY_BOXESa"><b>17.&mdash;THE PUZZLING MONEY-BOXES.</b></a></p>
+
+<p>Four brothers&mdash;named John, William, Charles, and Thomas&mdash;had each a
+money-box. The boxes were all given to them on the same day, and they
+at once put what money they had into them; only, as the boxes were not
+very large, they first changed the money into as few coins as
+possible. After they had done this, they told one another how much
+money they had saved, and it was found that if John had had 2<i>s</i>. more
+in his box than at present, if William had had 2<i>s</i>. less, if Charles
+had had twice as much, and if Thomas had had half as much, they would
+all have had exactly the same amount.</p>
+
+<p>Now, when I add that all four boxes together contained 45<i>s</i>., and that
+there were only six coins in all in them, it becomes an entertaining
+puzzle to discover just what coins were in each box.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_18_THE_MARKET_WOMEN" id="X_18_THE_MARKET_WOMEN"></a><a href="#X_18_THE_MARKET_WOMENa"><b>18.&mdash;THE MARKET WOMEN.</b></a></p>
+
+<p>A number of market women sold their various products at a certain
+price per pound (different in every case), and each received the same
+amount&mdash;2<i>s</i>. 2&frac12;d. What is the greatest number of women there could
+have been? The price per pound in every case must be such as could be
+paid in current money.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_19_THE_NEW_YEARS_EVE_SUPPERS" id="X_19_THE_NEW_YEARS_EVE_SUPPERS"></a><a href="#X_19_THE_NEW_YEARS_EVE_SUPPERSa"><b>19.&mdash;THE NEW YEAR'S EVE SUPPERS.</b></a></p>
+
+<p>The proprietor of a small London caf&eacute; has given me some interesting
+figures. He says that the ladies who come alone to his place for
+refreshment spend each on an average eighteenpence, that the
+unaccompanied men spend half a crown each, and that when a gentleman
+brings in a lady he spends half a guinea. On New Year's Eve he
+supplied suppers to twenty-five persons, and took five pounds in all.
+Now, assuming his averages to have held good in every case, how was
+his company made up on that occasion? Of course, only single
+gentlemen, single ladies, and pairs (a lady and gentleman) can be
+supposed to have been present, as we are not considering larger
+parties.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_20_BEEF_AND_SAUSAGES" id="X_20_BEEF_AND_SAUSAGES"></a><a href="#X_20_BEEF_AND_SAUSAGESa"><b>20.&mdash;BEEF AND SAUSAGES.</b></a></p>
+
+<p>"A neighbour of mine," said Aunt Jane, "bought a certain quantity of
+beef at two shillings a pound, and the same quantity of sausages at
+eighteenpence a pound. I pointed out to her that if she had divided
+the same money equally between beef and sausages she would have gained
+two pounds in the total weight. Can you tell me exactly how much she
+spent?"</p>
+
+<p>"Of course, it is no business of mine," said Mrs. Sunniborne; "but a
+lady who could pay such prices must be somewhat inexperienced in
+domestic economy."</p>
+
+<p>"I quite agree, my dear," Aunt Jane replied, "but you see that is not
+the precise point under discussion, any more than the name and morals
+of the tradesman."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_21_A_DEAL_IN_APPLES" id="X_21_A_DEAL_IN_APPLES"></a><a href="#X_21_A_DEAL_IN_APPLESa"><b>21.&mdash;A DEAL IN APPLES.</b></a></p>
+
+<p>I paid a man a shilling for some apples, but they were so small that I
+made him throw in two extra apples. I find that made them cost just a
+penny a dozen less than the first price he asked. How many apples did
+I get for my shilling?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_22_A_DEAL_IN_EGGS" id="X_22_A_DEAL_IN_EGGS"></a><a href="#X_22_A_DEAL_IN_EGGSa"><b>22.&mdash;A DEAL IN EGGS.</b></a></p>
+
+<p>A man went recently into a dairyman's shop to buy eggs. He wanted them
+of various qualities. <span class='pagenum'>Pg 4<a name="Page_4" id="Page_4"></a></span>The salesman had new-laid eggs at the high
+price of fivepence each, fresh eggs at one penny each, eggs at a
+halfpenny each, and eggs for electioneering purposes at a greatly
+reduced figure, but as there was no election on at the time the buyer
+had no use for the last. However, he bought some of each of the three
+other kinds and obtained exactly one hundred eggs for eight and
+fourpence. Now, as he brought away exactly the same number of eggs of
+two of the three qualities, it is an interesting puzzle to determine
+just how many he bought at each price.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_23_THE_CHRISTMAS-BOXES" id="X_23_THE_CHRISTMAS-BOXES"></a><a href="#X_23_THE_CHRISTMAS-BOXESa"><b>23.&mdash;THE CHRISTMAS-BOXES.</b></a></p>
+
+<p>Some years ago a man told me he had spent one hundred English silver
+coins in Christmas-boxes, giving every person the same amount, and it
+cost him exactly &pound;1, 10<i>s</i>. 1<i>d</i>. Can you tell just how many persons
+received the present, and how he could have managed the distribution?
+That odd penny looks queer, but it is all right.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_24_A_SHOPPING_PERPLEXITY" id="X_24_A_SHOPPING_PERPLEXITY"></a><a href="#X_24_A_SHOPPING_PERPLEXITYa"><b>24.&mdash;A SHOPPING PERPLEXITY.</b></a></p>
+
+<p>Two ladies went into a shop where, through some curious eccentricity,
+no change was given, and made purchases amounting together to less
+than five shillings. "Do you know," said one lady, "I find I shall
+require no fewer than six current coins of the realm to pay for what I
+have bought." The other lady considered a moment, and then exclaimed:
+"By a peculiar coincidence, I am exactly in the same dilemma." "Then
+we will pay the two bills together." But, to their astonishment, they
+still required six coins. What is the smallest possible amount of
+their purchases&mdash;both different?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_25_CHINESE_MONEY" id="X_25_CHINESE_MONEY"></a><a href="#X_25_CHINESE_MONEYa"><b>25.&mdash;CHINESE MONEY.</b></a></p>
+
+<p>The Chinese are a curious people, and have strange inverted ways of
+doing things. It is said that they use a saw with an upward pressure
+instead of a downward one, that they plane a deal board by pulling the
+tool toward them instead of pushing it, and that in building a house
+they first construct the roof and, having raised that into position,
+proceed to work downwards. In money the currency of the country
+consists of taels of fluctuating value. The tael became thinner and
+thinner until 2,000 of them piled together made less than three inches
+in height. The common cash consists of brass coins of varying
+thicknesses, with a round, square, or triangular hole in the centre,
+as in our illustration.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q025.png" width="400" height="134" alt="" title="" />
+</div>
+
+<p>These are strung on wires like buttons. Supposing that eleven coins
+with round holes are worth fifteen ching-changs, that eleven with
+square holes are worth sixteen ching-changs, and that eleven with
+triangular holes are worth seventeen ching-changs, how can a Chinaman
+give me change for half a crown, using no coins other than the three
+mentioned? A ching-chang is worth exactly twopence and four-fifteenths
+of a ching-chang.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_26_THE_JUNIOR_CLERKS_PUZZLE" id="X_26_THE_JUNIOR_CLERKS_PUZZLE"></a><a href="#X_26_THE_JUNIOR_CLERKS_PUZZLEa"><b>26.&mdash;THE JUNIOR CLERK'S PUZZLE.</b></a></p>
+
+<p>Two youths, bearing the pleasant names of Moggs and Snoggs, were
+employed as junior clerks by a merchant in Mincing Lane. They were
+both engaged at the same salary&mdash;that is, commencing at the rate of
+&pound;50 a year, payable half-yearly. Moggs had a yearly rise of &pound;10, and
+Snoggs was offered the same, only he asked, for reasons that do not
+concern our puzzle, that he might take his rise at &pound;2, 10<i>s</i>.
+half-yearly, to which his employer (not, perhaps, unnaturally!) had no
+objection.</p>
+
+<p>Now we come to the real point of the puzzle. Moggs put regularly into
+the Post Office Savings Bank a certain proportion of his salary, while
+Snoggs saved twice as great a proportion of his, and at the end of
+five years they had together saved &pound;268, 15<i>s</i>. How much had each saved?
+The question of interest can be ignored.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_27_GIVING_CHANGE" id="X_27_GIVING_CHANGE"></a><a href="#X_27_GIVING_CHANGEa"><b>27.&mdash;GIVING CHANGE.</b></a></p>
+
+<p>Every one is familiar with the difficulties that frequently arise over
+the giving of change, and how the assistance of a third person with a
+few coins in his pocket will sometimes help us to set the matter
+right. Here is an example. An Englishman went into a shop in New York
+and bought goods at a cost of thirty-four cents. The only money he had
+was a dollar, a three-cent piece, and a two-cent piece. The tradesman
+had only a half-dollar and a quarter-dollar. But another customer
+happened to be present, and when asked to help produced two dimes, a
+five-cent piece, a two-cent piece, and a one-cent piece. How did the
+tradesman manage to give change? For the benefit of those readers who
+are not familiar with the American coinage, it is only necessary to
+say that a dollar is a hundred cents and a dime ten cents. A puzzle of
+this kind should rarely cause any difficulty if attacked in a proper
+manner.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_28_DEFECTIVE_OBSERVATION" id="X_28_DEFECTIVE_OBSERVATION"></a><a href="#X_28_DEFECTIVE_OBSERVATIONa"><b>28.&mdash;DEFECTIVE OBSERVATION.</b></a></p>
+
+<p>Our observation of little things is frequently defective, and our
+memories very liable to lapse. A certain judge recently remarked in a
+case that he had no recollection whatever of putting the wedding-ring
+on his wife's finger. Can you correctly answer these questions without
+having the coins in sight? On which side of a penny is the date given?
+Some people are so unobservant that, although they are handling the
+coin nearly every day of their lives, they are at a loss to answer
+this simple question. If I lay a penny flat on the table, how many
+other pennies can I place around it, every one also lying flat on the
+table, so that they all touch the first one? The geometrician will, of
+course, give the answer at once, and not need to make any experiment.
+<span class='pagenum'>Pg 5<a name="Page_5" id="Page_5"></a></span>He will also know that, since all circles are similar, the same
+answer will necessarily apply to any coin. The next question is a most
+interesting one to ask a company, each person writing down his answer
+on a slip of paper, so that no one shall be helped by the answers of
+others. What is the greatest number of three-penny-pieces that may be
+laid flat on the surface of a half-crown, so that no piece lies on
+another or overlaps the surface of the half-crown? It is amazing what
+a variety of different answers one gets to this question. Very few
+people will be found to give the correct number. Of course the answer
+must be given without looking at the coins.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_29_THE_BROKEN_COINS" id="X_29_THE_BROKEN_COINS"></a><a href="#X_29_THE_BROKEN_COINSa"><b>29.&mdash;THE BROKEN COINS.</b></a></p>
+
+<p>A man had three coins&mdash;a sovereign, a shilling, and a penny&mdash;and he
+found that exactly the same fraction of each coin had been broken
+away. Now, assuming that the original intrinsic value of these coins
+was the same as their nominal value&mdash;that is, that the sovereign was
+worth a pound, the shilling worth a shilling, and the penny worth a
+penny&mdash;what proportion of each coin has been lost if the value of the
+three remaining fragments is exactly one pound?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_30_TWO_QUESTIONS_IN_PROBABILITIES" id="X_30_TWO_QUESTIONS_IN_PROBABILITIES"></a><a href="#X_30_TWO_QUESTIONS_IN_PROBABILITIESa"><b>30.&mdash;TWO QUESTIONS IN PROBABILITIES.</b></a></p>
+
+<p>There is perhaps no class of puzzle over which people so frequently
+blunder as that which involves what is called the theory of
+probabilities. I will give two simple examples of the sort of puzzle I
+mean. They are really quite easy, and yet many persons are tripped up
+by them. A friend recently produced five pennies and said to me: "In
+throwing these five pennies at the same time, what are the chances
+that at least four of the coins will turn up either all heads or all
+tails?" His own solution was quite wrong, but the correct answer ought
+not to be hard to discover. Another person got a wrong answer to the
+following little puzzle which I heard him propound: "A man placed
+three sovereigns and one shilling in a bag. How much should be paid
+for permission to draw one coin from it?" It is, of course, understood
+that you are as likely to draw any one of the four coins as another.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_31_DOMESTIC_ECONOMY" id="X_31_DOMESTIC_ECONOMY"></a><a href="#X_31_DOMESTIC_ECONOMYa"><b>31.&mdash;DOMESTIC ECONOMY.</b></a></p>
+
+<p>Young Mrs. Perkins, of Putney, writes to me as follows: "I should be
+very glad if you could give me the answer to a little sum that has
+been worrying me a good deal lately. Here it is: We have only been
+married a short time, and now, at the end of two years from the time
+when we set up housekeeping, my husband tells me that he finds we have
+spent a third of his yearly income in rent, rates, and taxes, one-half
+in domestic expenses, and one-ninth in other ways. He has a balance of
+&pound;190 remaining in the bank. I know this last, because he accidentally
+left out his pass-book the other day, and I peeped into it. Don't you
+think that a husband ought to give his wife his entire confidence in
+his money matters? Well, I do; and&mdash;will you believe it?&mdash;he has never
+told me what his income really is, and I want, very naturally, to find
+out. Can you tell me what it is from the figures I have given you?"</p>
+
+<p>Yes; the answer can certainly be given from the figures contained in
+Mrs. Perkins's letter. And my readers, if not warned, will be
+practically unanimous in declaring the income to be&mdash;something
+absurdly in excess of the correct answer!</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_32_THE_EXCURSION_TICKET_PUZZLE" id="X_32_THE_EXCURSION_TICKET_PUZZLE"></a><a href="#X_32_THE_EXCURSION_TICKET_PUZZLEa"><b>32.&mdash;THE EXCURSION TICKET PUZZLE.</b></a></p>
+
+<p>When the big flaming placards were exhibited at the little provincial
+railway station, announcing that the Great &mdash;&mdash; Company would run
+cheap excursion trains to London for the Christmas holidays, the
+inhabitants of Mudley-cum-Turmits were in quite a flutter of
+excitement. Half an hour before the train came in the little booking
+office was crowded with country passengers, all bent on visiting their
+friends in the great Metropolis. The booking clerk was unaccustomed to
+dealing with crowds of such a dimension, and he told me afterwards,
+while wiping his manly brow, that what caused him so much trouble was
+the fact that these rustics paid their fares in such a lot of small
+money.</p>
+
+<p>He said that he had enough farthings to supply a West End draper with
+change for a week, and a sufficient number of threepenny pieces for
+the congregations of three parish churches. "That excursion fare,"
+said he, "is nineteen shillings and ninepence, and I should like to
+know in just how many different ways it is possible for such an amount
+to be paid in the current coin of this realm."</p>
+
+<p>Here, then, is a puzzle: In how many different ways may nineteen
+shillings and ninepence be paid in our current coin? Remember that the
+fourpenny-piece is not now current.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_33_A_PUZZLE_IN_REVERSALS" id="X_33_A_PUZZLE_IN_REVERSALS"></a><a href="#X_33_A_PUZZLE_IN_REVERSALSa"><b>33.&mdash;A PUZZLE IN REVERSALS.</b></a></p>
+
+<p>Most people know that if you take any sum of money in pounds,
+shillings, and pence, in which the number of pounds (less than &pound;12)
+exceeds that of the pence, reverse it (calling the pounds pence and
+the pence pounds), find the difference, then reverse and add this
+difference, the result is always &pound;12, 18<i>s</i>. 11<i>d</i>. But if we omit the
+condition, "less than &pound;12," and allow nought to represent shillings or
+pence&mdash;(1) What is the lowest amount to which the rule will not apply?
+(2) What is the highest amount to which it will apply? Of course, when
+reversing such a sum as &pound;14, 15<i>s</i>. 3<i>d</i>. it may be written &pound;3, 16<i>s</i>. 2<i>d</i>.,
+which is the same as &pound;3, 15<i>s</i>. 14<i>d</i>.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_34_THE_GROCER_AND_DRAPER" id="X_34_THE_GROCER_AND_DRAPER"></a><a href="#X_34_THE_GROCER_AND_DRAPERa"><b>34.&mdash;THE GROCER AND DRAPER.</b></a></p>
+
+<p>A country "grocer and draper" had two rival assistants, who prided
+themselves on their rapidity in serving customers. The young man on
+the grocery side could weigh up two one-pound parcels of sugar per
+minute, while the drapery assistant could cut three one-yard lengths
+of cloth in the same time. Their employer, one slack day, set them a
+race, giving <span class='pagenum'>Pg 6<a name="Page_6" id="Page_6"></a></span>the grocer a barrel of sugar and telling him to weigh up
+forty-eight one-pound parcels of sugar While the draper divided a roll
+of forty-eight yards of cloth into yard pieces. The two men were
+interrupted together by customers for nine minutes, but the draper was
+disturbed seventeen times as long as the grocer. What was the result
+of the race?</p>
+<hr style="width: 30%;" />
+<p><a name="X_35_JUDKINSS_CATTLE" id="X_35_JUDKINSS_CATTLE"></a><a href="#X_35_JUDKINSS_CATTLEa"><b>35.&mdash;JUDKINS'S CATTLE.</b></a></p>
+
+<p>Hiram B. Judkins, a cattle-dealer of Texas, had five droves of
+animals, consisting of oxen, pigs, and sheep, with the same number of
+animals in each drove. One morning he sold all that he had to eight
+dealers. Each dealer bought the same number of animals, paying
+seventeen dollars for each ox, four dollars for each pig, and two
+dollars for each sheep; and Hiram received in all three hundred and
+one dollars. What is the greatest number of animals he could have had?
+And how many would there be of each kind?</p>
+<hr style="width: 30%;" />
+<p><a name="X_36_BUYING_APPLES" id="X_36_BUYING_APPLES"></a><a href="#X_36_BUYING_APPLESa"><b>36.&mdash;BUYING APPLES.</b></a></p>
+
+<p>As the purchase of apples in small quantities has always presented
+considerable difficulties, I think it well to offer a few remarks on
+this subject. We all know the story of the smart boy who, on being
+told by the old woman that she was selling her apples at four for
+threepence, said: "Let me see! Four for threepence; that's three for
+twopence, two for a penny, one for nothing&mdash;I'll take <i>one</i>!"</p>
+
+<p>There are similar cases of perplexity. For example, a boy once picked
+up a penny apple from a stall, but when he learnt that the woman's
+pears were the same price he exchanged it, and was about to walk off.
+"Stop!" said the woman. "You haven't paid me for the pear!" "No," said
+the boy, "of course not. I gave you the apple for it." "But you didn't
+pay for the apple!" "Bless the woman! You don't expect me to pay for
+the apple and the pear too!" And before the poor creature could get
+out of the tangle the boy had disappeared.</p>
+
+<p>Then, again, we have the case of the man who gave a boy sixpence and
+promised to repeat the gift as soon as the youngster had made it into
+ninepence. Five minutes later the boy returned. "I have made it into
+ninepence," he said, at the same time handing his benefactor
+threepence. "How do you make that out?" he was asked. "I bought
+threepennyworth of apples." "But that does not make it into
+ninepence!" "I should rather think it did," was the boy's reply. "The
+apple woman has threepence, hasn't she? Very well, I have
+threepennyworth of apples, and I have just given you the other
+threepence. What's that but ninepence?"</p>
+
+<p>I cite these cases just to show that the small boy really stands in
+need of a little instruction in the art of buying apples. So I will
+give a simple poser dealing with this branch of commerce.</p>
+
+<p>An old woman had apples of three sizes for sale&mdash;one a penny, two a
+penny, and three a penny. Of course two of the second size and three
+of the third size were respectively equal to one apple of the largest
+size. Now, a gentleman who had an equal number of boys and girls gave
+his children sevenpence to be spent amongst them all on these apples.
+The puzzle is to give each child an equal distribution of apples. How
+was the sevenpence spent, and how many children were there?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_37_BUYING_CHESTNUTS" id="X_37_BUYING_CHESTNUTS"></a><a href="#X_37_BUYING_CHESTNUTSa"><b>37.&mdash;BUYING CHESTNUTS.</b></a></p>
+
+<p>Though the following little puzzle deals with the purchase of
+chestnuts, it is not itself of the "chestnut" type. It is quite new.
+At first sight it has certainly the appearance of being of the
+"nonsense puzzle" character, but it is all right when properly
+considered.</p>
+
+<p>A man went to a shop to buy chestnuts. He said he wanted a pennyworth,
+and was given five chestnuts. "It is not enough; I ought to have a
+sixth," he remarked! "But if I give you one chestnut more." the
+shopman replied, "you will have five too many." Now, strange to say,
+they were both right. How many chestnuts should the buyer receive for
+half a crown?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_38_THE_BICYCLE_THIEF" id="X_38_THE_BICYCLE_THIEF"></a><a href="#X_38_THE_BICYCLE_THIEFa"><b>38.&mdash;THE BICYCLE THIEF.</b></a></p>
+
+<p>Here is a little tangle that is perpetually cropping up in various
+guises. A cyclist bought a bicycle for &pound;15 and gave in payment a
+cheque for &pound;25. The seller went to a neighbouring shopkeeper and got
+him to change the cheque for him, and the cyclist, having received his
+&pound;10 change, mounted the machine and disappeared. The cheque proved to
+be valueless, and the salesman was requested by his neighbour to
+refund the amount he had received. To do this, he was compelled to
+borrow the &pound;25 from a friend, as the cyclist forgot to leave his
+address, and could not be found. Now, as the bicycle cost the salesman
+&pound;11, how much money did he lose altogether?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_39_THE_COSTERMONGERS_PUZZLE" id="X_39_THE_COSTERMONGERS_PUZZLE"></a><a href="#X_39_THE_COSTERMONGERS_PUZZLEa"><b>39.&mdash;THE COSTERMONGER'S PUZZLE.</b></a></p>
+
+<p>"How much did yer pay for them oranges, Bill?"</p>
+
+<p>"I ain't a-goin' to tell yer, Jim. But I beat the old cove down
+fourpence a hundred."</p>
+
+<p>"What good did that do yer?"</p>
+
+<p>"Well, it meant five more oranges on every ten shillin's-worth."</p>
+
+<p>Now, what price did Bill actually pay for the oranges? There is only
+one rate that will fit in with his statements.</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="AGE_AND_KINSHIP_PUZZLES" id="AGE_AND_KINSHIP_PUZZLES"></a><a href="#CONTENTS">AGE AND KINSHIP PUZZLES.</a></h2>
+
+<p class='center'>
+<span style="margin-left: 0em;">"The days of our years are threescore years and ten."<br /></span>
+<span style="margin-left: 8em;">&mdash;<i>Psalm</i> xc. 10.<br /></span>
+</p>
+
+<p>For centuries it has been a favourite method of propounding
+arithmetical puzzles to pose them in the form of questions as to the
+age of an individual. They generally lend themselves to very easy
+solution by the use of algebra, though often the difficulty lies in
+stating them <span class='pagenum'>Pg 7<a name="Page_7" id="Page_7"></a></span>correctly. They may be made very complex and may demand
+considerable ingenuity, but no general laws can well be laid down for
+their solution. The solver must use his own sagacity. As for puzzles
+in relationship or kinship, it is quite curious how bewildering many
+people find these things. Even in ordinary conversation, some
+statement as to relationship, which is quite clear in the mind of the
+speaker, will immediately tie the brains of other people into knots.
+Such expressions as "He is my uncle's son-in-law's sister" convey
+absolutely nothing to some people without a detailed and laboured
+explanation. In such cases the best course is to sketch a brief
+genealogical table, when the eye comes immediately to the assistance
+of the brain. In these days, when we have a growing lack of respect
+for pedigrees, most people have got out of the habit of rapidly
+drawing such tables, which is to be regretted, as they would save a
+lot of time and brain racking on occasions.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_40_MAMMAS_AGE" id="X_40_MAMMAS_AGE"></a><a href="#X_40_MAMMAS_AGEa"><b>40.&mdash;MAMMA'S AGE.</b></a></p>
+
+<p>Tommy: "How old are you, mamma?"</p>
+
+<p>Mamma: "Let me think, Tommy. Well, our three ages add up to exactly
+seventy years."</p>
+
+<p>Tommy: "That's a lot, isn't it? And how old are you, papa?"</p>
+
+<p>Papa: "Just six times as old as you, my son."</p>
+
+<p>Tommy: "Shall I ever be half as old as you, papa?"</p>
+
+<p>Papa: "Yes, Tommy; and when that happens our three ages will add up to
+exactly twice as much as to-day."</p>
+
+<p>Tommy: "And supposing I was born before you, papa; and supposing mamma
+had forgot all about it, and hadn't been at home when I came; and
+supposing&mdash;&mdash;"</p>
+
+<p>Mamma: "Supposing, Tommy, we talk about bed. Come along, darling.
+You'll have a headache."</p>
+
+<p>Now, if Tommy had been some years older he might have calculated the
+exact ages of his parents from the information they had given him. Can
+you find out the exact age of mamma?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_41_THEIR_AGES" id="X_41_THEIR_AGES"></a><a href="#X_41_THEIR_AGESa"><b>41.&mdash;THEIR AGES.</b></a></p>
+
+<p>"My husband's age," remarked a lady the other day, "is represented by
+the figures of my own age reversed. He is my senior, and the
+difference between our ages is one-eleventh of their sum."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_42_THE_FAMILY_AGES" id="X_42_THE_FAMILY_AGES"></a><a href="#X_42_THE_FAMILY_AGESa"><b>42.&mdash;THE FAMILY AGES.</b></a></p>
+
+<p>When the Smileys recently received a visit from the favourite uncle,
+the fond parents had all the five children brought into his presence.
+First came Billie and little Gertrude, and the uncle was informed that
+the boy was exactly twice as old as the girl. Then Henrietta arrived,
+and it was pointed out that the combined ages of herself and Gertrude
+equalled twice the age of Billie. Then Charlie came running in, and
+somebody remarked that now the combined ages of the two boys were
+exactly twice the combined ages of the two girls. The uncle was
+expressing his astonishment at these coincidences when Janet came in.
+"Ah! uncle," she exclaimed, "you have actually arrived on my
+twenty-first birthday!" To this Mr. Smiley added the final staggerer:
+"Yes, and now the combined ages of the three girls are exactly equal
+to twice the combined ages of the two boys." Can you give the age of
+each child?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_43_MRS_TIMPKINSS_AGE" id="X_43_MRS_TIMPKINSS_AGE"></a><a href="#X_43_MRS_TIMPKINSS_AGEa"><b>43.&mdash;MRS. TIMPKINS'S AGE.</b></a></p>
+
+<p>Edwin: "Do you know, when the Timpkinses married eighteen years ago
+Timpkins was three times as old as his wife, and to-day he is just
+twice as old as she?"</p>
+
+<p>Angelina: "Then how old was Mrs. Timpkins on the wedding day?"</p>
+
+<p>Can you answer Angelina's question?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_44_A_CENSUS_PUZZLE" id="X_44_A_CENSUS_PUZZLE"></a><a href="#X_44_A_CENSUS_PUZZLEa"><b>44&mdash;A CENSUS PUZZLE.</b></a></p>
+
+<p>Mr. and Mrs. Jorkins have fifteen children, all born at intervals of
+one year and a half. Miss Ada Jorkins, the eldest, had an objection to
+state her age to the census man, but she admitted that she was just
+seven times older than little Johnnie, the youngest of all. What was
+Ada's age? Do not too hastily assume that you have solved this little
+poser. You may find that you have made a bad blunder!</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_45_MOTHER_AND_DAUGHTER" id="X_45_MOTHER_AND_DAUGHTER"></a><a href="#X_45_MOTHER_AND_DAUGHTERa"><b>45.&mdash;MOTHER AND DAUGHTER.</b></a></p>
+
+<p>"Mother, I wish you would give me a bicycle," said a girl of twelve
+the other day.</p>
+
+<p>"I do not think you are old enough yet, my dear," was the reply. "When
+I am only three times as old as you are you shall have one."</p>
+
+<p>Now, the mother's age is forty-five years. When may the young lady
+expect to receive her present?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_46_MARY_AND_MARMADUKE" id="X_46_MARY_AND_MARMADUKE"></a><a href="#X_46_MARY_AND_MARMADUKEa"><b>46.&mdash;MARY AND MARMADUKE.</b></a></p>
+
+<p>Marmaduke: "Do you know, dear, that in seven years' time our combined
+ages will be sixty-three years?"</p>
+
+<p>Mary: "Is that really so? And yet it is a fact that when you were my
+present age you were twice as old as I was then. I worked it out last
+night."</p>
+
+<p>Now, what are the ages of Mary and Marmaduke?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_47_ROVERS_AGE" id="X_47_ROVERS_AGE"></a><a href="#X_47_ROVERS_AGEa"><b>47&mdash;ROVER'S AGE.</b></a></p>
+
+<p>"Now, then, Tommy, how old is Rover?" Mildred's young man asked her
+brother.</p>
+
+<p>"Well, five years ago," was the youngster's reply, "sister was four
+times older than the dog, but now she is only three times as old."</p>
+
+<p>Can you tell Rover's age?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_48_CONCERNING_TOMMYS_AGE" id="X_48_CONCERNING_TOMMYS_AGE"></a><a href="#X_48_CONCERNING_TOMMYS_AGEa"><b>48.&mdash;CONCERNING TOMMY'S AGE.</b></a></p>
+
+<p>Tommy Smart was recently sent to a new school. On the first day of his
+arrival the teacher asked him his age, and this was his curious reply:
+"Well, you see, it is like this. At the time I was born&mdash;I forget the
+year&mdash;my only sister, Ann, happened to be just one-quarter the age <span class='pagenum'>Pg 8<a name="Page_8" id="Page_8"></a></span>of
+mother, and she is now one-third the age of father." "That's all very
+well," said the teacher, "but what I want is not the age of your
+sister Ann, but your own age." "I was just coming to that," Tommy
+answered; "I am just a quarter of mother's present age, and in four
+years' time I shall be a quarter the age of father. Isn't that funny?"</p>
+
+<p>This was all the information that the teacher could get out of Tommy
+Smart. Could you have told, from these facts, what was his precise
+age? It is certainly a little puzzling.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_49_NEXT-DOOR_NEIGHBOURS" id="X_49_NEXT-DOOR_NEIGHBOURS"></a><a href="#X_49_NEXT-DOOR_NEIGHBOURSa"><b>49.&mdash;NEXT-DOOR NEIGHBOURS.</b></a></p>
+
+<p>There were two families living next door to one another at Tooting
+Bec&mdash;the Jupps and the Simkins. The united ages of the four Jupps
+amounted to one hundred years, and the united ages of the Simkins also
+amounted to the same. It was found in the case of each family that the
+sum obtained by adding the squares of each of the children's ages to
+the square of the mother's age equalled the square of the father's
+age. In the case of the Jupps, however, Julia was one year older than
+her brother Joe, whereas Sophy Simkin was two years older than her
+brother Sammy. What was the age of each of the eight individuals?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_50_THE_BAG_OF_NUTS" id="X_50_THE_BAG_OF_NUTS"></a><a href="#X_50_THE_BAG_OF_NUTSa"><b>50.&mdash;THE BAG OF NUTS.</b></a></p>
+
+<p>Three boys were given a bag of nuts as a Christmas present, and it was
+agreed that they should be divided in proportion to their ages, which
+together amounted to 17� years. Now the bag contained 770 nuts, and
+as often as Herbert took four Robert took three, and as often as
+Herbert took six Christopher took seven. The puzzle is to find out how
+many nuts each had, and what were the boys' respective ages.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_51_HOW_OLD_WAS_MARY" id="X_51_HOW_OLD_WAS_MARY"></a><a href="#X_51_HOW_OLD_WAS_MARYa"><b>51.&mdash;HOW OLD WAS MARY?</b></a></p>
+
+<p>Here is a funny little age problem, by the late Sam Loyd, which has
+been very popular in the United States. Can you unravel the mystery?</p>
+
+<p>The combined ages of Mary and Ann are forty-four years, and Mary is
+twice as old as Ann was when Mary was half as old as Ann will be when
+Ann is three times as old as Mary was when Mary was three times as old
+as Ann. How old is Mary? That is all, but can you work it out? If not,
+ask your friends to help you, and watch the shadow of bewilderment
+creep over their faces as they attempt to grip the intricacies of the
+question.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_52_QUEER_RELATIONSHIPS" id="X_52_QUEER_RELATIONSHIPS"></a><a href="#X_52_QUEER_RELATIONSHIPSa"><b>52.&mdash;QUEER RELATIONSHIPS.</b></a></p>
+
+<p>"Speaking of relationships," said the Parson at a certain
+dinner-party, "our legislators are getting the marriage law into a
+frightful tangle, Here, for example, is a puzzling case that has come
+under my notice. Two brothers married two sisters. One man died and
+the other man's wife also died. Then the survivors married."</p>
+
+<p>"The man married his deceased wife's sister under the recent Act?" put
+in the Lawyer.</p>
+
+<p>"Exactly. And therefore, under the civil law, he is legally married
+and his child is legitimate. But, you see, the man is the woman's
+deceased husband's brother, and therefore, also under the civil law,
+she is not married to him and her child is illegitimate."</p>
+
+<p>"He is married to her and she is not married to him!" said the Doctor.</p>
+
+<p>"Quite so. And the child is the legitimate son of his father, but the
+illegitimate son of his mother."</p>
+
+<p>"Undoubtedly 'the law is a hass,'" the Artist exclaimed, "if I may be
+permitted to say so," he added, with a bow to the Lawyer.</p>
+
+<p>"Certainly," was the reply. "We lawyers try our best to break in the
+beast to the service of man. Our legislators are responsible for the
+breed."</p>
+
+<p>"And this reminds me," went on the Parson, "of a man in my parish who
+married the sister of his widow. This man&mdash;&mdash;"</p>
+
+<p>"Stop a moment, sir," said the Professor. "Married the sister of his
+widow? Do you marry dead men in your parish?"</p>
+
+<p>"No; but I will explain that later. Well, this man has a sister of his
+own. Their names are Stephen Brown and Jane Brown. Last week a young
+fellow turned up whom Stephen introduced to me as his nephew.
+Naturally, I spoke of Jane as his aunt, but, to my astonishment, the
+youth corrected me, assuring me that, though he was the nephew of
+Stephen, he was not the nephew of Jane, the sister of Stephen. This
+perplexed me a good deal, but it is quite correct."</p>
+
+<p>The Lawyer was the first to get at the heart of the mystery. What was
+his solution?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_53_HEARD_ON_THE_TUBE_RAILWAY" id="X_53_HEARD_ON_THE_TUBE_RAILWAY"></a><a href="#X_53_HEARD_ON_THE_TUBE_RAILWAYa"><b>53.&mdash;HEARD ON THE TUBE RAILWAY.</b></a></p>
+
+<p>First Lady: "And was he related to you, dear?"</p>
+
+<p>Second Lady: "Oh, yes. You see, that gentleman's mother was my
+mother's mother-in-law, but he is not on speaking terms with my papa."</p>
+
+<p>First Lady: "Oh, indeed!" (But you could see that she was not much
+wiser.)</p>
+
+<p>How was the gentleman related to the Second Lady?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_54_A_FAMILY_PARTY" id="X_54_A_FAMILY_PARTY"></a><a href="#X_54_A_FAMILY_PARTYa"><b>54.&mdash;A FAMILY PARTY.</b></a></p>
+
+<p>A certain family party consisted of 1 grandfather, 1 grandmother, 2
+fathers, 2 mothers, 4 children, 3 grandchildren, 1 brother, 2 sisters,
+2 sons, 2 daughters, 1 father-in-law, 1 mother-in-law, and 1
+daughter-in-law. Twenty-three people, you will say. No; there were
+only seven persons present. Can you show how this might be?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_55_A_MIXED_PEDIGREE" id="X_55_A_MIXED_PEDIGREE"></a><a href="#X_55_A_MIXED_PEDIGREEa"><b>55.&mdash;A MIXED PEDIGREE.</b></a></p>
+
+<p>Joseph Bloggs: "I can't follow it, my dear boy. It makes me dizzy!"</p>
+
+<p>John Snoggs: "It's very simple. Listen again! You happen to be my
+father's brother-in-law, my brother's father-in-law, and also my
+father-in-law's brother. You see, my father was&mdash;&mdash;"</p>
+
+<p><span class='pagenum'>Pg 9<a name="Page_9" id="Page_9"></a></span>But Mr. Bloggs refused to hear any more. Can the reader show how this
+extraordinary triple relationship might have come about?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_56_WILSONS_POSER" id="X_56_WILSONS_POSER"></a><a href="#X_56_WILSONS_POSERa"><b>56.&mdash;WILSON'S POSER.</b></a></p>
+
+<p>"Speaking of perplexities&mdash;&mdash;" said Mr. Wilson, throwing down a
+magazine on the table in the commercial room of the Railway Hotel.</p>
+
+<p>"Who was speaking of perplexities?" inquired Mr. Stubbs.</p>
+
+<p>"Well, then, reading about them, if you want to be exact&mdash;it just
+occurred to me that perhaps you three men may be interested in a
+little matter connected with myself."</p>
+
+<p>It was Christmas Eve, and the four commercial travellers were spending
+the holiday at Grassminster. Probably each suspected that the others
+had no homes, and perhaps each was conscious of the fact that he was
+in that predicament himself. In any case they seemed to be perfectly
+comfortable, and as they drew round the cheerful fire the conversation
+became general.</p>
+
+<p>"What is the difficulty?" asked Mr. Packhurst.</p>
+
+<p>"There's no difficulty in the matter, when you rightly understand it.
+It is like this. A man named Parker had a flying-machine that would
+carry two. He was a venturesome sort of chap&mdash;reckless, I should call
+him&mdash;and he had some bother in finding a man willing to risk his life
+in making an ascent with him. However, an uncle of mine thought he
+would chance it, and one fine morning he took his seat in the machine
+and she started off well. When they were up about a thousand feet, my
+nephew suddenly&mdash;&mdash;"</p>
+
+<p>"Here, stop, Wilson! What was your nephew doing there? You said your
+uncle," interrupted Mr. Stubbs.</p>
+
+<p>"Did I? Well, it does not matter. My nephew suddenly turned to Parker
+and said that the engine wasn't running well, so Parker called out to
+my uncle&mdash;&mdash;"</p>
+
+<p>"Look here," broke in Mr. Waterson, "we are getting mixed. Was it your
+uncle or your nephew? Let's have it one way or the other."</p>
+
+<p>"What I said is quite right. Parker called out to my uncle to do
+something or other, when my nephew&mdash;&mdash;"</p>
+
+<p>"There you are again, Wilson," cried Mr. Stubbs; "once for all, are we
+to understand that both your uncle and your nephew were on the
+machine?"</p>
+
+<p>"Certainly. I thought I made that clear. Where was I? Well, my nephew
+shouted back to Parker&mdash;&mdash;"</p>
+
+<p>"Phew! I'm sorry to interrupt you again, Wilson, but we can't get on
+like this. Is it true that the machine would only carry two?"</p>
+
+<p>"Of course. I said at the start that it only carried two."</p>
+
+<p>"Then what in the name of aerostation do you mean by saying that there
+were three persons on board?" shouted Mr. Stubbs.</p>
+
+<p>"Who said there were three?"</p>
+
+<p>"You have told us that Parker, your uncle, and your nephew went up on
+this blessed flying-machine."</p>
+
+<p>"That's right."</p>
+
+<p>"And the thing would only carry two!"</p>
+
+<p>"Right again."</p>
+
+<p>"Wilson, I have known you for some time as a truthful man and a
+temperate man," said Mr. Stubbs, solemnly. "But I am afraid since you
+took up that new line of goods you have overworked yourself."</p>
+
+<p>"Half a minute, Stubbs," interposed Mr. Waterson. "I see clearly where
+we all slipped a cog. Of course, Wilson, you meant us to understand
+that Parker is either your uncle or your nephew. Now we shall be all
+right if you will just tell us whether Parker is your uncle or
+nephew."</p>
+
+<p>"He is no relation to me whatever."</p>
+
+<p>The three men sighed and looked anxiously at one another. Mr. Stubbs
+got up from his chair to reach the matches, Mr. Packhurst proceeded to
+wind up his watch, and Mr. Waterson took up the poker to attend to the
+fire. It was an awkward moment, for at the season of goodwill nobody
+wished to tell Mr. Wilson exactly what was in his mind.</p>
+
+<p>"It's curious," said Mr. Wilson, very deliberately, "and it's rather
+sad, how thick-headed some people are. You don't seem to grip the
+facts. It never seems to have occurred to either of you that my uncle
+and my nephew are one and the same man."</p>
+
+<p>"What!" exclaimed all three together.</p>
+
+<p>"Yes; David George Linklater is my uncle, and he is also my nephew.
+Consequently, I am both his uncle and nephew. Queer, isn't it? I'll
+explain how it comes about."</p>
+
+<p>Mr. Wilson put the case so very simply that the three men saw how it
+might happen without any marriage within the prohibited degrees.
+Perhaps the reader can work it out for himself.</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="CLOCK_PUZZLES" id="CLOCK_PUZZLES"></a><a href="#CONTENTS">CLOCK PUZZLES.</a></h2>
+
+<p class='center'>
+<span style="margin-left: 0em;">"Look at the clock!"</span><br />
+<span style="margin-left: 8em;"><i>Ingoldsby Legends</i>.</span><br />
+</p>
+
+
+<p>In considering a few puzzles concerning clocks and watches, and the
+times recorded by their hands under given conditions, it is well that
+a particular convention should always be kept in mind. It is
+frequently the case that a solution requires the assumption that the
+hands can actually record a time involving a minute fraction of a
+second. Such a time, of course, cannot be really indicated. Is the
+puzzle, therefore, impossible of solution? The conclusion deduced from
+a logical syllogism depends for its truth on the two premises assumed,
+and it is the same in mathematics. Certain things are antecedently
+assumed, and the answer depends entirely on the truth of those
+assumptions.</p>
+
+<p>"If two horses," says Lagrange, "can pull a load of a certain weight,
+it is natural to suppose that four horses could pull a load of double
+that weight, six horses a load of three times that weight. Yet,
+strictly speaking, such is not the <span class='pagenum'>Pg 10<a name="Page_10" id="Page_10"></a></span>case. For the inference is based
+on the assumption that the four horses pull alike in amount and
+direction, which in practice can scarcely ever be the case. It so
+happens that we are frequently led in our reckonings to results which
+diverge widely from reality. But the fault is not the fault of
+mathematics; for mathematics always gives back to us exactly what we
+have put into it. The ratio was constant according to that
+supposition. The result is founded upon that supposition. If the
+supposition is false the result is necessarily false."</p>
+
+<p>If one man can reap a field in six days, we say two men will reap it
+in three days, and three men will do the work in two days. We here
+assume, as in the case of Lagrange's horses, that all the men are
+exactly equally capable of work. But we assume even more than this.
+For when three men get together they may waste time in gossip or play;
+or, on the other hand, a spirit of rivalry may spur them on to greater
+diligence. We may assume any conditions we like in a problem, provided
+they be clearly expressed and understood, and the answer will be in
+accordance with those conditions.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_57_WHAT_WAS_THE_TIME" id="X_57_WHAT_WAS_THE_TIME"></a><a href="#X_57_WHAT_WAS_THE_TIMEa"><b>57.&mdash;WHAT WAS THE TIME?</b></a></p>
+
+<p>"I say, Rackbrane, what is the time?" an acquaintance asked our friend
+the professor the other day. The answer was certainly curious.</p>
+
+<p>"If you add one quarter of the time from noon till now to half the
+time from now till noon to-morrow, you will get the time exactly."</p>
+
+<p>What was the time of day when the professor spoke?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_58_A_TIME_PUZZLE" id="X_58_A_TIME_PUZZLE"></a><a href="#X_58_A_TIME_PUZZLEa"><b>58.&mdash;A TIME PUZZLE.</b></a></p>
+
+<p>How many minutes is it until six o'clock if fifty minutes ago it was
+four times as many minutes past three o'clock?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_59_A_PUZZLING_WATCH" id="X_59_A_PUZZLING_WATCH"></a><a href="#X_59_A_PUZZLING_WATCHa"><b>59.&mdash;A PUZZLING WATCH.</b></a></p>
+
+<p>A friend pulled out his watch and said, "This watch of mine does not
+keep perfect time; I must have it seen to. I have noticed that the
+minute hand and the hour hand are exactly together every sixty-five
+minutes." Does that watch gain or lose, and how much per hour?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_60_THE_WAPSHAWS_WHARF_MYSTERY" id="X_60_THE_WAPSHAWS_WHARF_MYSTERY"></a><a href="#X_60_THE_WAPSHAWS_WHARF_MYSTERYa"><b>60.&mdash;THE WAPSHAW'S WHARF MYSTERY.</b></a></p>
+
+<p>There was a great commotion in Lower Thames Street on the morning of
+January 12, 1887. When the early members of the staff arrived at
+Wapshaw's Wharf they found that the safe had been broken open, a
+considerable sum of money removed, and the offices left in great
+disorder. The night watchman was nowhere to be found, but nobody who
+had been acquainted with him for one moment suspected him to be guilty
+of the robbery. In this belief the proprietors were confirmed when,
+later in the day, they were informed that the poor fellow's body had
+been picked up by the River Police. Certain marks of violence pointed
+to the fact that he had been brutally attacked and thrown into the
+river. A watch found in his pocket had stopped, as is invariably the
+case in such circumstances, and this was a valuable clue to the time
+of the outrage. But a very stupid officer (and we invariably find one
+or two stupid individuals in the most intelligent bodies of men) had
+actually amused himself by turning the hands round and round, trying
+to set the watch going again. After he had been severely reprimanded
+for this serious indiscretion, he was asked whether he could remember
+the time that was indicated by the watch when found. He replied that
+he could not, but he recollected that the hour hand and minute hand
+were exactly together, one above the other, and the second hand had
+just passed the forty-ninth second. More than this he could not
+remember.</p>
+
+<p>What was the exact time at which the watchman's watch stopped? The
+watch is, of course, assumed to have been an accurate one.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_61_CHANGING_PLACES" id="X_61_CHANGING_PLACES"></a><a href="#X_61_CHANGING_PLACESa"><b>61.&mdash;CHANGING PLACES.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q061.png" width="400" height="397" alt="" title="" />
+</div>
+
+<p>The above clock face indicates a little before 42 minutes past 4. The
+hands will again point at exactly the same spots a little after 23
+minutes past 8. In fact, the hands will have changed places. How many
+times do the hands of a clock change places between three o'clock p.m.
+and midnight? And out of all the pairs of times indicated by these
+changes, what is the exact time when the minute hand will be nearest
+to the point IX?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_62_THE_CLUB_CLOCK" id="X_62_THE_CLUB_CLOCK"></a><a href="#X_62_THE_CLUB_CLOCKa"><b>62.&mdash;THE CLUB CLOCK.</b></a></p>
+
+<p>One of the big clocks in the Cogitators' Club was found the other
+night to have stopped just when, as will be seen in the illustration,
+the second hand was exactly midway between the other two hands. One of
+the members proposed to some of his friends that they should tell him
+the exact time when (if the clock had not <span class='pagenum'>Pg 11<a name="Page_11" id="Page_11"></a></span>stopped) the second hand
+would next again have been midway between the minute hand and the hour
+hand. Can you find the correct time that it would happen?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q062.png" width="400" height="375" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_63_THE_STOP-WATCH" id="X_63_THE_STOP-WATCH"></a><a href="#X_63_THE_STOP-WATCHa"><b>63.&mdash;THE STOP-WATCH.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q063.png" width="400" height="532" alt="" title="" />
+</div>
+
+<p>We have here a stop-watch with three hands. The second hand, which
+travels once round the face in a minute, is the one with the little
+ring at its end near the centre. Our dial indicates the exact time
+when its owner stopped the watch. You will notice that the three hands
+are nearly equidistant. The hour and minute hands point to spots that
+are exactly a third of the circumference apart, but the second hand is
+a little too advanced. An exact equidistance for the three hands is
+not possible. Now, we want to know what the time will be when the
+three hands are next at exactly the same distances as shown from one
+another. Can you state the time?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_64_THE_THREE_CLOCKS" id="X_64_THE_THREE_CLOCKS"></a><a href="#X_64_THE_THREE_CLOCKSa"><b>64.&mdash;THE THREE CLOCKS.</b></a></p>
+
+<p>On Friday, April 1, 1898, three new clocks were all set going
+precisely at the same time&mdash;twelve noon. At noon on the following day
+it was found that clock A had kept perfect time, that clock B had
+gained exactly one minute, and that clock C had lost exactly one
+minute. Now, supposing that the clocks B and C had not been regulated,
+but all three allowed to go on as they had begun, and that they
+maintained the same rates of progress without stopping, on what date
+and at what time of day would all three pairs of hands again point at
+the same moment at twelve o'clock?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_65_THE_RAILWAY_STATION_CLOCK" id="X_65_THE_RAILWAY_STATION_CLOCK"></a><a href="#X_65_THE_RAILWAY_STATION_CLOCKa"><b>65.&mdash;THE RAILWAY STATION CLOCK.</b></a></p>
+
+<p>A clock hangs on the wall of a railway station, 71 ft. 9 in. long and
+10 ft. 4 in. high. Those are the dimensions of the wall, not of the
+clock! While waiting for a train we noticed that the hands of the
+clock were pointing in opposite directions, and were parallel to one
+of the diagonals of the wall. What was the exact time?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_66_THE_VILLAGE_SIMPLETON" id="X_66_THE_VILLAGE_SIMPLETON"></a><a href="#X_66_THE_VILLAGE_SIMPLETONa"><b>66.&mdash;THE VILLAGE SIMPLETON.</b></a></p>
+
+<p>A facetious individual who was taking a long walk in the country came
+upon a yokel sitting on a stile. As the gentleman was not quite sure
+of his road, he thought he would make inquiries of the local
+inhabitant; but at the first glance he jumped too hastily to the
+conclusion that he had dropped on the village idiot. He therefore
+decided to test the fellow's intelligence by first putting to him the
+simplest question he could think of, which was, "What day of the week
+is this, my good man?" The following is the smart answer that he
+received:&mdash;</p>
+
+<p>"When the day after to-morrow is yesterday, to-day will be as far from
+Sunday as to-day was from Sunday when the day before yesterday was
+to-morrow."</p>
+
+<p>Can the reader say what day of the week it was? It is pretty evident
+that the countryman was not such a fool as he looked. The gentleman
+went on his road a puzzled but a wiser man.</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="LOCOMOTION_AND_SPEED_PUZZLES" id="LOCOMOTION_AND_SPEED_PUZZLES"></a><a href="#CONTENTS">LOCOMOTION AND SPEED PUZZLES.</a></h2>
+
+<p class='center'>"The race is not to the swift."&mdash;<i>Ecclesiastes</i> ix. II.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_67_AVERAGE_SPEED" id="X_67_AVERAGE_SPEED"></a><a href="#X_67_AVERAGE_SPEEDa"><b>67.&mdash;AVERAGE SPEED.</b></a></p>
+
+<p>In a recent motor ride it was found that we had gone at the rate of
+ten miles an hour, but we did the return journey over the same route,
+owing to the roads being more clear of traffic, at fifteen miles an
+hour. What was our average speed? Do not be too hasty in your answer
+to this simple little question, or it is pretty certain that you will
+be wrong.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_68_THE_TWO_TRAINS" id="X_68_THE_TWO_TRAINS"></a><a href="#X_68_THE_TWO_TRAINSa"><b>68.&mdash;THE TWO TRAINS.</b></a></p>
+
+<p>I put this little question to a stationmaster, and his correct answer
+was so prompt that I am <span class='pagenum'>Pg 12<a name="Page_12" id="Page_12"></a></span>convinced there is no necessity to seek
+talented railway officials in America or elsewhere.</p>
+
+<p>Two trains start at the same time, one from London to Liverpool, the
+other from Liverpool to London. If they arrive at their destinations
+one hour and four hours respectively after passing one another, how
+much faster is one train running than the other?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_69_THE_THREE_VILLAGES" id="X_69_THE_THREE_VILLAGES"></a><a href="#X_69_THE_THREE_VILLAGESa"><b>69.&mdash;THE THREE VILLAGES.</b></a></p>
+
+<p>I set out the other day to ride in a motor-car from Acrefield to
+Butterford, but by mistake I took the road going <i>via</i> Cheesebury,
+which is nearer Acrefield than Butterford, and is twelve miles to the
+left of the direct road I should have travelled. After arriving at
+Butterford I found that I had gone thirty-five miles. What are the
+three distances between these villages, each being a whole number of
+miles? I may mention that the three roads are quite straight.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_70_DRAWING_HER_PENSION" id="X_70_DRAWING_HER_PENSION"></a><a href="#X_70_DRAWING_HER_PENSIONa"><b>70.&mdash;DRAWING HER PENSION.</b></a></p>
+
+<p>"Speaking of odd figures," said a gentleman who occupies some post in
+a Government office, "one of the queerest characters I know is an old
+lame widow who climbs up a hill every week to draw her pension at the
+village post office. She crawls up at the rate of a mile and a half an
+hour and comes down at the rate of four and a half miles an hour, so
+that it takes her just six hours to make the double journey. Can any
+of you tell me how far it is from the bottom of the hill to the top?"</p>
+
+
+<hr style="width: 30%;" />
+<p><a name="X_71_SIR_EDWYN_DE_TUDOR" id="X_71_SIR_EDWYN_DE_TUDOR"></a><a href="#X_71_SIR_EDWYN_DE_TUDORa"><b>71.&mdash;SIR EDWYN DE TUDOR.</b></a></p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/q071.png" width="500" height="432" alt="" title="" />
+</div>
+
+<p>In the illustration we have a sketch of Sir Edwyn de Tudor going to
+rescue his lady-love, the fair Isabella, who was held a captive by a
+neighbouring wicked baron. Sir Edwyn calculated that if he rode
+fifteen miles an hour he would arrive at the castle an hour too soon,
+while if he rode ten miles an hour he would get there just an hour too
+late. Now, it was of the first importance that he should arrive at the
+exact time appointed, in order that the rescue that he had planned
+should be a success, and the time of the tryst was five o'clock, when
+the captive lady would be taking her afternoon tea. The puzzle is to
+discover exactly how far Sir Edwyn de Tudor had to ride.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_72_THE_HYDROPLANE_QUESTION" id="X_72_THE_HYDROPLANE_QUESTION"></a><a href="#X_72_THE_HYDROPLANE_QUESTIONa"><b>72.&mdash;THE HYDROPLANE QUESTION.</b></a></p>
+
+<p>The inhabitants of Slocomb-on-Sea were greatly excited over the visit
+of a certain flying man. All the town turned out to see the flight of
+the wonderful hydroplane, and, of course, Dobson and his family were
+there. Master Tommy was in good form, and informed his father that
+Englishmen made better airmen than Scotsmen <span class='pagenum'>Pg 13<a name="Page_13" id="Page_13"></a></span>and Irishmen because they
+are not so heavy. "How do you make that out?" asked Mr. Dobson. "Well,
+you see," Tommy replied, "it is true that in Ireland there are men of
+Cork and in Scotland men of Ayr, which is better still, but in England
+there are lightermen." Unfortunately it had to be explained to Mrs.
+Dobson, and this took the edge off the thing. The hydroplane flight
+was from Slocomb to the neighbouring watering-place Poodleville&mdash;five
+miles distant. But there was a strong wind, which so helped the airman
+that he made the outward journey in the short time of ten minutes,
+though it took him an hour to get back to the starting point at
+Slocomb, with the wind dead against him. Now, how long would the ten
+miles have taken him if there had been a perfect calm? Of course, the
+hydroplane's engine worked uniformly throughout.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_73_DONKEY_RIDING" id="X_73_DONKEY_RIDING"></a><a href="#X_73_DONKEY_RIDINGa"><b>73.&mdash;DONKEY RIDING.</b></a></p>
+
+<p>During a visit to the seaside Tommy and Evangeline insisted on having
+a donkey race over the mile course on the sands. Mr. Dobson and some
+of his friends whom he had met on the beach acted as judges, but, as
+the donkeys were familiar acquaintances and declined to part company
+the whole way, a dead heat was unavoidable. However, the judges, being
+stationed at different points on the course, which was marked off in
+quarter-miles, noted the following results:&mdash;The first three-quarters
+were run in six and three-quarter minutes, the first half-mile took
+the same time as the second half, and the third quarter was run in
+exactly the same time as the last quarter. From these results Mr.
+Dobson amused himself in discovering just how long it took those two
+donkeys to run the whole mile. Can you give the answer?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_74_THE_BASKET_OF_POTATOES" id="X_74_THE_BASKET_OF_POTATOES"></a><a href="#X_74_THE_BASKET_OF_POTATOESa"><b>74.&mdash;THE BASKET OF POTATOES.</b></a></p>
+
+<p>A man had a basket containing fifty potatoes. He proposed to his son,
+as a little recreation, that he should place these potatoes on the
+ground in a straight line. The distance between the first and second
+potatoes was to be one yard, between the second and third three yards,
+between the third and fourth five yards, between the fourth and fifth
+seven yards, and so on&mdash;an increase of two yards for every successive
+potato laid down. Then the boy was to pick them up and put them in the
+basket one at a time, the basket being placed beside the first potato.
+How far would the boy have to travel to accomplish the feat of picking
+them all up? We will not consider the journey involved in placing the
+potatoes, so that he starts from the basket with them all laid out.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_75_THE_PASSENGERS_FARE" id="X_75_THE_PASSENGERS_FARE"></a><a href="#X_75_THE_PASSENGERS_FAREa"><b>75.&mdash;THE PASSENGER'S FARE.</b></a></p>
+
+<p>At first sight you would hardly think there was matter for dispute in
+the question involved in the following little incident, yet it took
+the two persons concerned some little time to come to an agreement.
+Mr. Smithers hired a motor-car to take him from Addleford to
+Clinkerville and back again for &pound;3. At Bakenham, just midway, he
+picked up an acquaintance, Mr. Tompkins, and agreed to take him on to
+Clinkerville and bring him back to Bakenham on the return journey. How
+much should he have charged the passenger? That is the question. What
+was a reasonable fare for Mr. Tompkins?</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="DIGITAL_PUZZLES" id="DIGITAL_PUZZLES"></a><a href="#CONTENTS">DIGITAL PUZZLES.</a></h2>
+
+<p class='center'>
+<span style="margin-left: 0em;">"Nine worthies were they called."<br /></span>
+<span style="margin-left: 8em;">DRYDEN: <i>The Flower and the Leaf.</i><br /></span>
+</p>
+
+<p>I give these puzzles, dealing with the nine digits, a class to
+themselves, because I have always thought that they deserve more
+consideration than they usually receive. Beyond the mere trick of
+"casting out nines," very little seems to be generally known of the
+laws involved in these problems, and yet an acquaintance with the
+properties of the digits often supplies, among other uses, a certain
+number of arithmetical checks that are of real value in the saving of
+labour. Let me give just one example&mdash;the first that occurs to me.</p>
+
+<p>If the reader were required to determine whether or not
+15,763,530,163,289 is a square number, how would he proceed? If the
+number had ended with a 2, 3, 7, or 8 in the digits place, of course
+he would know that it could not be a square, but there is nothing in
+its apparent form to prevent its being one. I suspect that in such a
+case he would set to work, with a sigh or a groan, at the laborious
+task of extracting the square root. Yet if he had given a little
+attention to the study of the digital properties of numbers, he would
+settle the question in this simple way. The sum of the digits is 59,
+the sum of which is 14, the sum of which is 5 (which I call the
+"digital root"), and therefore I know that the number cannot be a
+square, and for this reason. The digital root of successive square
+numbers from 1 upwards is always 1, 4, 7, or 9, and can never be
+anything else. In fact, the series, 1, 4, 9, 7, 7, 9, 4, 1, 9, is
+repeated into infinity. The analogous series for triangular numbers is
+1, 3, 6, 1, 6, 3, 1, 9, 9. So here we have a similar negative check,
+for a number cannot be triangular (that is, (n&sup2;+n)/2) if its digital
+root be 2, 4, 5, 7, or 8.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_76_THE_BARREL_OF_BEER" id="X_76_THE_BARREL_OF_BEER"></a><a href="#X_76_THE_BARREL_OF_BEERa"><b>76.&mdash;THE BARREL OF BEER.</b></a></p>
+
+<p>A man bought an odd lot of wine in barrels and one barrel containing
+beer. These are shown in the illustration, marked with the number of
+gallons that each barrel contained. He sold a quantity of the wine to
+one man and twice the quantity to another, but kept the beer to
+himself. The puzzle is to point out which barrel contains beer. Can
+you say which one it is? Of course, the man sold the barrels just as
+he <span class='pagenum'>Pg 14<a name="Page_14" id="Page_14"></a></span>bought them, without manipulating in any way the contents.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q076.png" width="400" height="317" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_77_DIGITS_AND_SQUARES" id="X_77_DIGITS_AND_SQUARES"></a><a href="#X_77_DIGITS_AND_SQUARESa"><b>77.&mdash;DIGITS AND SQUARES.</b></a></p>
+
+<div class="figcenter" style="width: 300px;">
+<img src="images/q077.png" width="300" height="268" alt="" title="" />
+</div>
+
+<p>It will be seen in the diagram that we have so arranged the nine
+digits in a square that the number in the second row is twice that in
+the first row, and the number in the bottom row three times that in
+the top row. There are three other ways of arranging the digits so as
+to produce the same result. Can you find them?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_78_ODD_AND_EVEN_DIGITS" id="X_78_ODD_AND_EVEN_DIGITS"></a><a href="#X_78_ODD_AND_EVEN_DIGITSa"><b>78.&mdash;ODD AND EVEN DIGITS.</b></a></p>
+
+<p>The odd digits, 1, 3, 5, 7, and 9, add up 25, while the even figures,
+2, 4, 6, and 8, only add up 20. Arrange these figures so that the odd
+ones and the even ones add up alike. Complex and improper fractions
+and recurring decimals are not allowed.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_79_THE_LOCKERS_PUZZLE" id="X_79_THE_LOCKERS_PUZZLE"></a><a href="#X_79_THE_LOCKERS_PUZZLEa"><b>79&mdash;THE LOCKERS PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/q079.png" width="500" height="180" alt="" title="" />
+</div>
+
+<p>A man had in his office three cupboards, each containing nine lockers,
+as shown in the diagram. He told his clerk to place a different
+one-figure number on each locker of cupboard A, and to do the same in
+the case of B, and of C. As we are here allowed to call nought a
+digit, and he was not prohibited from using nought as a number, he
+clearly had the option of omitting any one of ten digits from each
+cupboard.</p>
+
+<p>Now, the employer did not say the lockers were to be numbered in any
+numerical order, and he was surprised to find, when the work was done,
+that the figures had apparently been mixed up indiscriminately.
+Calling upon his clerk for an explanation, the eccentric lad stated
+that the notion had occurred to him so to arrange the figures that in
+each case they formed a simple addition sum, the two upper rows of
+figures producing the sum in the lowest row. But the most surprising
+point was this: that he had so arranged them that the addition in A
+gave the smallest possible sum, that the addition in C gave the
+largest possible sum, and that all the nine digits in the three totals
+were different. The puzzle is to show how this could be done. No
+decimals are allowed and the nought may not appear in the hundreds
+place.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_80_THE_THREE_GROUPS" id="X_80_THE_THREE_GROUPS"></a><a href="#X_80_THE_THREE_GROUPSa"><b>80.&mdash;THE THREE GROUPS.</b></a></p>
+
+<p>There appeared in "Nouvelles Annales de Math&eacute;matiques" the following
+puzzle as a modification of one of my "Canterbury Puzzles." Arrange
+the nine digits in three groups of two, three, and four digits, so
+that the first two numbers when multiplied together make the third.
+Thus, 12&nbsp;&times;&nbsp;483&nbsp;=&nbsp;5,796. I now also propose to include the cases where
+there are one, four, and four digits, such as 4&nbsp;&times;&nbsp;1,738&nbsp;=&nbsp;6,952. Can
+you find all the possible solutions in both cases?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_81_THE_NINE_COUNTERS" id="X_81_THE_NINE_COUNTERS"></a><a href="#X_81_THE_NINE_COUNTERSa"><b>81.&mdash;THE NINE COUNTERS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q081.png" width="400" height="137" alt="" title="" />
+</div>
+
+<p>I have nine counters, each bearing one of the nine digits, 1, 2, 3, 4,
+5, 6, 7, 8 and 9. I arranged them on the table in two groups, as shown
+in the illustration, so as to form two multiplication sums, and found
+that both sums gave the same product. You will find that 158
+multiplied by 23 is 3,634, and that 79 multiplied by 46 is also 3,634.
+Now, the puzzle I propose is to rearrange the counters so as to get as
+large a product as possible. What is the best way of placing them?
+Remember both groups must multiply to the same amount, and there must
+be three counters multiplied by two in one case, and two multiplied by
+two counters in the other, just as at present.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 15<a name="Page_15" id="Page_15"></a></span><a name="X_82_THE_TEN_COUNTERS" id="X_82_THE_TEN_COUNTERS"></a><a href="#X_82_THE_TEN_COUNTERSa"><b>82.&mdash;THE TEN COUNTERS.</b></a></p>
+
+<p>In this case we use the nought in addition to the 1, 2, 3, 4, 5, 6, 7,
+8, 9. The puzzle is, as in the last case, so to arrange the ten
+counters that the products of the two multiplications shall be the
+same, and you may here have one or more figures in the multiplier, as
+you choose. The above is a very easy feat; but it is also required to
+find the two arrangements giving pairs of the highest and lowest
+products possible. Of course every counter must be used, and the
+cipher may not be placed to the left of a row of figures where it
+would have no effect. Vulgar fractions or decimals are not allowed.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_83_DIGITAL_MULTIPLICATION" id="X_83_DIGITAL_MULTIPLICATION"></a><a href="#X_83_DIGITAL_MULTIPLICATIONa"><b>83.&mdash;DIGITAL MULTIPLICATION.</b></a></p>
+
+<p>Here is another entertaining problem with the nine digits, the nought
+being excluded. Using each figure once, and only once, we can form two
+multiplication sums that have the same product, and this may be done
+in many ways. For example, 7x658 and 14x329 contain all the digits
+once, and the product in each case is the same&mdash;4,606. Now, it will be
+seen that the sum of the digits in the product is 16, which is neither
+the highest nor the lowest sum so obtainable. Can you find the
+solution of the problem that gives the lowest possible sum of digits
+in the common product? Also that which gives the highest possible sum?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_84_THE_PIERROTS_PUZZLE" id="X_84_THE_PIERROTS_PUZZLE"></a><a href="#X_84_THE_PIERROTS_PUZZLEa"><b>84.&mdash;THE PIERROT'S PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/q084.png" width="500" height="209" alt="" title="" />
+</div>
+
+<p>The Pierrot in the illustration is standing in a posture that
+represents the sign of multiplication. He is indicating the peculiar
+fact that 15 multiplied by 93 produces exactly the same figures
+(1,395), differently arranged. The puzzle is to take any four digits
+you like (all different) and similarly arrange them so that the number
+formed on one side of the Pierrot when multiplied by the number on the
+other side shall produce the same figures. There are very few ways of
+doing it, and I shall give all the cases possible. Can you find them
+all? You are allowed to put two figures on each side of the Pierrot as
+in the example shown, or to place a single figure on one side and
+three figures on the other. If we only used three digits instead of
+four, the only possible ways are these: 3 multiplied by 51 equals 153,
+and 6 multiplied by 21 equals 126.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_85_THE_CAB_NUMBERS" id="X_85_THE_CAB_NUMBERS"></a><a href="#X_85_THE_CAB_NUMBERSa"><b>85.&mdash;THE CAB NUMBERS.</b></a></p>
+
+<p>A London policeman one night saw two cabs drive off in opposite
+directions under suspicious circumstances. This officer was a
+particularly careful and wide-awake man, and he took out his
+pocket-book to make an entry of the numbers of the cabs, but
+discovered that he had lost his pencil. Luckily, however, he found a
+small piece of chalk, with which he marked the two numbers on the
+gateway of a wharf close by. When he returned to the same spot on his
+beat he stood and looked again at the numbers, and noticed this
+peculiarity, that all the nine digits (no nought) were used and that
+no figure was repeated, but that if he multiplied the two numbers
+together they again produced the nine digits, all once, and once only.
+When one of the clerks arrived at the wharf in the early morning, he
+observed the chalk marks and carefully rubbed them out. As the
+policeman could not remember them, certain mathematicians were then
+consulted as to whether there was any known method for discovering all
+the pairs of numbers that have the peculiarity that the officer had
+noticed; but they knew of none. The investigation, however, was
+interesting, and the following question out of many was proposed: What
+two numbers, containing together all the nine digits, will, when
+multiplied together, produce another number (the <i>highest possible</i>)
+containing also all the nine digits? The nought is not allowed
+anywhere.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_86_QUEER_MULTIPLICATION" id="X_86_QUEER_MULTIPLICATION"></a><a href="#X_86_QUEER_MULTIPLICATIONa"><b>86.&mdash;QUEER MULTIPLICATION.</b></a></p>
+
+<p>If I multiply 51,249,876 by 3 (thus using all the nine digits once,
+and once only), I get 153,749,628 (which again contains all the nine
+digits once). Similarly, if I multiply 16,583,742 by 9 the <span class='pagenum'>Pg 16<a name="Page_16" id="Page_16"></a></span>result is
+149,253,678, where in each case all the nine digits are used. Now,
+take 6 as your multiplier and try to arrange the remaining eight
+digits so as to produce by multiplication a number containing all nine
+once, and once only. You will find it far from easy, but it can be
+done.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_87_THE_NUMBER_CHECKS_PUZZLE" id="X_87_THE_NUMBER_CHECKS_PUZZLE"></a><a href="#X_87_THE_NUMBER_CHECKS_PUZZLEa"><b>87.&mdash;THE NUMBER CHECKS PUZZLE.</b></a></p>
+
+<p>Where a large number of workmen are employed on a building it is
+customary to provide every man with a little disc bearing his number.
+These are hung on a board by the men as they arrive, and serve as a
+check on punctuality. Now, I once noticed a foreman remove a number of
+these checks from his board and place them on a split-ring which he
+carried in his pocket. This at once gave me the idea for a good
+puzzle. In fact, I will confide to my readers that this is just how
+ideas for puzzles arise. You cannot really create an idea: it
+happens&mdash;and you have to be on the alert to seize it when it does so
+happen.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q087.png" width="400" height="335" alt="" title="" />
+</div>
+
+<p>It will be seen from the illustration that there are ten of these
+checks on a ring, numbered 1 to 9 and 0. The puzzle is to divide them
+into three groups without taking any off the ring, so that the first
+group multiplied by the second makes the third group. For example, we
+can divide them into the three groups, 2&mdash;8 9 7&mdash;1 5 4 6 3, by
+bringing the 6 and the 3 round to the 4, but unfortunately the first
+two when multiplied together do not make the third. Can you separate
+them correctly? Of course you may have as many of the checks as you
+like in any group. The puzzle calls for some ingenuity, unless you
+have the luck to hit on the answer by chance.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_88_DIGITAL_DIVISION" id="X_88_DIGITAL_DIVISION"></a><a href="#X_88_DIGITAL_DIVISIONa"><b>88.&mdash;DIGITAL DIVISION.</b></a></p>
+
+<p>It is another good puzzle so to arrange the nine digits (the nought
+excluded) into two groups so that one group when divided by the other
+produces a given number without remainder. For example, 1 3 4 5 8
+divided by 6 7 2 9 gives 2. Can the reader find similar arrangements
+producing 3, 4, 5, 6, 7, 8, and 9 respectively? Also, can he find the
+pairs of smallest possible numbers in each case? Thus, 1 4 6 5 8
+divided by 7 3 2 9 is just as correct for 2 as the other example we
+have given, but the numbers are higher.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_89_ADDING_THE_DIGITS" id="X_89_ADDING_THE_DIGITS"></a><a href="#X_89_ADDING_THE_DIGITSa"><b>89.&mdash;ADDING THE DIGITS.</b></a></p>
+
+<p>If I write the sum of money, &pound;987, 5<i>s</i>. 4�<i>d</i>.., and add up the digits,
+they sum to 36. No digit has thus been used a second time in the
+amount or addition. This is the largest amount possible under the
+conditions. Now find the smallest possible amount, pounds, shillings,
+pence, and farthings being all represented. You need not use more of
+the nine digits than you choose, but no digit may be repeated
+throughout. The nought is not allowed.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_90_THE_CENTURY_PUZZLE" id="X_90_THE_CENTURY_PUZZLE"></a><a href="#X_90_THE_CENTURY_PUZZLEa"><b>90.&mdash;THE CENTURY PUZZLE.</b></a></p>
+
+<p>Can you write 100 in the form of a mixed number, using all the nine
+digits once, and only once? The late distinguished French
+mathematician, Edouard Lucas, found seven different ways of doing it,
+and expressed his doubts as to there being any other ways. As a matter
+of fact there are just eleven ways and no more. Here is one of them,
+91 <sup>5742</sup>/<sub>638</sub>. Nine of the other ways have similarly two figures in the
+integral part of the number, but the eleventh expression has only one
+figure there. Can the reader find this last form?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_91_MORE_MIXED_FRACTIONS" id="X_91_MORE_MIXED_FRACTIONS"></a><a href="#X_91_MORE_MIXED_FRACTIONSa"><b>91.&mdash;MORE MIXED FRACTIONS.</b></a></p>
+
+<p>When I first published my solution to the last puzzle, I was led to
+attempt the expression of all numbers in turn up to 100 by a mixed
+fraction containing all the nine digits. Here are twelve numbers for
+the reader to try his hand at: 13, 14, 15, 16, 18, 20, 27, 36, 40, 69,
+72, 94. Use every one of the nine digits once, and only once, in every
+case.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_92_DIGITAL_SQUARE_NUMBERS" id="X_92_DIGITAL_SQUARE_NUMBERS"></a><a href="#X_92_DIGITAL_SQUARE_NUMBERSa"><b>92.&mdash;DIGITAL SQUARE NUMBERS.</b></a></p>
+
+<p>Here are the nine digits so arranged that they form four square
+numbers: 9, 81, 324, 576. Now, can you put them all together so as to
+form a single square number&mdash;(I) the smallest possible, and (II) the
+largest possible?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_93_THE_MYSTIC_ELEVEN" id="X_93_THE_MYSTIC_ELEVEN"></a><a href="#X_93_THE_MYSTIC_ELEVENa"><b>93.&mdash;THE MYSTIC ELEVEN.</b></a></p>
+
+<p>Can you find the largest possible number containing any nine of the
+ten digits (calling nought a digit) that can be divided by 11 without
+a remainder? Can you also find the smallest possible number produced
+in the same way that is divisible by 11? Here is an example, where the
+digit 5 has been omitted: 896743012. This number contains nine of the
+digits and is divisible by 11, but it is neither the largest nor the
+smallest number that will work.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_94_THE_DIGITAL_CENTURY" id="X_94_THE_DIGITAL_CENTURY"></a><a href="#X_94_THE_DIGITAL_CENTURYa"><b>94.&mdash;THE DIGITAL CENTURY.</b></a></p>
+
+<p>1 2 3 4 5 6 7 8 9&nbsp;=&nbsp;100.</p>
+
+<p>It is required to place arithmetical signs between the nine figures so
+that they shall equal <span class='pagenum'>Pg 17<a name="Page_17" id="Page_17"></a></span>100. Of course, you must not alter the present
+numerical arrangement of the figures. Can you give a correct solution
+that employs (1) the fewest possible signs, and (2) the fewest
+possible separate strokes or dots of the pen? That is, it is necessary
+to use as few signs as possible, and those signs should be of the
+simplest form. The signs of addition and multiplication (+ and &times;) will
+thus count as two strokes, the sign of subtraction (-) as one stroke,
+the sign of division (&divide;) as three, and so on.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_95_THE_FOUR_SEVENS" id="X_95_THE_FOUR_SEVENS"></a><a href="#X_95_THE_FOUR_SEVENSa"><b>95.&mdash;THE FOUR SEVENS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q095.png" width="400" height="500" alt="" title="" />
+</div>
+
+<p>In the illustration Professor Rackbrane is seen demonstrating one of
+the little posers with which he is accustomed to entertain his class.
+He believes that by taking his pupils off the beaten tracks he is the
+better able to secure their attention, and to induce original and
+ingenious methods of thought. He has, it will be seen, just shown how
+four 5's may be written with simple arithmetical signs so as to
+represent 100. Every juvenile reader will see at a glance that his
+example is quite correct. Now, what he wants you to do is this:
+Arrange four 7's (neither more nor less) with arithmetical signs so
+that they shall represent 100. If he had said we were to use four 9's
+we might at once have written 99<sup>9</sup>/<sub>9</sub>, but the four 7's call for rather
+more ingenuity. Can you discover the little trick?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_96_THE_DICE_NUMBERS" id="X_96_THE_DICE_NUMBERS"></a><a href="#X_96_THE_DICE_NUMBERSa"><b>96.&mdash;THE DICE NUMBERS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q096.png" width="400" height="140" alt="" title="" />
+</div>
+
+<p>I have a set of four dice, not marked with spots in the ordinary way,
+but with Arabic figures, as shown in the illustration. Each die, of
+course, bears the numbers 1 to 6. When put together they will form a
+good many, different numbers. As represented they make the number
+1246. Now, if I make all the different four-figure numbers that are
+possible with these dice (never putting the same figure more than once
+in any number), what will they all add up to? You are allowed to turn
+the 6 upside down, so as to represent a 9. I do not ask, or expect,
+the reader to go to all the labour of writing out the full list of
+numbers and then adding them up. Life is not long enough for such
+wasted energy. Can you get at the answer in any other way?</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="VARIOUS_ARITHMETICAL_AND_ALGEBRAICAL_PROBLEMS" id="VARIOUS_ARITHMETICAL_AND_ALGEBRAICAL_PROBLEMS"></a><a href="#CONTENTS">VARIOUS ARITHMETICAL AND ALGEBRAICAL PROBLEMS.</a></h2>
+
+<p class='center'>
+<span style="margin-left: 0em">"Variety's the very spice of life,<br /></span>
+<span style="margin-left: 0em">That gives it all its flavour."<br /></span>
+<span style="margin-left: 8em">COWPER: <i>The Task.</i><br /></span>
+</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_97_THE_SPOT_ON_THE_TABLE" id="X_97_THE_SPOT_ON_THE_TABLE"></a><a href="#X_97_THE_SPOT_ON_THE_TABLEa"><b>97.&mdash;THE SPOT ON THE TABLE.</b></a></p>
+
+<p>A boy, recently home from school, wished to give his father an
+exhibition of his precocity. He pushed a large circular table into the
+corner of the room, as shown in the illustration, so that it touched
+both walls, and he then pointed to a spot of ink on the extreme edge.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q097.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>"Here is a little puzzle for you, pater," said the youth. "That spot
+is exactly eight inches from one wall and nine inches from the other.
+Can you tell me the diameter of the table without measuring it?"</p>
+
+<p>The boy was overheard to tell a friend, "It <span class='pagenum'>Pg 18<a name="Page_18" id="Page_18"></a></span>fairly beat the guv'nor;"
+but his father is known to have remarked to a City acquaintance that
+he solved the thing in his head in a minute. I often wonder which
+spoke the truth.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_98_ACADEMIC_COURTESIES" id="X_98_ACADEMIC_COURTESIES"></a><a href="#X_98_ACADEMIC_COURTESIESa"><b>98.&mdash;ACADEMIC COURTESIES.</b></a></p>
+
+<p>In a certain mixed school, where a special feature was made of the
+inculcation of good manners, they had a curious rule on assembling
+every morning. There were twice as many girls as boys. Every girl made
+a bow to every other girl, to every boy, and to the teacher. Every boy
+made a bow to every other boy, to every girl, and to the teacher. In
+all there were nine hundred bows made in that model academy every
+morning. Now, can you say exactly how many boys there were in the
+school? If you are not very careful, you are likely to get a good deal
+out in your calculation.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_99_THE_THIRTY-THREE_PEARLS" id="X_99_THE_THIRTY-THREE_PEARLS"></a><a href="#X_99_THE_THIRTY-THREE_PEARLSa"><b>99.&mdash;THE THIRTY-THREE PEARLS.</b></a></p>
+
+<p>"A man I know," said Teddy Nicholson at a certain family party,
+"possesses a string of thirty-three pearls. The middle pearl is the
+largest and best of all, and the others are so selected and arranged
+that, starting from one end, each successive pearl is worth &pound;100 more
+than the preceding one, right up to the big pearl. From the other end
+the pearls increase in value by &pound;150 up to the large pearl. The whole
+string is worth &pound;65,000. What is the value of that large pearl?"</p>
+
+<div class="figcenter" style="width: 300px;">
+<img src="images/q099.png" width="300" height="668" alt="" title="" />
+</div>
+
+<p>"Pearls and other articles of clothing," said Uncle Walter, when the
+price of the precious gem had been discovered, "remind me of Adam and
+Eve. Authorities, you may not know, differ as to the number of apples
+that were eaten by Adam and Eve. It is the opinion of some that Eve 8
+(ate) and Adam 2 (too), a total of 10 only. But certain mathematicians
+have figured it out differently, and hold that Eve 8 and Adam a total
+of 16. Yet the most recent investigators think the above figures
+entirely wrong, for if Eve 8 and Adam 8 2, the total must be 90."</p>
+
+<p>"Well," said Harry, "it seems to me that if there were giants in those
+days, probably Eve 8 1 and Adam 8 2, which would give a total of 163."</p>
+
+<p>"I am not at all satisfied," said Maud. "It seems to me that if Eve 8
+1 and Adam 8 1 2, they together consumed 893."</p>
+
+<p>"I am sure you are all wrong," insisted Mr. Wilson, "for I consider
+that Eve 8 1 4 Adam, and Adam 8 1 2 4 Eve, so we get a total of
+8,938."</p>
+
+<p>"But, look here," broke in Herbert. "If Eve 8 1 4 Adam and Adam 8 1 2
+4 2 oblige Eve, surely the total must have been 82,056!"</p>
+
+<p>At this point Uncle Walter suggested that they might let the matter
+rest. He declared it to be clearly what mathematicians call an
+indeterminate problem.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_100_THE_LABOURERS_PUZZLE" id="X_100_THE_LABOURERS_PUZZLE"></a><a href="#X_100_THE_LABOURERS_PUZZLEa"><b>100.&mdash;THE LABOURER'S PUZZLE.</b></a></p>
+
+<p>Professor Rackbrane, during one of his rambles, chanced to come upon a
+man digging a deep hole.</p>
+
+<p>"Good morning," he said. "How deep is that hole?"</p>
+
+<p>"Guess," replied the labourer. "My height is exactly five feet ten
+inches."</p>
+
+<p>"How much deeper are you going?" said the professor.</p>
+
+<p>"I am going twice as deep," was the answer, "and then my head will be
+twice as far below ground as it is now above ground."</p>
+
+<p>Rackbrane now asks if you could tell how deep that hole would be when
+finished.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_101_THE_TRUSSES_OF_HAY" id="X_101_THE_TRUSSES_OF_HAY"></a><a href="#X_101_THE_TRUSSES_OF_HAYa"><b>101.&mdash;THE TRUSSES OF HAY.</b></a></p>
+
+<p>Farmer Tompkins had five trusses of hay, which he told his man Hodge
+to weigh before delivering them to a customer. The stupid fellow
+weighed them two at a time in all possible ways, and informed his
+master that the weights in pounds were 110, 112, 113, 114, 115, 116,
+117, 118, 120, and 121. Now, how was Farmer Tompkins to find out from
+these figures how much every one of the five trusses weighed singly?
+The reader may at first think that he ought to be told "which pair is
+which pair," or something of that sort, but it is quite unnecessary.
+Can you give the five correct weights?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_102_MR_GUBBINS_IN_A_FOG" id="X_102_MR_GUBBINS_IN_A_FOG"></a><a href="#X_102_MR_GUBBINS_IN_A_FOGa"><b>102.&mdash;MR. GUBBINS IN A FOG.</b></a></p>
+
+<p>Mr. Gubbins, a diligent man of business, was much inconvenienced by a
+London fog. The electric light happened to be out of order and he had
+to manage as best he could with two candles. His clerk assured him
+that though <span class='pagenum'>Pg 19<a name="Page_19" id="Page_19"></a></span>both were of the same length one candle would burn for
+four hours and the other for five hours. After he had been working
+some time he put the candles out as the fog had lifted, and he then
+noticed that what remained of one candle was exactly four times the
+length of what was left of the other.</p>
+
+<p>When he got home that night Mr. Gubbins, who liked a good puzzle, said
+to himself, "Of course it is possible to work out just how long those
+two candles were burning to-day. I'll have a shot at it." But he soon
+found himself in a worse fog than the atmospheric one. Could you have
+assisted him in his dilemma? How long were the candles burning?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_103_PAINTING_THE_LAMP-POSTS" id="X_103_PAINTING_THE_LAMP-POSTS"></a><a href="#X_103_PAINTING_THE_LAMP-POSTSa"><b>103.&mdash;PAINTING THE LAMP-POSTS.</b></a></p>
+
+<p>Tim Murphy and Pat Donovan were engaged by the local authorities to
+paint the lamp-posts in a certain street. Tim, who was an early riser,
+arrived first on the job, and had painted three on the south side when
+Pat turned up and pointed out that Tim's contract was for the north
+side. So Tim started afresh on the north side and Pat continued on the
+south. When Pat had finished his side he went across the street and
+painted six posts for Tim, and then the job was finished. As there was
+an equal number of lamp-posts on each side of the street, the simple
+question is: Which man painted the more lamp-posts, and just how many
+more?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_104_CATCHING_THE_THIEF" id="X_104_CATCHING_THE_THIEF"></a><a href="#X_104_CATCHING_THE_THIEFa"><b>104.&mdash;CATCHING THE THIEF.</b></a></p>
+
+<p>"Now, constable," said the defendant's counsel in cross-examination,"
+you say that the prisoner was exactly twenty-seven steps ahead of you
+when you started to run after him?"</p>
+
+<p>"Yes, sir."</p>
+
+<p>"And you swear that he takes eight steps to your five?"</p>
+
+<p>"That is so."</p>
+
+<p>"Then I ask you, constable, as an intelligent man, to explain how you
+ever caught him, if that is the case?"</p>
+
+<p>"Well, you see, I have got a longer stride. In fact, two of my steps
+are equal in length to five of the prisoner's. If you work it out, you
+will find that the number of steps I required would bring me exactly
+to the spot where I captured him."</p>
+
+<p>Here the foreman of the jury asked for a few minutes to figure out the
+number of steps the constable must have taken. Can you also say how
+many steps the officer needed to catch the thief?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_105_THE_PARISH_COUNCIL_ELECTION" id="X_105_THE_PARISH_COUNCIL_ELECTION"></a><a href="#X_105_THE_PARISH_COUNCIL_ELECTIONa"><b>105.&mdash;THE PARISH COUNCIL ELECTION.</b></a></p>
+
+<p>Here is an easy problem for the novice. At the last election of the
+parish council of Tittlebury-in-the-Marsh there were twenty-three
+candidates for nine seats. Each voter was qualified to vote for nine
+of these candidates or for any less number. One of the electors wants
+to know in just how many different ways it was possible for him to
+vote.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_106_THE_MUDDLETOWN_ELECTION" id="X_106_THE_MUDDLETOWN_ELECTION"></a><a href="#X_106_THE_MUDDLETOWN_ELECTIONa"><b>106.&mdash;THE MUDDLETOWN ELECTION.</b></a></p>
+
+<p>At the last Parliamentary election at Muddletown 5,473 votes were
+polled. The Liberal was elected by a majority of 18 over the
+Conservative, by 146 over the Independent, and by 575 over the
+Socialist. Can you give a simple rule for figuring out how many votes
+were polled for each candidate?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_107_THE_SUFFRAGISTS_MEETING" id="X_107_THE_SUFFRAGISTS_MEETING"></a><a href="#X_107_THE_SUFFRAGISTS_MEETINGa"><b>107.&mdash;THE SUFFRAGISTS' MEETING.</b></a></p>
+
+<p>At a recent secret meeting of Suffragists a serious difference of
+opinion arose. This led to a split, and a certain number left the
+meeting. "I had half a mind to go myself," said the chair-woman, "and
+if I had done so, two-thirds of us would have retired." "True," said
+another member; "but if I had persuaded my friends Mrs. Wild and
+Christine Armstrong to remain we should only have lost half our
+number." Can you tell how many were present at the meeting at the
+start?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_108_THE_LEAP-YEAR_LADIES" id="X_108_THE_LEAP-YEAR_LADIES"></a><a href="#X_108_THE_LEAP-YEAR_LADIESa"><b>108.&mdash;THE LEAP-YEAR LADIES.</b></a></p>
+
+<p>Last leap-year ladies lost no time in exercising the privilege of
+making proposals of marriage. If the figures that reached me from an
+occult source are correct, the following represents the state of
+affairs in this country.</p>
+
+<p>A number of women proposed once each, of whom one-eighth were widows.
+In consequence, a number of men were to be married of whom
+one-eleventh were widowers. Of the proposals made to widowers,
+one-fifth were declined. All the widows were accepted. Thirty-five
+forty-fourths of the widows married bachelors. One thousand two
+hundred and twenty-one spinsters were declined by bachelors. The
+number of spinsters accepted by bachelors was seven times the number
+of widows accepted by bachelors. Those are all the particulars that I
+was able to obtain. Now, how many women proposed?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_109_THE_GREAT_SCRAMBLE" id="X_109_THE_GREAT_SCRAMBLE"></a><a href="#X_109_THE_GREAT_SCRAMBLEa"><b>109.&mdash;THE GREAT SCRAMBLE.</b></a></p>
+
+<p>After dinner, the five boys of a household happened to find a parcel
+of sugar-plums. It was quite unexpected loot, and an exciting scramble
+ensued, the full details of which I will recount with accuracy, as it
+forms an interesting puzzle.</p>
+
+<p>You see, Andrew managed to get possession of just two-thirds of the
+parcel of sugar-plums. Bob at once grabbed three-eighths of these, and
+Charlie managed to seize three-tenths also. Then young David dashed
+upon the scene, and captured all that Andrew had left, except
+one-seventh, which Edgar artfully secured for himself by a cunning
+trick. Now the fun began in real earnest, for Andrew and Charlie
+jointly set upon Bob, who stumbled against the fender and dropped half
+of all that he had, which were equally picked up by David and Edgar,
+who had crawled under a table and were waiting. Next, Bob sprang on
+Charlie from a chair, and upset all the latter's collection on to the
+floor. Of this prize Andrew got just a quarter, Bob <span class='pagenum'>Pg 20<a name="Page_20" id="Page_20"></a></span>gathered up
+one-third, David got two-sevenths, while Charlie and Edgar divided
+equally what was left of that stock.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q109.png" width="400" height="454" alt="" title="" />
+</div>
+
+<p>They were just thinking the fray was over when David suddenly struck
+out in two directions at once, upsetting three-quarters of what Bob
+and Andrew had last acquired. The two latter, with the greatest
+difficulty, recovered five-eighths of it in equal shares, but the
+three others each carried off one-fifth of the same. Every sugar-plum
+was now accounted for, and they called a truce, and divided equally
+amongst them the remainder of the parcel. What is the smallest number
+of sugar-plums there could have been at the start, and what proportion
+did each boy obtain?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_110_THE_ABBOTS_PUZZLE" id="X_110_THE_ABBOTS_PUZZLE"></a><a href="#X_110_THE_ABBOTS_PUZZLEa"><b>110.&mdash;THE ABBOT'S PUZZLE.</b></a></p>
+
+<p>The first English puzzlist whose name has come down to us was a
+Yorkshireman&mdash;no other than Alcuin, Abbot of Canterbury (A.D.
+735-804). Here is a little puzzle from his works, which is at least
+interesting on account of its antiquity. "If 100 bushels of corn were
+distributed among 100 people in such a manner that each man received
+three bushels, each woman two, and each child half a bushel, how many
+men, women, and children were there?"</p>
+
+<p>Now, there are six different correct answers, if we exclude a case
+where there would be no women. But let us say that there were just
+five times as many women as men, then what is the correct solution?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_111_REAPING_THE_CORN" id="X_111_REAPING_THE_CORN"></a><a href="#X_111_REAPING_THE_CORNa"><b>111.&mdash;REAPING THE CORN.</b></a></p>
+
+<p>A farmer had a square cornfield. The corn was all ripe for reaping,
+and, as he was short of men, it was arranged that he and his son
+should share the work between them. The farmer first cut one rod wide
+all round the square, thus leaving a smaller square of standing corn
+in the middle of the field. "Now," he said to his son, "I have cut my
+half of the field, and you can do your share." The son was not quite
+satisfied as to the proposed division of labour, and as the village
+schoolmaster happened to be passing, he appealed to that person to
+decide the matter. He found the farmer was quite correct, provided
+there was no dispute as to the size of the field, and on this point
+they were agreed. Can you tell the area of the field, as that
+ingenious schoolmaster succeeded in doing?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_112_A_PUZZLING_LEGACY" id="X_112_A_PUZZLING_LEGACY"></a><a href="#X_112_A_PUZZLING_LEGACYa"><b>112.&mdash;A PUZZLING LEGACY.</b></a></p>
+
+<p>A man left a hundred acres of land to be divided among his three
+sons&mdash;Alfred, Benjamin, and Charles&mdash;in the proportion of one-third,
+one-fourth, and one-fifth respectively. But Charles died. How was the
+land to be divided fairly between Alfred and Benjamin?</p>
+<hr style="width: 30%;" />
+<p><a name="X_113_THE_TORN_NUMBER" id="X_113_THE_TORN_NUMBER"></a><a href="#X_113_THE_TORN_NUMBERa"><b>113.&mdash;THE TORN NUMBER.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q113.png" width="400" height="165" alt="" title="" />
+</div>
+
+<p>I had the other day in my possession a label bearing the number 3 0 2 5
+in large figures. This got accidentally torn in half, so that 3 0 was on
+one piece and 2 5 on the other, as shown on the illustration. On
+looking at these pieces I began to make a calculation, scarcely
+conscious of what I was doing, when I discovered this little
+peculiarity. If we add the 3 and the 2 5 together and square the sum
+we get as the result the complete original number on the label! Thus,
+30 added to 25 is 55, and 55 multiplied by 55 is 3025. Curious, is it
+not? Now, the puzzle is to find another number, composed of four
+figures, all different, which may be divided in the middle and produce
+the same result.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_114_CURIOUS_NUMBERS" id="X_114_CURIOUS_NUMBERS"></a><a href="#X_114_CURIOUS_NUMBERSa"><b>114.&mdash;CURIOUS NUMBERS.</b></a></p>
+
+<p>The number 48 has this peculiarity, that if you add 1 to it the result
+is a square number (49, the square of 7), and if you add 1 to its
+half, you also get a square number (25, the square of 5). Now, there
+is no limit to the numbers that have this peculiarity, and it is an
+interesting puzzle to find three more of them&mdash;the smallest possible
+numbers. What are they?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_115_A_PRINTERS_ERROR" id="X_115_A_PRINTERS_ERROR"></a><a href="#X_115_A_PRINTERS_ERRORa"><b>115.&mdash;A PRINTER'S ERROR.</b></a></p>
+
+<p>In a certain article a printer had to set up the figures 5<sup>4</sup>&nbsp;&times;&nbsp;2<sup>3</sup>,
+which, of course, means that the fourth power of 5 (625) is to be
+multiplied by the cube of 2 (8), the product of which is 5,000. But he
+printed 5<sup>4</sup>&nbsp;&times;&nbsp;2<sup>3</sup> as 5 4 2 3, which is not correct. Can you place four
+digits in the manner shown, so that it will be equally correct if the
+printer sets it up aright or makes the same blunder?</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 21<a name="Page_21" id="Page_21"></a></span><a name="X_116_THE_CONVERTED_MISER" id="X_116_THE_CONVERTED_MISER"></a><a href="#X_116_THE_CONVERTED_MISERa"><b>116.&mdash;THE CONVERTED MISER.</b></a></p>
+
+<p>Mr. Jasper Bullyon was one of the very few misers who have ever been
+converted to a sense of their duty towards their less fortunate
+fellow-men. One eventful night he counted out his accumulated wealth,
+and resolved to distribute it amongst the deserving poor.</p>
+
+<p>He found that if he gave away the same number of pounds every day in
+the year, he could exactly spread it over a twelvemonth without there
+being anything left over; but if he rested on the Sundays, and only
+gave away a fixed number of pounds every weekday, there would be one
+sovereign left over on New Year's Eve. Now, putting it at the lowest
+possible, what was the exact number of pounds that he had to
+distribute?</p>
+
+<p>Could any question be simpler? A sum of pounds divided by one number
+of days leaves no remainder, but divided by another number of days
+leaves a sovereign over. That is all; and yet, when you come to tackle
+this little question, you will be surprised that it can become so
+puzzling.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_117_A_FENCE_PROBLEM" id="X_117_A_FENCE_PROBLEM"></a><a href="#X_117_A_FENCE_PROBLEMa"><b>117.&mdash;A FENCE PROBLEM.</b></a></p>
+
+<p>The practical usefulness of puzzles is a point that we are liable to
+overlook. Yet, as a matter of fact, I have from time to time received
+quite a large number of letters from individuals who have found that
+the mastering of some little principle upon which a puzzle was built
+has proved of considerable value to them in a most unexpected way.
+Indeed, it may be accepted as a good maxim that a puzzle is of little
+real value unless, as well as being amusing and perplexing, it
+conceals some instructive and possibly useful feature. It is, however,
+very curious how these little bits of acquired knowledge dovetail into
+the occasional requirements of everyday life, and equally curious to
+what strange and mysterious uses some of our readers seem to apply
+them. What, for example, can be the object of Mr. Wm. Oxley, who
+writes to me all the way from Iowa, in wishing to ascertain the
+dimensions of a field that he proposes to enclose, containing just as
+many acres as there shall be rails in the fence?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q117.png" width="400" height="341" alt="" title="" />
+</div>
+
+<p>The man wishes to fence in a perfectly square field which is to
+contain just as many acres as there are rails in the required fence.
+Each hurdle, or portion of fence, is seven rails high, and two lengths
+would extend one pole (16� ft.): that is to say, there are fourteen
+rails to the pole, lineal measure. Now, what must be the size of the
+field?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_118_CIRCLING_THE_SQUARES" id="X_118_CIRCLING_THE_SQUARES"></a><a href="#X_118_CIRCLING_THE_SQUARESa"><b>118.&mdash;CIRCLING THE SQUARES.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q118.png" width="400" height="412" alt="" title="" />
+</div>
+
+<p>The puzzle is to place a different number in each of the ten squares
+so that the sum of the squares of any two adjacent numbers shall be
+equal to the sum of the squares of the two numbers diametrically
+opposite to them. The four numbers placed, as examples, must stand as
+they are. The square of 16 is 256, and the square of 2 is 4. Add these
+together, and the result is 260. Also&mdash;the square of 14 is 196, and
+the square of 8 is 64. These together also make 260. Now, in precisely
+the same way, B and C should be equal to G and H (the sum will not
+necessarily be 260), A and K to F and E, H and I to C and D, and so
+on, with any two adjoining squares in the circle.</p>
+
+<p>All you have to do is to fill in the remaining six numbers. Fractions
+are not allowed, and I shall show that no number need contain more
+than two figures.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_119_RACKBRANES_LITTLE_LOSS" id="X_119_RACKBRANES_LITTLE_LOSS"></a><a href="#X_119_RACKBRANES_LITTLE_LOSSa"><b>119.&mdash;RACKBRANE'S LITTLE LOSS.</b></a></p>
+
+<p>Professor Rackbrane was spending an evening with his old friends, Mr.
+and Mrs. Potts, and they engaged in some game (he does not say what
+game) of cards. The professor lost the first game, which resulted in
+doubling the money that both Mr. and Mrs. Potts had laid on the table.
+The second game was lost by Mrs. Potts, which doubled the money then
+held by her husband and the professor. Curiously enough, the third
+game was lost by Mr. Potts, and had the <span class='pagenum'>Pg 22<a name="Page_22" id="Page_22"></a></span>effect of doubling the money
+then held by his wife and the professor. It was then found that each
+person had exactly the same money, but the professor had lost five
+shillings in the course of play. Now, the professor asks, what was the
+sum of money with which he sat down at the table? Can you tell him?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_120_THE_FARMER_AND_HIS_SHEEP" id="X_120_THE_FARMER_AND_HIS_SHEEP"></a><a href="#X_120_THE_FARMER_AND_HIS_SHEEPa"><b>120.&mdash;THE FARMER AND HIS SHEEP.</b></a></p>
+
+<div class="figcenter" style="width: 550px;">
+<img src="images/q120.png" width="550" height="391" alt="" title="" />
+</div>
+
+<p>Farmer Longmore had a curious aptitude for arithmetic, and was known
+in his district as the "mathematical farmer." The new vicar was not
+aware of this fact when, meeting his worthy parishioner one day in the
+lane, he asked him in the course of a short conversation, "Now, how
+many sheep have you altogether?" He was therefore rather surprised at
+Longmore's answer, which was as follows: "You can divide my sheep into
+two different parts, so that the difference between the two numbers is
+the same as the difference between their squares. Maybe, Mr. Parson,
+you will like to work out the little sum for yourself."</p>
+
+<p>Can the reader say just how many sheep the farmer had? Supposing he
+had possessed only twenty sheep, and he divided them into the two
+parts 12 and 8. Now, the difference between their squares, 144 and 64,
+is 80. So that will not do, for 4 and 80 are certainly not the same.
+If you can find numbers that work out correctly, you will know exactly
+how many sheep Farmer Longmore owned.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_121_HEADS_OR_TAILS" id="X_121_HEADS_OR_TAILS"></a><a href="#X_121_HEADS_OR_TAILSa"><b>121.&mdash;HEADS OR TAILS.</b></a></p>
+
+<p>Crooks, an inveterate gambler, at Goodwood recently said to a friend,
+"I'll bet you half the money in my pocket on the toss of a coin&mdash;heads
+I win, tails I lose." The coin was tossed and the money handed over.
+He repeated the offer again and again, each time betting half the
+money then in his possession. We are not told how long the game went
+on, or how many times the coin was tossed, but this we know, that the
+number of times that Crooks lost was exactly equal to the number of
+times that he won. Now, did he gain or lose by this little venture?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_122_THE_SEE-SAW_PUZZLE" id="X_122_THE_SEE-SAW_PUZZLE"></a><a href="#X_122_THE_SEE-SAW_PUZZLEa"><b>122.&mdash;THE SEE-SAW PUZZLE.</b></a></p>
+
+<p>Necessity is, indeed, the mother of invention. I was amused the other
+day in watching a boy who wanted to play see-saw and, in his failure
+to find another child to share the sport with him, had been driven
+back upon the ingenious resort of tying a number of bricks to one end
+of the plank to balance his weight at the other.</p>
+
+<p>As a matter of fact, he just balanced against sixteen bricks, when
+these were fixed to the short end of plank, but if he fixed them to
+the long end of plank he only needed eleven as balance.</p>
+
+<p>Now, what was that boy's weight, if a brick <span class='pagenum'>Pg 23<a name="Page_23" id="Page_23"></a></span>weighs equal to a
+three-quarter brick and three-quarters of a pound?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_123_A_LEGAL_DIFFICULTY" id="X_123_A_LEGAL_DIFFICULTY"></a><a href="#X_123_A_LEGAL_DIFFICULTYa"><b>123.&mdash;A LEGAL DIFFICULTY.</b></a></p>
+
+<p>"A client of mine," said a lawyer, "was on the point of death when his
+wife was about to present him with a child. I drew up his will, in
+which he settled two-thirds of his estate upon his son (if it should
+happen to be a boy) and one-third on the mother. But if the child
+should be a girl, then two-thirds of the estate should go to the
+mother and one-third to the daughter. As a matter of fact, after his
+death twins were born&mdash;a boy and a girl. A very nice point then arose.
+How was the estate to be equitably divided among the three in the
+closest possible accordance with the spirit of the dead man's will?"</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_124_A_QUESTION_OF_DEFINITION" id="X_124_A_QUESTION_OF_DEFINITION"></a><a href="#X_124_A_QUESTION_OF_DEFINITIONa"><b>124.&mdash;A QUESTION OF DEFINITION.</b></a></p>
+
+<p>"My property is exactly a mile square," said one landowner to another.</p>
+
+<p>"Curiously enough, mine is a square mile," was the reply.</p>
+
+<p>"Then there is no difference?"</p>
+
+<p>Is this last statement correct?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_125_THE_MINERS_HOLIDAY" id="X_125_THE_MINERS_HOLIDAY"></a><a href="#X_125_THE_MINERS_HOLIDAYa"><b>125.&mdash;THE MINERS' HOLIDAY.</b></a></p>
+
+<p>Seven coal-miners took a holiday at the seaside during a big strike.
+Six of the party spent exactly half a sovereign each, but Bill Harris
+was more extravagant. Bill spent three shillings more than the average
+of the party. What was the actual amount of Bill's expenditure?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_126_SIMPLE_MULTIPLICATION" id="X_126_SIMPLE_MULTIPLICATION"></a><a href="#X_126_SIMPLE_MULTIPLICATIONa"><b>126.&mdash;SIMPLE MULTIPLICATION.</b></a></p>
+
+<p>If we number six cards 1, 2, 4, 5, 7, and 8, and arrange them on the
+table in this order:&mdash;</p>
+
+<p>
+<span style="margin-left: 2em;">1&nbsp; &nbsp;  4&nbsp; &nbsp;  2&nbsp; &nbsp;  8
+&nbsp; &nbsp;  5&nbsp; &nbsp;  7</span><br />
+</p>
+
+<p>We can demonstrate that in order to multiply by 3 all that is
+necessary is to remove the 1 to the other end of the row, and the
+thing is done. The answer is 428571. Can you find a number that, when
+multiplied by 3 and divided by 2, the answer will be the same as if we
+removed the first card (which in this case is to be a 3) From the
+beginning of the row to the end?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_127_SIMPLE_DIVISION" id="X_127_SIMPLE_DIVISION"></a><a href="#X_127_SIMPLE_DIVISIONa"><b>127.&mdash;SIMPLE DIVISION.</b></a></p>
+
+<p>Sometimes a very simple question in elementary arithmetic will cause a
+good deal of perplexity. For example, I want to divide the four
+numbers, 701, 1,059, 1,417, and 2,312, by the largest number possible
+that will leave the same remainder in every case. How am I to set to
+work Of course, by a laborious system of trial one can in time
+discover the answer, but there is quite a simple method of doing it if
+you can only find it.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_128_A_PROBLEM_IN_SQUARES" id="X_128_A_PROBLEM_IN_SQUARES"></a><a href="#X_128_A_PROBLEM_IN_SQUARESa"><b>128.&mdash;A PROBLEM IN SQUARES.</b></a></p>
+
+<p>We possess three square boards. The surface of the first contains five
+square feet more than the second, and the second contains five square
+feet more than the third. Can you give exact measurements for the
+sides of the boards? If you can solve this little puzzle, then try to
+find three squares in arithmetical progression, with a common
+difference of 7 and also of 13.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_129_THE_BATTLE_OF_HASTINGS" id="X_129_THE_BATTLE_OF_HASTINGS"></a><a href="#X_129_THE_BATTLE_OF_HASTINGSa"><b>129.&mdash;THE BATTLE OF HASTINGS.</b></a></p>
+
+<p>All historians know that there is a great deal of mystery and
+uncertainty concerning the details of the ever-memorable battle on
+that fatal day, October 14, 1066. My puzzle deals with a curious
+passage in an ancient monkish chronicle that may never receive the
+attention that it deserves, and if I am unable to vouch for the
+authenticity of the document it will none the less serve to furnish us
+with a problem that can hardly fail to interest those of my readers
+who have arithmetical predilections. Here is the passage in question.</p>
+
+<p>"The men of Harold stood well together, as their wont was, and formed
+sixty and one squares, with a like number of men in every square
+thereof, and woe to the hardy Norman who ventured to enter their
+redoubts; for a single blow of a Saxon war-hatchet would break his
+lance and cut through his coat of mail.... When Harold threw himself
+into the fray the Saxons were one mighty square of men, shouting the
+battle-cries, 'Ut!' 'Olicrosse!' 'Godemit&egrave;!'"</p>
+
+<p>Now, I find that all the contemporary authorities agree that the
+Saxons did actually fight in this solid order. For example, in the
+"Carmen de Bello Hastingensi," a poem attributed to Guy, Bishop of
+Amiens, living at the time of the battle, we are told that "the Saxons
+stood fixed in a dense mass," and Henry of Huntingdon records that
+"they were like unto a castle, impenetrable to the Normans;" while
+Robert Wace, a century after, tells us the same thing. So in this
+respect my newly-discovered chronicle may not be greatly in error. But
+I have reason to believe that there is something wrong with the actual
+figures. Let the reader see what he can make of them.</p>
+
+<p>The number of men would be sixty-one times a square number; but when
+Harold himself joined in the fray they were then able to form one
+large square. What is the smallest possible number of men there could
+have been?</p>
+
+<p>In order to make clear to the reader the simplicity of the question, I
+will give the lowest solutions in the case of 60 and 62, the numbers
+immediately preceding and following 61. They are 60&nbsp;&times;&nbsp;4<sup>2</sup>&nbsp;+&nbsp;1 =
+31<sup>2</sup>, and 62&nbsp;&times;&nbsp;8<sup>2</sup>&nbsp;+&nbsp;1&nbsp;=&nbsp;63<sup>2</sup>. That is, 60 squares of 16 men
+each would be 960 men, and when Harold joined them they would be 961
+in number, and so form a square with 31 men on every side. Similarly
+in the case of the figures I have given for 62. Now, find the lowest
+answer for 61.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_130_THE_SCULPTORS_PROBLEM" id="X_130_THE_SCULPTORS_PROBLEM"></a><a href="#X_130_THE_SCULPTORS_PROBLEMa"><b>130.&mdash;THE SCULPTOR'S PROBLEM.</b></a></p>
+
+<p>An ancient sculptor was commissioned to supply two statues, each on a
+cubical pedestal. It is with these pedestals that we are concerned.
+They were of unequal sizes, as will be seen in the illustration, and
+when the time arrived for <span class='pagenum'>Pg 24<a name="Page_24" id="Page_24"></a></span>payment a dispute arose as to whether the
+agreement was based on lineal or cubical measurement. But as soon as
+they came to measure the two pedestals the matter was at once settled,
+because, curiously enough, the number of lineal feet was exactly the
+same as the number of cubical feet. The puzzle is to find the
+dimensions for two pedestals having this peculiarity, in the smallest
+possible figures. You see, if the two pedestals, for example, measure
+respectively 3 ft. and 1 ft. on every side, then the lineal
+measurement would be 4 ft. and the cubical contents 28 ft., which are
+not the same, so these measurements will not do.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q130.png" width="400" height="405" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_131_THE_SPANISH_MISER" id="X_131_THE_SPANISH_MISER"></a><a href="#X_131_THE_SPANISH_MISERa"><b>131.&mdash;THE SPANISH MISER.</b></a></p>
+
+<p>There once lived in a small town in New Castile a noted miser named
+Don Manuel Rodriguez. His love of money was only equalled by a strong
+passion for arithmetical problems. These puzzles usually dealt in some
+way or other with his accumulated treasure, and were propounded by him
+solely in order that he might have the pleasure of solving them
+himself. Unfortunately very few of them have survived, and when
+travelling through Spain, collecting material for a proposed work on
+"The Spanish Onion as a Cause of National Decadence," I only
+discovered a very few. One of these concerns the three boxes that
+appear in the accompanying authentic portrait.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q131.png" width="400" height="472" alt="" title="" />
+</div>
+
+<p>Each box contained a different number of golden doubloons. The
+difference between the number of doubloons in the upper box and the
+number in the middle box was the same as the difference between the
+number in the middle box and the number in the bottom box. And if the
+contents of any two of the boxes were united they would form a square
+number. What is the smallest number of doubloons that there could have
+been in any one of the boxes?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_132_THE_NINE_TREASURE_BOXES" id="X_132_THE_NINE_TREASURE_BOXES"></a><a href="#X_132_THE_NINE_TREASURE_BOXESa"><b>132.&mdash;THE NINE TREASURE BOXES.</b></a></p>
+
+<p>The following puzzle will illustrate the importance on occasions of
+being able to fix the minimum and maximum limits of a required number.
+This can very frequently be done. For example, it has not yet been
+ascertained in how many different ways the knight's tour can be
+performed on the chess board; but we know that it is fewer than the
+number of combinations of 168 things taken 63 at a time and is greater
+than 31,054,144&mdash;for the latter is the number of routes of a
+particular type. Or, to take a more familiar case, if you ask a man
+how many coins he has in his pocket, he may tell you that he has not
+the slightest idea. But on further questioning you will get out of him
+some such statement as the following: "Yes, I am positive that I have
+more than three coins, and equally certain that there are not so many
+as twenty-five." Now, the knowledge that a certain number lies between
+2 and 12 in my puzzle will enable the solver to find the exact answer;
+without that information there would be an infinite number of answers,
+from which it would be impossible to select the correct one.</p>
+
+<p>This is another puzzle received from my friend Don Manuel Rodriguez,
+the cranky miser of New Castile. On New Year's Eve in 1879 he showed
+me nine treasure boxes, and after informing me that every box
+contained a square number of golden doubloons, and that the difference
+between the contents of A and B was the same as between B and C, D and
+E, E and F, G and H, or H and I, he requested me to tell him the
+number of coins in every one of the boxes. At first I thought this was
+impossible, as there would be an infinite number of different answers,
+but on consideration I found that this was not the case. I discovered
+that while every box contained coins, the contents of A, B, C
+in<span class='pagenum'>Pg 25<a name="Page_25" id="Page_25"></a></span>creased in weight in alphabetical order; so did D, E, F; and so
+did G, H, I; but D or E need not be heavier than C, nor G or H heavier
+than F. It was also perfectly certain that box A could not contain
+more than a dozen coins at the outside; there might not be half that
+number, but I was positive that there were not more than twelve. With
+this knowledge I was able to arrive at the correct answer.</p>
+
+<p>In short, we have to discover nine square numbers such that A, B, C;
+and D, E, F; and G, H, I are three groups in arithmetical progression,
+the common difference being the same in each group, and A being less
+than 12. How many doubloons were there in every one of the nine boxes?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_133_THE_FIVE_BRIGANDS" id="X_133_THE_FIVE_BRIGANDS"></a><a href="#X_133_THE_FIVE_BRIGANDSa"><b>133.&mdash;THE FIVE BRIGANDS.</b></a></p>
+
+<p>The five Spanish brigands, Alfonso, Benito, Carlos, Diego, and
+Esteban, were counting their spoils after a raid, when it was found
+that they had captured altogether exactly 200 doubloons. One of the
+band pointed out that if Alfonso had twelve times as much, Benito
+three times as much, Carlos the same amount, Diego half as much, and
+Esteban one-third as much, they would still have altogether just 200
+doubloons. How many doubloons had each?</p>
+
+<p>There are a good many equally correct answers to this question. Here
+is one of them:</p>
+
+<div class='center'>
+<table border="0" cellpadding="2" cellspacing="0" summary="">
+<tr><td align='left'>A</td><td align='right'>6</td><td align='center'>&times;</td><td align='right'>12</td><td align='center'>=</td><td align='right'>72</td></tr>
+<tr><td align='left'>B</td><td align='right'>12</td><td align='center'>&times;</td><td align='right'>3</td><td align='center'>=</td><td align='right'>36</td></tr>
+<tr><td align='left'>C</td><td align='right'>17</td><td align='center'>&times;</td><td align='right'>1</td><td align='center'>=</td><td align='right'>17</td></tr>
+<tr><td align='left'>D</td><td align='right'>120</td><td align='center'>&times;</td><td align='right'>�</td><td align='center'>=</td><td align='right'>60</td></tr>
+<tr><td align='left'>E</td><td class='bb' align='right'>45</td><td align='center'>&times;</td><td align='right'><sup>1</sup>/<sub>3</sub></td><td align='center'>=</td><td class='bb' align='right'>15</td></tr>
+<tr><td>&nbsp;</td><td align='right'>200</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td align='center'>200</td></tr>
+</table></div>
+
+<p>The puzzle is to discover exactly how many different answers there
+are, it being understood that every man had something and that there
+is to be no fractional money&mdash;only doubloons in every case.</p>
+
+<p>This problem, worded somewhat differently, was propounded by Tartaglia
+(died 1559), and he flattered himself that he had found one solution;
+but a French mathematician of note (M.A. Labosne), in a recent work,
+says that his readers will be astonished when he assures them that
+there are 6,639 different correct answers to the question. Is this so?
+How many answers are there?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_134_THE_BANKERS_PUZZLE" id="X_134_THE_BANKERS_PUZZLE"></a><a href="#X_134_THE_BANKERS_PUZZLEa"><b>134.&mdash;THE BANKER'S PUZZLE.</b></a></p>
+
+<p>A banker had a sporting customer who was always anxious to wager on
+anything. Hoping to cure him of his bad habit, he proposed as a wager
+that the customer would not be able to divide up the contents of a box
+containing only sixpences into an exact number of equal piles of
+sixpences. The banker was first to put in one or more sixpences (as
+many as he liked); then the customer was to put in one or more (but in
+his case not more than a pound in value), neither knowing what the
+other put in. Lastly, the customer was to transfer from the banker's
+counter to the box as many sixpences as the banker desired him to put
+in. The puzzle is to find how many sixpences the banker should first
+put in and how many he should ask the customer to transfer, so that he
+may have the best chance of winning.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_135_THE_STONEMASONS_PROBLEM" id="X_135_THE_STONEMASONS_PROBLEM"></a><a href="#X_135_THE_STONEMASONS_PROBLEMa"><b>135.&mdash;THE STONEMASON'S PROBLEM.</b></a></p>
+
+<p>A stonemason once had a large number of cubic blocks of stone in his
+yard, all of exactly the same size. He had some very fanciful little
+ways, and one of his queer notions was to keep these blocks piled in
+cubical heaps, no two heaps containing the same number of blocks. He
+had discovered for himself (a fact that is well known to
+mathematicians) that if he took all the blocks contained in any number
+of heaps in regular order, beginning with the single cube, he could
+always arrange those on the ground so as to form a perfect square.
+This will be clear to the reader, because one block is a square, 1&nbsp;+&nbsp;8
+= 9 is a square, 1&nbsp;+&nbsp;8&nbsp;+&nbsp;27&nbsp;=&nbsp;36 is a square, 1&nbsp;+&nbsp;8&nbsp;+&nbsp;27&nbsp;+&nbsp;64&nbsp;=&nbsp;100 is
+a square, and so on. In fact, the sum of any number of consecutive
+cubes, beginning always with 1, is in every case a square number.</p>
+
+<p>One day a gentleman entered the mason's yard and offered him a certain
+price if he would supply him with a consecutive number of these
+cubical heaps which should contain altogether a number of blocks that
+could be laid out to form a square, but the buyer insisted on more
+than three heaps and <i>declined to take the single block</i> because it
+contained a flaw. What was the smallest possible number of blocks of
+stone that the mason had to supply?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_136_THE_SULTANS_ARMY" id="X_136_THE_SULTANS_ARMY"></a><a href="#X_136_THE_SULTANS_ARMYa"><b>136.&mdash;THE SULTAN'S ARMY.</b></a></p>
+
+<p>A certain Sultan wished to send into battle an army that could be
+formed into two perfect squares in twelve different ways. What is the
+smallest number of men of which that army could be composed? To make
+it clear to the novice, I will explain that if there were 130 men,
+they could be formed into two squares in only two different ways&mdash;81
+and 49, or 121 and 9. Of course, all the men must be used on every
+occasion.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_137_A_STUDY_IN_THRIFT" id="X_137_A_STUDY_IN_THRIFT"></a><a href="#X_137_A_STUDY_IN_THRIFTa"><b>137.&mdash;A STUDY IN THRIFT.</b></a></p>
+
+<p>Certain numbers are called triangular, because if they are taken to
+represent counters or coins they may be laid out on the table so as to
+form triangles. The number 1 is always regarded as triangular, just as
+1 is a square and a cube number. Place one counter on the table&mdash;that
+is, the first triangular number. Now place two more counters beneath
+it, and you have a triangle of three counters; therefore 3 is
+triangular. Next place a row of three more counters, and you have a
+triangle of six counters; therefore 6 is triangular. We see that every
+row of counters that we add, containing just one more counter than the
+row above it, makes a larger triangle.</p>
+
+<p><span class='pagenum'>Pg 26<a name="Page_26" id="Page_26"></a></span>Now, half the sum of any number and its square is always a triangular
+number. Thus half of 2&nbsp;+&nbsp;2<sup>2</sup>&nbsp;=&nbsp;3; half of 3&nbsp;+&nbsp;3<sup>2</sup>&nbsp;=&nbsp;6; half of 4 +
+4<sup>2</sup>&nbsp;=&nbsp;10; half of 5&nbsp;+&nbsp;5<sup>2</sup>= 15; and so on. So if we want to form a
+triangle with 8 counters on each side we shall require half of 8 +
+8<sup>2</sup>, or 36 counters. This is a pretty little property of numbers.
+Before going further, I will here say that if the reader refers to the
+"Stonemason's Problem" (No. 135) he will remember that the sum of any
+number of consecutive cubes beginning with 1 is always a square, and
+these form the series 1<sup>2</sup>, 3<sup>2</sup>, 6<sup>2</sup>, 10<sup>2</sup>, etc. It will now be
+understood when I say that one of the keys to the puzzle was the fact
+that these are always the squares of triangular numbers&mdash;that is, the
+squares of 1, 3, 6, 10, 15, 21, 28, etc., any of which numbers we have
+seen will form a triangle.</p>
+
+<p>Every whole number is either triangular, or the sum of two triangular
+numbers or the sum of three triangular numbers. That is, if we take
+any number we choose we can always form one, two, or three triangles
+with them. The number 1 will obviously, and uniquely, only form one
+triangle; some numbers will only form two triangles (as 2, 4, 11,
+etc.); some numbers will only form three triangles (as 5, 8, 14,
+etc.). Then, again, some numbers will form both one and two triangles
+(as 6), others both one and three triangles (as 3 and 10), others both
+two and three triangles (as 7 and 9), while some numbers (like 21)
+will form one, two, or three triangles, as we desire. Now for a little
+puzzle in triangular numbers.</p>
+
+<p>Sandy McAllister, of Aberdeen, practised strict domestic economy, and
+was anxious to train his good wife in his own habits of thrift. He
+told her last New Year's Eve that when she had saved so many
+sovereigns that she could lay them all out on the table so as to form
+a perfect square, or a perfect triangle, or two triangles, or three
+triangles, just as he might choose to ask he would add five pounds to
+her treasure. Soon she went to her husband with a little bag of &pound;36 in
+sovereigns and claimed her reward. It will be found that the
+thirty-six coins will form a square (with side 6), that they will form
+a single triangle (with side 8), that they will form two triangles
+(with sides 5 and 6), and that they will form three triangles (with
+sides 3, 5, and 5). In each of the four cases all the thirty-six coins
+are used, as required, and Sandy therefore made his wife the promised
+present like an honest man.</p>
+
+<p>The Scotsman then undertook to extend his promise for five more years,
+so that if next year the increased number of sovereigns that she has
+saved can be laid out in the same four different ways she will receive
+a second present; if she succeeds in the following year she will get a
+third present, and so on until she has earned six presents in all.
+Now, how many sovereigns must she put together before she can win the
+sixth present?</p>
+
+<p>What you have to do is to find five numbers, the smallest possible,
+higher than 36, that can be displayed in the four ways&mdash;to form a
+square, to form a triangle, to form two triangles, and to form three
+triangles. The highest of your five numbers will be your answer.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_138_THE_ARTILLERYMENS_DILEMMA" id="X_138_THE_ARTILLERYMENS_DILEMMA"></a><a href="#X_138_THE_ARTILLERYMENS_DILEMMAa"><b>138.&mdash;THE ARTILLERYMEN'S DILEMMA.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q138.png" width="400" height="346" alt="" title="" />
+</div>
+
+
+<p>"All cannon-balls are to be piled in square pyramids," was the order
+issued to the regiment. This was done. Then came the further order,
+"All pyramids are to contain a square number of balls." Whereupon the
+trouble arose. "It can't be done," said the major. "Look at this
+pyramid, for example; there are sixteen balls at the base, then nine,
+then four, then one at the top, making thirty balls in all. But there
+must be six more balls, or five fewer, to make a square number." "It
+<i>must</i> be done," insisted the general. "All you have to do is to put
+the right number of balls in your pyramids." "I've got it!" said a
+lieutenant, the mathematical genius of the regiment. "Lay the balls
+out singly." "Bosh!" exclaimed the general. "You can't <i>pile</i> one
+ball into a pyramid!" Is it really possible to obey both orders?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_139_THE_DUTCHMENS_WIVES" id="X_139_THE_DUTCHMENS_WIVES"></a><a href="#X_139_THE_DUTCHMENS_WIVESa"><b>139.&mdash;THE DUTCHMEN'S WIVES.</b></a></p>
+
+<p>I wonder how many of my readers are acquainted with the puzzle of the
+"Dutchmen's Wives"&mdash;in which you have to determine the names of three
+men's wives, or, rather, which wife belongs to each husband. Some
+thirty years ago it was "going the rounds," as something quite new,
+but I recently discovered it in the <i>Ladies' Diary</i> for 1739-40, so it
+was clearly familiar to the fair sex over one hundred and seventy
+years ago. How many of our mothers, wives, sisters, daughters, and
+aunts could solve the puzzle to-day? A far greater proportion than
+then, let us hope.</p>
+
+<p>Three Dutchmen, named Hendrick, Elas, and Cornelius, and their wives,
+Gurtr&uuml;n, Katr&uuml;n, and Anna, purchase hogs. Each buys as many as he (or
+she) gives shillings for one. Each husband pays altogether three
+guineas more than his wife. Hendrick buys twenty-three more hogs than
+Katr&uuml;n, and Elas eleven more <span class='pagenum'>Pg 27<a name="Page_27" id="Page_27"></a></span>than Gurtr&uuml;n. Now, what was the name of
+each man's wife?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q139.png" width="400" height="365" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_140_FIND_ADAS_SURNAME" id="X_140_FIND_ADAS_SURNAME"></a><a href="#X_140_FIND_ADAS_SURNAMEa"><b>140.&mdash;FIND ADA'S SURNAME.</b></a></p>
+
+<p>This puzzle closely resembles the last one, my remarks on the solution
+of which the reader may like to apply in another case. It was recently
+submitted to a Sydney evening newspaper that indulges in "intellect
+sharpeners," but was rejected with the remark that it is childish and
+that they only published problems capable of solution! Five ladies,
+accompanied by their daughters, bought cloth at the same shop. Each of
+the ten paid as many farthings per foot as she bought feet, and each
+mother spent 8<i>s</i>. 5&frac14;d. more than her daughter. Mrs. Robinson spent 6<i>s</i>.
+more than Mrs. Evans, who spent about a quarter as much as Mrs. Jones.
+Mrs. Smith spent most of all. Mrs. Brown bought 21 yards more than
+Bessie&mdash;one of the girls. Annie bought 16 yards more than Mary and
+spent &pound;3, 0<i>s</i>. 8<i>d</i>. more than Emily. The Christian name of the other
+girl was Ada. Now, what was her surname?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_141_SATURDAY_MARKETING" id="X_141_SATURDAY_MARKETING"></a><a href="#X_141_SATURDAY_MARKETINGa"><b>141.&mdash;SATURDAY MARKETING.</b></a></p>
+
+<p>Here is an amusing little case of marketing which, although it deals
+with a good many items of money, leads up to a question of a totally
+different character. Four married couples went into their village on a
+recent Saturday night to do a little marketing. They had to be very
+economical, for among them they only possessed forty shilling coins.
+The fact is, Ann spent 1<i>s</i>., Mary spent 2<i>s</i>., Jane spent 3<i>s</i>., and Kate
+spent 4<i>s</i>. The men were rather more extravagant than their wives, for
+Ned Smith spent as much as his wife, Tom Brown twice as much as his
+wife, Bill Jones three times as much as his wife, and Jack Robinson
+four times as much as his wife. On the way home somebody suggested
+that they should divide what coin they had left equally among them.
+This was done, and the puzzling question is simply this: What was the
+surname of each woman? Can you pair off the four couples?</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="GEOMETRICAL_PROBLEMS" id="GEOMETRICAL_PROBLEMS"></a><a href="#CONTENTS">GEOMETRICAL PROBLEMS.</a></h2>
+
+<p class='center'>
+<span style="margin-left: 0em;">"God geometrizes continually."<br /></span>
+<span style="margin-left: 8em;">PLATO.<br /></span>
+</p>
+
+<p>"There is no study," said Augustus de Morgan, "which presents so
+simple a beginning as that of geometry; there is none in which
+difficulties grow more rapidly as we proceed." This will be found when
+the reader comes to consider the following puzzles, though they are
+not arranged in strict order of difficulty. And the fact that they
+have interested and given pleasure to man for untold ages is no doubt
+due in some measure to the appeal they make to the eye as well as to
+the brain. Sometimes an algebraical formula or theorem seems to give
+pleasure to the mathematician's eye, but it is probably only an
+intellectual pleasure. But there can be no doubt that in the case of
+certain geometrical problems, notably dissection or superposition
+puzzles, the &aelig;sthetic faculty in man contributes to the delight. For
+example, there are probably few readers who will examine the various
+cuttings of the Greek cross in the following pages without being in
+some degree stirred by a sense of beauty. Law and order in Nature are
+always pleasing to contemplate, but when they come under the very eye
+they seem to make a specially strong appeal. Even the person with no
+geometrical knowledge whatever is induced after the inspection of such
+things to exclaim, "How very pretty!" In fact, I have known more than
+one person led on to a study of geometry by the fascination of
+cutting-out puzzles. I have, therefore, thought it well to keep these
+dissection puzzles distinct from the geometrical problems on more
+general lines.</p>
+
+
+<h2><a name="DISSECTION_PUZZLES" id="DISSECTION_PUZZLES"></a><a href="#CONTENTS">DISSECTION PUZZLES.</a></h2>
+
+
+<p class='center'>
+"Take him and cut him out in little stars."<br />
+<span style="margin-left: 8em;"><i>Romeo and Juliet</i>, iii. 2.<br /></span>
+</p>
+
+<p>Puzzles have infinite variety, but perhaps there is no class more
+ancient than dissection, cutting-out, or superposition puzzles. They
+were certainly known to the Chinese several thousand years before the
+Christian era. And they are just as fascinating to-day as they can
+have been at any period of their history. It is supposed by those who
+have investigated the matter that the ancient Chinese philosophers
+used these <span class='pagenum'>Pg 28<a name="Page_28" id="Page_28"></a></span>puzzles as a sort of kindergarten method of imparting the
+principles of geometry. Whether this was so or not, it is certain that
+all good dissection puzzles (for the nursery type of jig-saw puzzle,
+which merely consists in cutting up a picture into pieces to be put
+together again, is not worthy of serious consideration) are really
+based on geometrical laws. This statement need not, however, frighten
+off the novice, for it means little more than this, that geometry will
+give us the "reason why," if we are interested in knowing it, though
+the solutions may often be discovered by any intelligent person after
+the exercise of patience, ingenuity, and common sagacity.</p>
+
+<p>If we want to cut one plane figure into parts that by readjustment
+will form another figure, the first thing is to find a way of doing it
+at all, and then to discover how to do it in the fewest possible
+pieces. Often a dissection problem is quite easy apart from this
+limitation of pieces. At the time of the publication in the <i>Weekly
+Dispatch</i>, in 1902, of a method of cutting an equilateral triangle
+into four parts that will form a square (see No. 26, "Canterbury
+Puzzles"), no geometrician would have had any difficulty in doing what
+is required in five pieces: the whole point of the discovery lay in
+performing the little feat in four pieces only.</p>
+
+<p>Mere approximations in the case of these problems are valueless; the
+solution must be geometrically exact, or it is not a solution at all.
+Fallacies are cropping up now and again, and I shall have occasion to
+refer to one or two of these. They are interesting merely as
+fallacies. But I want to say something on two little points that are
+always arising in cutting-out puzzles&mdash;the questions of "hanging by a
+thread" and "turning over." These points can best be illustrated by a
+puzzle that is frequently to be found in the old books, but invariably
+with a false solution. The puzzle is to cut the figure shown in Fig. 1
+into three pieces that will fit together and form a half-square
+triangle. The answer that is invariably given is that shown in Figs. 1
+and 2. Now, it is claimed that the four pieces marked C are really
+only one piece, because they may be so cut that they are left "hanging
+together by a mere thread." But no serious puzzle lover will ever
+admit this. If the cut is made so as to leave the four pieces joined
+in one, then it cannot result in a perfectly exact solution. If, on
+the other hand, the solution is to be exact, then there will be four
+pieces&mdash;or six pieces in all. It is, therefore, not a solution in
+three pieces.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectiona.png" width="400" height="215" alt="" title="" />
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectionb.png" width="400" height="222" alt="" title="" />
+</div>
+
+<p>If, however, the reader will look at the solution in Figs. 3 and 4, he
+will see that no such fault can be found with it. There is no question
+whatever that there are three pieces, and the solution is in this
+respect quite satisfactory. But another question arises. It will be
+found on inspection that the piece marked F, in Fig. 3, is turned over
+in Fig. 4&mdash;that is to say, a different side has necessarily to be
+presented. If the puzzle were merely to be cut out of cardboard or
+wood, there might be no objection to this reversal, but it is quite
+possible that the material would not admit of being reversed. There
+might be a pattern, a polish, a difference of texture, that prevents
+it. But it is generally understood that in dissection puzzles you are
+allowed to turn pieces over unless it is distinctly stated that you
+may not do so. And very often a puzzle is greatly improved by the
+added condition, "no piece may be turned over." I have often made
+puzzles, too, in which the diagram has a small repeated pattern, and
+the pieces have then so to be cut that not only is there no turning
+over, but the pattern has to be matched, which cannot be done if the
+pieces are turned round, even with the proper side uppermost.</p>
+
+<p>Before presenting a varied series of cutting-out puzzles, some very
+easy and others difficult, I propose to consider one family
+alone&mdash;those problems involving what is known as the Greek cross with
+the square. This will exhibit a great variety of curious
+transpositions, and, by having the solutions as we go along, the
+reader will be saved the trouble of perpetually turning to another
+part of the book, and will have everything under his eye. It is hoped
+that in this way the article may prove somewhat instructive to the
+novice and interesting to others.</p>
+
+
+<h2><a name="GREEK_CROSS_PUZZLES" id="GREEK_CROSS_PUZZLES"></a><a href="#CONTENTS">GREEK CROSS PUZZLES.</a></h2>
+
+<p class='center'>
+"To fret thy soul with crosses."<br />
+<span style="margin-left: 8em;">SPENSER.<br /></span>
+</p>
+
+<p class='center'>
+"But, for my part, it was Greek to me."<br />
+<span style="margin-left: 8em;"><i>Julius C&aelig;sar</i>, i. 2.<br /></span>
+</p>
+
+<p>Many people are accustomed to consider the cross as a wholly Christian
+symbol. This is erroneous: it is of very great antiquity. The ancient
+Egyptians employed it as a sacred <span class='pagenum'>Pg 29<a name="Page_29" id="Page_29"></a></span>symbol, and on Greek sculptures we
+find representations of a cake (the supposed real origin of our hot
+cross buns) bearing a cross. Two such cakes were discovered at
+Herculaneum. Cecrops offered to Jupiter Olympus a sacred cake or
+<i>boun</i> of this kind. The cross and ball, so frequently found on
+Egyptian figures, is a circle and the <i>tau</i> cross. The circle
+signified the eternal preserver of the world, and the T, named from
+the Greek letter <i>tau</i>, is the monogram of Thoth, the Egyptian
+Mercury, meaning wisdom. This <i>tau</i> cross is also called by Christians
+the cross of St. Anthony, and is borne on a badge in the bishop's
+palace at Exeter. As for the Greek or mundane cross, the cross with
+four equal arms, we are told by competent antiquaries that it was
+regarded by ancient occultists for thousands of years as a sign of the
+dual forces of Nature&mdash;the male and female spirit of everything that
+was everlasting.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectionfig5.png" width="400" height="370" alt="" title="" />
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectionc.png" width="400" height="238" alt="" title="" />
+</div>
+
+<p>The Greek cross, as shown in Fig. 5, is formed by the assembling
+together of five equal squares. We will start with what is known as
+the Hindu problem, supposed to be upwards of three thousand years old.
+It appears in the seal of Harvard College, and is often given in old
+works as symbolical of mathematical science and exactitude. Cut the
+cross into five pieces to form a square. Figs. 6 and 7 show how this
+is done. It was not until the middle of the nineteenth century that we
+found that the cross might be transformed into a square in only four
+pieces. Figs. 8 and 9 will show how to do it, if we further require
+the four pieces to be all of the same size and shape. This Fig. 9 is
+remarkable because, according to Dr. Le Plongeon and others, as
+expounded in a work by Professor Wilson of the Smithsonian Institute,
+here we have the great Swastika, or sign, of "good luck to you "&mdash;the
+most ancient symbol of the human race of which there is any record.
+Professor Wilson's work gives some four hundred illustrations of this
+curious sign as found in the Aztec mounds of Mexico, the pyramids of
+Egypt, the ruins of Troy, and the ancient lore of India and China. One
+might almost say there is a curious affinity between the Greek cross
+and Swastika! If, however, we require that the four pieces shall be
+produced by only two clips of the scissors (assuming the puzzle is in
+paper form), then we must cut as in Fig. 10 to form Fig. 11, the first
+clip of the scissors being from <i>a</i> to <i>b</i>. Of course folding the
+paper, or holding the pieces together after the first cut, would not
+in this case be allowed. But there is an infinite number of different
+ways of making the cuts to solve the puzzle in four pieces. To this
+point I propose to return.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectiond.png" width="400" height="239" alt="" title="" />
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectione.png" width="400" height="249" alt="" title="" />
+</div>
+
+<p>It will be seen that every one of these puzzles has its reverse
+puzzle&mdash;to cut a square into pieces to form a Greek cross. But as a
+square has not so many angles as the cross, it is not always equally
+easy to discover the true directions of the cuts. Yet in the case of
+the examples given, I will leave the reader to determine their
+direction for himself, as they are rather obvious from the diagrams.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectionf.png" width="400" height="206" alt="" title="" />
+</div>
+
+<p><span class='pagenum'>Pg 30<a name="Page_30" id="Page_30"></a></span>Cut a square into five pieces that will form two separate Greek
+crosses of <i>different sizes</i>. This is quite an easy puzzle. As will be
+seen in Fig. 12, we have only to divide our square into 25 little
+squares and then cut as shown. The cross A is cut out entire, and the
+pieces B, C, D, and E form the larger cross in Fig. 13. The reader may
+here like to cut the single piece, B, into four pieces all similar in
+shape to itself, and form a cross with them in the manner shown in
+Fig. 13. I hardly need give the solution.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectiong.png" width="400" height="216" alt="" title="" />
+</div>
+
+
+<p>Cut a square into five pieces that will form two separate Greek
+crosses of exactly the <i>same size</i>. This is more difficult. We make
+the cuts as in Fig. 14, where the cross A comes out entire and the
+other four pieces form the cross in Fig. 15. The direction of the cuts
+is pretty obvious. It will be seen that the sides of the square in
+Fig. 14 are marked off into six equal parts. The sides of the cross
+are found by ruling lines from certain of these points to others.</p>
+
+
+<p>I will now explain, as I promised, why a Greek cross may be cut into
+four pieces in an infinite number of different ways to make a square.
+Draw a cross, as in Fig. 16. Then draw on transparent paper the square
+shown in Fig. 17, taking care that the distance <i>c</i> to <i>d</i> is exactly
+the same as the distance <i>a</i> to <i>b</i> in the cross. Now place the
+transparent paper over the cross and slide it about into different
+positions, only be very careful always to keep the square at the same
+angle to the cross as shown, where <i>a</i> <i>b</i> is parallel to <i>c</i> <i>d</i>. If
+you place the point <i>c</i> exactly over <i>a</i> the lines will indicate the
+solution (Figs. 10 and 11). If you place <i>c</i> in the very centre of the
+dotted square, it will give the solution in Figs. 8 and 9. You will
+now see that by sliding the square about so that the point <i>c</i> is
+always within the dotted square you may get as many different
+solutions as you like; because, since an infinite number of different
+points may theoretically be placed within this square, there must be
+an infinite number of different solutions. But the point <i>c</i> need not
+necessarily be placed within the dotted square. It may be placed, for
+example, at point <i>e</i> to give a solution in four pieces. Here the
+joins at <i>a</i> and <i>f</i> may be as slender as you like. Yet if you once
+get over the edge at <i>a</i> or <i>f</i> you no longer have a solution in four
+pieces. This proof will be found both entertaining and instructive. If
+you do not happen to have any transparent paper at hand, any thin
+paper will of course do if you hold the two sheets against a pane of
+glass in the window.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectionh.png" width="400" height="279" alt="" title="" />
+</div>
+
+<p>It may have been noticed from the solutions of the puzzles that I have
+given that the side of the square formed from the cross is always
+equal to the distance <i>a</i> to <i>b</i> in Fig. 16. This must necessarily be
+so, and I will presently try to make the point quite clear.</p>
+
+<p>We will now go one step further. I have already said that the ideal
+solution to a cutting-out puzzle is always that which requires the
+fewest possible pieces. We have just seen that two crosses of the same
+size may be cut out of a square in five pieces. The reader who
+succeeded in solving this perhaps asked himself: "Can it be done in
+fewer pieces?" This is just the sort of question that the true puzzle
+lover is always asking, and it is the right attitude for him to adopt.
+The answer to the question is that the puzzle may be solved in four
+pieces&mdash;the fewest possible. This, then, is a new puzzle. Cut a square
+into four pieces that will form two Greek crosses of the same size.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectioni.png" width="400" height="169" alt="" title="" />
+</div>
+
+<p>The solution is very beautiful. If you divide by points the sides of
+the square into three equal parts, the directions of the lines in Fig.
+18 will be quite obvious. If you cut along these lines, <span class='pagenum'>Pg 31<a name="Page_31" id="Page_31"></a></span>the pieces A
+and B will form the cross in Fig. 19 and the pieces C and D the
+similar cross in Fig. 20. In this square we have another form of
+Swastika.</p>
+
+<p>The reader will here appreciate the truth of my remark to the effect
+that it is easier to find the directions of the cuts when transforming
+a cross to a square than when converting a square into a cross. Thus,
+in Figs. 6, 8, and 10 the directions of the cuts are more obvious than
+in Fig. 14, where we had first to divide the sides of the square into
+six equal parts, and in Fig. 18, where we divide them into three equal
+parts. Then, supposing you were required to cut two equal Greek
+crosses, each into two pieces, to form a square, a glance at Figs. 19
+and 20 will show how absurdly more easy this is than the reverse
+puzzle of cutting the square to make two crosses.</p>
+
+<p>Referring to my remarks on "fallacies," I will now give a little
+example of these "solutions" that are not solutions. Some years ago a
+young correspondent sent me what he evidently thought was a brilliant
+new discovery&mdash;the transforming of a square into a Greek cross in four
+pieces by cuts all parallel to the sides of the square. I give his
+attempt in Figs. 21 and 22, where it will be seen that the four pieces
+do not form a symmetrical Greek cross, because the four arms are not
+really squares but oblongs. To make it a true Greek cross we should
+require the additions that I have indicated with dotted lines. Of
+course his solution produces a cross, but it is not the symmetrical
+Greek variety required by the conditions of the puzzle. My young
+friend thought his attempt was "near enough" to be correct; but if he
+bought a penny apple with a sixpence he probably would not have
+thought it "near enough" if he had been given only fourpence change.
+As the reader advances he will realize the importance of this question
+of exactitude.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectionj.png" width="400" height="251" alt="" title="" />
+</div>
+<p>In these cutting-out puzzles it is necessary not only to get the
+directions of the cutting lines as correct as possible, but to
+remember that these lines have no width. If after cutting up one of
+the crosses in a manner indicated in these articles you find that the
+pieces do not exactly fit to form a square, you may be certain that
+the fault is entirely your own. Either your cross was not exactly
+drawn, or your cuts were not made quite in the right directions, or
+(if you used wood and a fret-saw) your saw was not sufficiently fine.
+If you cut out the puzzles in paper with scissors, or in cardboard
+with a penknife, no material is lost; but with a saw, however fine,
+there is a certain loss. In the case of most puzzles this slight loss
+is not sufficient to be appreciable, if the puzzle is cut out on a
+large scale, but there have been instances where I have found it
+desirable to draw and cut out each part separately&mdash;not from one
+diagram&mdash;in order to produce a perfect result.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectionk.png" width="400" height="213" alt="" title="" />
+</div>
+
+<p>Now for another puzzle. If you have cut out the five pieces indicated
+in Fig. 14, you will find that these can be put together so as to form
+the curious cross shown in Fig. 23. So if I asked you to cut Fig. 24
+into five pieces to form either a square or two equal Greek crosses
+you would know how to do it. You would make the cuts as in Fig. 23,
+and place them together as in Figs. 14 and 15. But I want something
+better than that, and it is this. Cut Fig. 24 into only four pieces
+that will fit together and form a square.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectionl.png" width="400" height="208" alt="" title="" />
+</div>
+
+<p>The solution to the puzzle is shown in Figs. 25 and 26. The direction
+of the cut dividing A and C in the first diagram is very obvious, and
+the second cut is made at right angles to it. That the four pieces
+should fit together and form a square will surprise the novice, who
+will do well to study the puzzle with some care, as it is most
+instructive.</p>
+
+<p>I will now explain the beautiful rule by which we determine the size
+of a square that shall have the same area as a Greek cross, for it is
+applicable, and necessary, to the solution of almost every dissection
+puzzle that we meet with. It was first discovered by the philosopher
+Pythagoras, who died 500 B.C., and is the 47th proposition of Euclid.
+The young reader who knows nothing of the elements of geometry will
+get some idea of the fascinating character of that science. The
+triangle ABC in Fig. 27 is what we call a right-angled triangle,
+because the side BC is at right angles to the side AB. Now if we build
+up a square on each side of the tri<span class='pagenum'>Pg 32<a name="Page_32" id="Page_32"></a></span>angle, the squares on AB and BC
+will together be exactly equal to the square on the long side AC,
+which we call the hypotenuse. This is proved in the case I have given
+by subdividing the three squares into cells of equal dimensions.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectionm.png" width="400" height="213" alt="" title="" />
+</div>
+
+<p>It will be seen that 9 added to 16 equals 25, the number of cells in
+the large square. If you make triangles with the sides 5, 12 and 13,
+or with 8, 15 and 17, you will get similar arithmetical proofs, for
+these are all "rational" right-angled triangles, but the law is
+equally true for all cases. Supposing we cut off the lower arm of a
+Greek cross and place it to the left of the upper arm, as in Fig. 28,
+then the square on EF added to the square on DE exactly equals a
+square on DF. Therefore we know that the square of DF will contain the
+same area as the cross. This fact we have proved practically by the
+solutions of the earlier puzzles of this series. But whatever length
+we give to DE and EF, we can never give the exact length of DF in
+numbers, because the triangle is not a "rational" one. But the law is
+none the less geometrically true.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectionn.png" width="400" height="198" alt="" title="" />
+</div>
+
+<p>Now look at Fig. 29, and you will see an elegant method for cutting a
+piece of wood of the shape of two squares (of any relative dimensions)
+into three pieces that will fit together and form a single square. If
+you mark off the distance <i>ab</i> equal to the side <i>cd</i> the directions
+of the cuts are very evident. From what we have just been considering,
+you will at once see why <i>bc</i> must be the length of the side of the
+new square. Make the experiment as often as you like, taking different
+relative proportions for the two squares, and you will find the rule
+always come true. If you make the two squares of exactly the same
+size, you will see that the diagonal of any square is always the side
+of a square that is twice the size. All this, which is so simple that
+anybody can understand it, is very essential to the solving of
+cutting-out puzzles. It is in fact the key to most of them. And it is
+all so beautiful that it seems a pity that it should not be familiar
+to everybody.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectiono.png" width="400" height="250" alt="" title="" />
+</div>
+
+<p>We will now go one step further and deal with the half-square. Take a
+square and cut it in half diagonally. Now try to discover how to cut
+this triangle into four pieces that will form a Greek cross. The
+solution is shown in Figs. 31 and 32. In this case it will be seen
+that we divide two of the sides of the triangle into three equal parts
+and the long side into four equal parts. Then the direction of the
+cuts will be easily found. It is a pretty puzzle, and a little more
+difficult than some of the others that I have given. It should be
+noted again that it would have been much easier to locate the cuts in
+the reverse puzzle of cutting the cross to form a half-square
+triangle.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectionp.png" width="400" height="230" alt="" title="" />
+</div>
+
+<p>Another ideal that the puzzle maker always keeps in mind is to
+contrive that there shall, if possible, be only one correct solution.
+Thus, in the case of the first puzzle, if we only require that a Greek
+cross shall be cut into four pieces to form a square, there is, as I
+have shown, an infinite number of different solutions. It makes a
+better puzzle to add the condition that all the four pieces shall be
+of the same size and shape, because it can then be solved in only one
+way, as in Figs. 8 and 9. In this way, too, a puzzle that is too easy
+to be interesting may be improved by such an addition. Let us take an
+example. We have seen in Fig. 28 that Fig. 33 can be cut into two
+pieces to form a Greek cross. I suppose an intelligent child would do
+it in five minutes. But suppose we say that the puzzle has to be
+solved with a piece of wood that has <span class='pagenum'>Pg 33<a name="Page_33" id="Page_33"></a></span>a bad knot in the position shown
+in Fig. 33&mdash;a knot that we must not attempt to cut through&mdash;then a
+solution in two pieces is barred out, and it becomes a more
+interesting puzzle to solve it in three pieces. I have shown in Figs.
+33 and 34 one way of doing this, and it will be found entertaining to
+discover other ways of doing it. Of course I could bar out all these
+other ways by introducing more knots, and so reduce the puzzle to a
+single solution, but it would then be overloaded with conditions.</p>
+
+<p>And this brings us to another point in seeking the ideal. Do not
+overload your conditions, or you will make your puzzle too complex to
+be interesting. The simpler the conditions of a puzzle are, the
+better. The solution may be as complex and difficult as you like, or
+as happens, but the conditions ought to be easily understood, or
+people will not attempt a solution.</p>
+
+<p>If the reader were now asked "to cut a half-square into as few pieces
+as possible to form a Greek cross," he would probably produce our
+solution, Figs. 31-32, and confidently claim that he had solved the
+puzzle correctly. In this way he would be wrong, because it is not now
+stated that the square is to be divided diagonally. Although we should
+always observe the exact conditions of a puzzle we must not read into
+it conditions that are not there. Many puzzles are based entirely on
+the tendency that people have to do this.</p>
+
+<p>The very first essential in solving a puzzle is to be sure that you
+understand the exact conditions. Now, if you divided your square in
+half so as to produce Fig. 35 it is possible to cut it into as few as
+three pieces to form a Greek cross. We thus save a piece.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectionq.png" width="400" height="263" alt="" title="" />
+</div>
+
+<p>I give another puzzle in Fig. 36. The dotted lines are added merely to
+show the correct proportions of the figure&mdash;a square of 25 cells with
+the four corner cells cut out. The puzzle is to cut this figure into
+five pieces that will form a Greek cross (entire) and a square.</p>
+
+<p>The solution to the first of the two puzzles last given&mdash;to cut a
+rectangle of the shape of a half-square into three pieces that will
+form a Greek cross&mdash;is shown in Figs. 37 and 38. It will be seen that
+we divide the long sides of the oblong into six equal parts and the
+short sides into three equal parts, in order to get the points that
+will indicate the direction of the cuts. The reader should compare
+this solution with some of the previous illustrations. He will see,
+for example, that if we continue the cut that divides B and C in the
+cross, we get Fig. 15.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectionr.png" width="400" height="257" alt="" title="" />
+</div>
+
+<p>The other puzzle, like the one illustrated in Figs. 12 and 13, will
+show how useful a little arithmetic may sometimes prove to be in the
+solution of dissection puzzles. There are twenty-one of those little
+square cells into which our figure is subdivided, from which we have
+to form both a square and a Greek cross. Now, as the cross is built up
+of five squares, and 5 from 21 leaves 16&mdash;a square number&mdash;we ought
+easily to be led to the solution shown in Fig. 39. It will be seen
+that the cross is cut out entire, while the four remaining pieces form
+the square in Fig. 40.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disections.png" width="400" height="211" alt="" title="" />
+</div>
+
+<p>Of course a half-square rectangle is the same as a double square, or
+two equal squares joined together. Therefore, if you want to solve the
+puzzle of cutting a Greek cross into four pieces to form two separate
+squares of the same size, all you have to do is to continue the short
+cut in Fig. 38 right across the cross, and you will have four pieces
+of the same size and shape. Now divide Fig. 37 into two equal squares
+by a horizontal cut midway and you will see the four pieces forming
+the two squares.</p>
+
+<div class="figcenter" style="width: 300px;">
+<img src="images/disectiont.png" width="300" height="338" alt="" title="" />
+</div>
+
+<p><span class='pagenum'>Pg 34<a name="Page_34" id="Page_34"></a></span>Cut a Greek cross into five
+pieces that will form two separate squares, one of which shall contain
+half the area of one of the arms of the cross. In further illustration
+of what I have already written, if the two squares of the same size A
+B C D and B C F E, in Fig. 41, are cut in the manner indicated by the
+dotted lines, the four pieces will form the large square A G E C. We
+thus see that the diagonal A C is the side of a square twice the size
+of A B C D. It is also clear that half the diagonal of any square is
+equal to the side of a square of half the area. Therefore, if the
+large square in the diagram is one of the arms of your cross, the
+small square is the size of one of the squares required in the
+puzzle.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/disectionu.png" width="400" height="237" alt="" title="" />
+</div>
+
+
+<p>The solution is shown in Figs. 42 and 43. It will be seen that the
+small square is cut out whole and the large square composed of the
+four pieces B, C, D, and E. After what I have written, the reader will
+have no difficulty in seeing that the square A is half the size of one
+of the arms of the cross, because the length of the diagonal of the
+former is clearly the same as the side of the latter. The thing is now
+self-evident. I have thus tried to show that some of these puzzles
+that many people are apt to regard as quite wonderful and bewildering,
+are really not difficult if only we use a little thought and judgment.
+In conclusion of this particular subject I will give four Greek cross
+puzzles, with detached solutions.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_142_THE_SILK_PATCHWORK" id="X_142_THE_SILK_PATCHWORK"></a><a href="#X_142_THE_SILK_PATCHWORKa"><b>142.&mdash;THE SILK PATCHWORK.</b></a></p>
+
+<p>The lady members of the Wilkinson family had made a simple patchwork
+quilt, as a small Christmas present, all composed of square pieces of
+the same size, as shown in the illustration. It only lacked the four
+corner pieces to make it complete. Somebody pointed out to them that
+if you unpicked the Greek cross in the middle and then cut the
+stitches along the dark joins, the four pieces all of the same size
+and shape would fit together and form a square. This the reader knows,
+from the solution in Fig. 39, is quite easily done. But George
+Wilkinson suddenly suggested to them this poser. He said, "Instead of
+picking out the cross entire, and forming the square from four equal
+pieces, can you cut out a square entire and four equal pieces that
+will form a perfect Greek cross?" The puzzle is, of course, now quite
+easy.</p>
+
+<div class="figcenter" style="width: 550px;">
+<img src="images/q142.png" width="550" height="456" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 35<a name="Page_35" id="Page_35"></a></span><a name="X_143_TWO_CROSSES_FROM_ONE" id="X_143_TWO_CROSSES_FROM_ONE"></a><b><a href="#X_143_TWO_CROSSES_FROM_ONEa">143.&mdash;TWO CROSSES FROM ONE.</a></b></p>
+
+<p>Cut a Greek cross into five pieces that will form two such crosses,
+both of the same size. The solution of this puzzle is very beautiful.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_144_THE_CROSS_AND_THE_TRIANGLE" id="X_144_THE_CROSS_AND_THE_TRIANGLE"></a><a href="#X_144_THE_CROSS_AND_THE_TRIANGLEa"><b>144.&mdash;THE CROSS AND THE TRIANGLE.</b></a></p>
+
+<p>Cut a Greek cross into six pieces that will form an equilateral
+triangle. This is another hard problem, and I will state here that a
+solution is practically impossible without a previous knowledge of my
+method of transforming an equilateral triangle into a square (see No.
+26, "Canterbury Puzzles").</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_145_THE_FOLDED_CROSS" id="X_145_THE_FOLDED_CROSS"></a><a href="#X_145_THE_FOLDED_CROSSa"><b>145.&mdash;THE FOLDED CROSS.</b></a></p>
+
+<p>Cut out of paper a Greek cross; then so fold it that with a single
+straight cut of the scissors the four pieces produced will form a
+square.</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="VARIOUS_DISSECTION_PUZZLES" id="VARIOUS_DISSECTION_PUZZLES"></a><a href="#CONTENTS">VARIOUS DISSECTION PUZZLES.</a></h2>
+
+
+<p>We will now consider a small miscellaneous selection of cutting-out
+puzzles, varying in degrees of difficulty.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_146_AN_EASY_DISSECTION_PUZZLE" id="X_146_AN_EASY_DISSECTION_PUZZLE"></a><a href="#X_146_AN_EASY_DISSECTION_PUZZLEa"><b>146.&mdash;AN EASY DISSECTION PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q146.png" width="400" height="211" alt="" title="" />
+</div>
+
+<p>First, cut out a piece of paper or cardboard of the shape shown in the
+illustration. It will be seen at once that the proportions are simply
+those of a square attached to half of another similar square, divided
+diagonally. The puzzle is to cut it into four pieces all of precisely
+the same size and shape.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_147_AN_EASY_SQUARE_PUZZLE" id="X_147_AN_EASY_SQUARE_PUZZLE"></a><a href="#X_147_AN_EASY_SQUARE_PUZZLEa"><b>147.&mdash;AN EASY SQUARE PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q147.png" width="400" height="302" alt="" title="" />
+</div>
+
+<p>If you take a rectangular piece of cardboard, twice as long as it is
+broad, and cut it in half diagonally, you will get two of the pieces
+shown in the illustration. The puzzle is with five such pieces of
+equal size to form a square. One of the pieces may be cut in two, but
+the others must be used intact.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_148_THE_BUN_PUZZLE" id="X_148_THE_BUN_PUZZLE"></a><a href="#X_148_THE_BUN_PUZZLEa"><b>148.&mdash;THE BUN PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q148.png" width="400" height="167" alt="" title="" />
+</div>
+
+<p>The three circles represent three buns, and it is simply required to
+show how these may be equally divided among four boys. The buns must
+be regarded as of equal thickness throughout and of equal thickness to
+each other. Of course, they must be cut into as few pieces as
+possible. To simplify it I will state the rather surprising fact that
+only five pieces are necessary, from which it will be seen that one
+boy gets his share in two pieces and the other three receive theirs in
+a single piece. I am aware that this statement "gives away" the
+puzzle, but it should not destroy its interest to those who like to
+discover the "reason why."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_149_THE_CHOCOLATE_SQUARES" id="X_149_THE_CHOCOLATE_SQUARES"></a><a href="#X_149_THE_CHOCOLATE_SQUARESa"><b>149.&mdash;THE CHOCOLATE SQUARES.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q149.png" width="400" height="390" alt="" title="" />
+</div>
+
+<p>Here is a slab of chocolate, indented at the dotted lines so that the
+twenty squares can be easily separated. Make a copy of the slab in
+paper or cardboard and then try to cut it into nine pieces so that
+they will form four perfect squares all of exactly the same size.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_150_DISSECTING_A_MITRE" id="X_150_DISSECTING_A_MITRE"></a><a href="#X_150_DISSECTING_A_MITREa"><b>150.&mdash;DISSECTING A MITRE.</b></a></p>
+
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/q150a.png" width="500" height="402" alt="" title="" />
+</div>
+
+<p>The figure that is perplexing the carpenter in the illustration
+represents a mitre. It will be seen that its proportions are those of
+a square with one quarter removed. The puzzle is to cut it into five
+pieces that will fit together and form a perfect square. I show an
+attempt, published in America, to perform the feat in <span class='pagenum'>Pg 36<a name="Page_36" id="Page_36"></a></span>four pieces,
+based on what is known as the "step principle," but it is a fallacy.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q150b.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>We are told first to cut oft the pieces 1 and 2 and pack them into the
+triangular space marked off by the dotted line, and so form a
+rectangle.</p>
+
+<p>So far, so good. Now, we are directed to apply the old step principle,
+as shown, and, by moving down the piece 4 one step, form the required
+square. But, unfortunately, it does <i>not</i> produce a square: only an
+oblong. Call the three long sides of the mitre 84 in. each. Then,
+before cutting the steps, our rectangle in three pieces will be 84&times;63.
+The steps must be 10� in. in height and 12 in. in breadth.
+Therefore, by moving down a step we reduce by 12 in. the side 84 in.
+and increase by 10� in. the side 63 in. Hence our final rectangle
+must be 72 in.&nbsp;&times;&nbsp;73� in., which certainly is not a square! The fact
+is, the step principle can only be applied to rectangles with sides of
+particular relative lengths. For example, if the shorter side in this
+case were 61<sup>5</sup>/<sub>7</sub> (instead of 63), then the step method would apply.
+For the steps would then be 10<sup>2</sup>/<sub>7</sub> in. in height and 12 in. in
+breadth. Note that 61<sup>5</sup>/<sub>7</sub>&nbsp;&times;&nbsp;84= the square of 72. At present no
+solution has been found in four pieces, and I do not believe one
+possible.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_151_THE_JOINERS_PROBLEM" id="X_151_THE_JOINERS_PROBLEM"></a><a href="#X_151_THE_JOINERS_PROBLEMa"><b>151.&mdash;THE JOINER'S PROBLEM.</b></a></p>
+
+<p>I have often had occasion to remark on the practical utility of
+puzzles, arising out of an application to the ordinary affairs of life
+of the little tricks and "wrinkles" that we learn while solving
+recreation problems.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q151.png" width="400" height="471" alt="" title="" />
+</div>
+
+<p>The joiner, in the illustration, wants to cut the piece of wood into
+as few pieces as possible to form a square table-top, without any
+waste <span class='pagenum'>Pg 37<a name="Page_37" id="Page_37"></a></span>of material. How should he go to work? How many pieces would
+you require?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_152_ANOTHER_JOINERS_PROBLEM" id="X_152_ANOTHER_JOINERS_PROBLEM"></a><a href="#X_152_ANOTHER_JOINERS_PROBLEMa"><b>152.&mdash;ANOTHER JOINER'S PROBLEM.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q152.png" width="400" height="322" alt="" title="" />
+</div>
+
+<p>A joiner had two pieces of wood of the shapes and relative proportions
+shown in the diagram. He wished to cut them into as few pieces as
+possible so that they could be fitted together, without waste, to form
+a perfectly square table-top. How should he have done it? There is no
+necessity to give measurements, for if the smaller piece (which is
+half a square) be made a little too large or a little too small it
+will not affect the method of solution.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_153_A_CUTTING-OUT_PUZZLE" id="X_153_A_CUTTING-OUT_PUZZLE"></a><a href="#X_153_A_CUTTING-OUT_PUZZLEa"><b>153.&mdash;A CUTTING-OUT PUZZLE.</b></a></p>
+
+<p>Here is a little cutting-out poser. I take a strip of paper, measuring
+five inches by one inch, and, by cutting it into five pieces, the
+parts fit together and form a square, as shown in the illustration.
+Now, it is quite an interesting puzzle to discover how we can do this
+in only four pieces.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q153.png" width="400" height="131" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_154_MRS_HOBSONS_HEARTHRUG" id="X_154_MRS_HOBSONS_HEARTHRUG"></a><a href="#X_154_MRS_HOBSONS_HEARTHRUGa"><b>154.&mdash;MRS. HOBSON'S HEARTHRUG.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q154.png" width="400" height="289" alt="" title="" />
+</div>
+
+<p>Mrs. Hobson's boy had an accident when playing with the fire, and
+burnt two of the corners of a pretty hearthrug. The damaged corners
+have been cut away, and it now has the appearance and proportions
+shown in my diagram. How is Mrs. Hobson to cut the rug into the fewest
+possible pieces that will fit together and form a perfectly square
+rug? It will be seen that the rug is in the proportions 36&nbsp;&times;&nbsp;27 (it
+does not matter whether we say inches or yards), and each piece cut
+away measured 12 and 6 on the outside.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_155_THE_PENTAGON_AND_SQUARE" id="X_155_THE_PENTAGON_AND_SQUARE"></a><a href="#X_155_THE_PENTAGON_AND_SQUAREa"><b>155.&mdash;THE PENTAGON AND SQUARE.</b></a></p>
+
+<p>I wonder how many of my readers, amongst those who have not given any
+close attention to the elements of geometry, could draw a regular
+pentagon, or five-sided figure, if they suddenly required to do so. A
+regular hexagon, or six-sided figure, is easy enough, for everybody
+knows that all you have to do is to describe a circle and then, taking
+the radius as the length of one of the sides, mark off the six points
+round the circumference. But a pentagon is quite another matter. So,
+as my puzzle has to do with the cutting up of a regular pentagon, it
+will perhaps be well if I first show my less experienced readers how
+this figure is to be correctly drawn. Describe a circle and draw the
+two lines H B and D G, in the diagram, through the centre at right
+angles. Now find the point A, midway between C and B. Next place the
+point of your compasses at A and with <span class='pagenum'>Pg 38<a name="Page_38" id="Page_38"></a></span>the distance A D describe the
+arc cutting H B at E. Then place the point of your compasses at D and
+with the distance D E describe the arc cutting the circumference at F.
+Now, D F is one of the sides of your pentagon, and you have simply to
+mark off the other sides round the circle. Quite simple when you know
+how, but otherwise somewhat of a poser.</p>
+
+<div class="figcenter" style="width: 499px;">
+<img src="images/q155.png" width="499" height="491" alt="" title="" />
+</div>
+
+<p>Having formed your pentagon, the puzzle is to cut it into the fewest
+possible pieces that will fit together and form a perfect square.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_156_THE_DISSECTED_TRIANGLE" id="X_156_THE_DISSECTED_TRIANGLE"></a><a href="#X_156_THE_DISSECTED_TRIANGLEa"><b>156.&mdash;THE DISSECTED TRIANGLE.</b></a></p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/q156.png" width="500" height="340" alt="" title="" />
+</div>
+
+<p>A good puzzle is that which the gentleman in the illustration is
+showing to his friends. He has simply cut out of paper an equilateral
+triangle&mdash;that is, a triangle with all its three sides of the same
+length. He proposes that it shall be cut into five pieces in such a
+way that they will fit together and form either two or three smaller
+equilateral triangles, using all the material in each case. Can you
+discover how the cuts should be made?</p>
+
+<p>Remember that when you have made your five pieces, you must be able,
+as desired, to put them together to form either the single original
+triangle or to form two triangles or to form three triangles&mdash;all
+equilateral.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_157_THE_TABLE-TOP_AND_STOOLS" id="X_157_THE_TABLE-TOP_AND_STOOLS"></a><a href="#X_157_THE_TABLE-TOP_AND_STOOLSa"><b>157.&mdash;THE TABLE-TOP AND STOOLS.</b></a></p>
+
+<p>I have frequently had occasion to show that the published answers to a
+great many of the oldest and most widely known puzzles are either
+quite incorrect or capable of improvement. I propose to consider the
+old poser of the table-top and stools that most of my readers have
+probably seen in some form or another in books compiled for the
+recreation of childhood.</p>
+
+<p>The story is told that an economical and ingenious schoolmaster once
+wished to convert a circular table-top, for which he had no use, into
+seats for two oval stools, each with a hand-hole in the centre. He
+instructed the carpenter to make the cuts as in the illustration and
+then join the eight pieces together in the manner shown. So impressed
+was he with the ingenuity of his performance that he set the puzzle to
+his geometry class as a little study in dissection. But the remainder
+of the story has never been published, because, so it is said, it was
+a characteristic of the principals of academies that they would never
+admit that they could err. I get my information from a de<span class='pagenum'>Pg 39<a name="Page_39" id="Page_39"></a></span>scendant of
+the original boy who had most reason to be interested in the matter.</p>
+
+<p>The clever youth suggested modestly to the master that the hand-holes
+were too big, and that a small boy might perhaps fall through them. He
+therefore proposed another way of making the cuts that would get over
+this objection. For his impertinence he received such severe
+chastisement that he became convinced that the larger the hand-hole in
+the stools the more comfortable might they be.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q157.png" width="600" height="263" alt="" title="" />
+</div>
+
+<p>Now what was the method the boy proposed?</p>
+
+<p>Can you show how the circular table-top may be cut into eight pieces
+that will fit together and form two oval seats for stools (each of
+exactly the same size and shape) and each having similar hand-holes of
+smaller dimensions than in the case shown above? Of course, all the
+wood must be used.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_158_THE_GREAT_MONAD" id="X_158_THE_GREAT_MONAD"></a><a href="#X_158_THE_GREAT_MONADa"><b>158.&mdash;THE GREAT MONAD.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q158.png" width="400" height="405" alt="" title="" />
+</div>
+
+<p>Here is a symbol of tremendous antiquity which is worthy of notice. It
+is borne on the Korean ensign and merchant flag, and has been adopted
+as a trade sign by the Northern Pacific Railroad Company, though
+probably few are aware that it is the Great Monad, as shown in the
+sketch below. This sign is to the Chinaman what the cross is to the
+Christian. It is the sign of Deity and eternity, while the two parts
+into which the circle is divided are called the Yin and the Yan&mdash;the
+male and female forces of nature. A writer on the subject more than
+three thousand years ago is reported to have said in reference to it:
+"The illimitable produces the great extreme. The great extreme
+produces the two principles. The two principles produce the four
+quarters, and from the four quarters we develop the quadrature of the
+eight diagrams of Feuh-hi." I hope readers will not ask me to explain
+this, for I have not the slightest idea what it means. Yet I am
+persuaded that for ages the symbol has had occult and probably
+mathematical meanings for the esoteric student.</p>
+
+<p>I will introduce the Monad in its elementary form. Here are three easy
+questions respecting this great symbol:&mdash;</p>
+
+<p>(I.) Which has the greater area, the inner circle containing the Yin
+and the Yan, or the outer ring?</p>
+
+<p>(II.) Divide the Yin and the Yan into four pieces of the same size and
+shape by one cut.</p>
+
+<p>(III.) Divide the Yin and the Yan into four pieces of the same size,
+but different shape, by one straight cut.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_159_THE_SQUARE_OF_VENEER" id="X_159_THE_SQUARE_OF_VENEER"></a><a href="#X_159_THE_SQUARE_OF_VENEERa"><b>159.&mdash;THE SQUARE OF VENEER.</b></a></p>
+
+<p>The following represents a piece of wood in my possession, 5 in.
+square. By markings on the surface it is divided into twenty-five
+square inches. I want to discover a way of cutting this piece of wood
+into the fewest possible pieces that will fit together and form two
+perfect squares of different sizes and of known dimensions. But,
+unfortunately, at every one of the sixteen intersections of the cross
+lines a small nail has been driven in at some time or other, and my
+fret-saw will be injured if it comes in <span class='pagenum'>Pg 40<a name="Page_40" id="Page_40"></a></span>contact with any of these. I
+have therefore to find a method of doing the work that will not
+necessitate my cutting through any of those sixteen points. How is it
+to be done? Remember, the exact dimensions of the two squares must be
+given.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q159.png" width="400" height="394" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_160_THE_TWO_HORSESHOES" id="X_160_THE_TWO_HORSESHOES"></a><a href="#X_160_THE_TWO_HORSESHOESa"><b>160.&mdash;THE TWO HORSESHOES.</b></a></p>
+
+<p>Why horseshoes should be considered "lucky" is one of those things
+which no man can understand. It is a very old superstition, and John
+Aubrey (1626-1700) says, "Most houses at the West End of London have a
+horseshoe on the threshold." In Monmouth Street there were seventeen
+in 1813 and seven so late as 1855. Even Lord Nelson had one nailed to
+the mast of the ship <i>Victory</i>. To-day we find it more conducive to
+"good luck" to see that they are securely nailed on the feet of the
+horse we are about to drive.</p>
+
+<p>Nevertheless, so far as the horseshoe, like the Swastika and other
+emblems that I have had occasion at times to deal with, has served to
+symbolize health, prosperity, and goodwill towards men, we may well
+treat it with a certain amount of respectful interest. May there not,
+moreover, be some esoteric or lost mathematical mystery concealed in
+the form of a horseshoe? I have been looking into this matter, and I
+wish to draw my readers' attention to the very remarkable fact that
+the pair of horseshoes shown in my illustration are related in a
+striking and beautiful manner to the circle, which is the symbol of
+eternity. I present this fact in the form of a simple problem, so that
+it may be seen how subtly this relation has been concealed for ages
+and ages. My readers will, I know, be pleased when they find the key
+to the mystery.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q160.png" width="400" height="238" alt="" title="" />
+</div>
+
+<p>Cut out the two horseshoes carefully round the outline and then cut
+them into four pieces, all different in shape, that will fit together
+and form a perfect circle. Each shoe must be cut into two pieces and
+all the part of the horse's hoof contained within the outline is to be
+used and regarded as part of the area.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_161_THE_BETSY_ROSS_PUZZLE" id="X_161_THE_BETSY_ROSS_PUZZLE"></a><a href="#X_161_THE_BETSY_ROSS_PUZZLEa"><b>161.&mdash;THE BETSY ROSS PUZZLE.</b></a></p>
+
+<p>A correspondent asked me to supply him with the solution to an old
+puzzle that is attributed to a certain Betsy Ross, of Philadelphia,
+who showed it to George Washington. It consists in so folding a piece
+of paper that with one clip of the scissors a five-pointed star of
+Freedom may be produced. Whether the story of the puzzle's origin is a
+true one or not I cannot say, but I have a print of the old house in
+Philadelphia where the lady is said to have lived, and I believe it
+still stands there. But my readers will doubtless be interested in the
+little poser.</p>
+
+<p>Take a circular piece of paper and so fold it that with one cut of the
+scissors you can produce a perfect five-pointed star.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_162_THE_CARDBOARD_CHAIN" id="X_162_THE_CARDBOARD_CHAIN"></a><a href="#X_162_THE_CARDBOARD_CHAINa"><b>162.&mdash;THE CARDBOARD CHAIN.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q162.png" width="400" height="119" alt="" title="" />
+</div>
+
+<p>Can you cut this chain out of a piece of cardboard without any join
+whatever? Every link is solid; without its having been split and
+afterwards joined at any place. It is an interesting old puzzle that I
+learnt as a child, but I have no knowledge as to its inventor.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_163_THE_PAPER_BOX" id="X_163_THE_PAPER_BOX"></a><a href="#X_163_THE_PAPER_BOX"><b>163.&mdash;THE PAPER BOX.</b></a></p>
+
+<p>It may be interesting to introduce here, though it is not strictly a
+puzzle, an ingenious method for making a paper box.</p>
+
+<p>Take a square of stout paper and by successive foldings make all the
+creases indicated by the dotted lines in the illustration. Then cut
+away the eight little triangular pieces that are shaded, and cut
+through the paper along the dark lines. The second illustration shows
+the box half folded up, and the reader will have no difficulty in
+effecting its completion. Before folding up, the reader might cut out
+the circular piece indicated in the diagram, for a purpose I will now
+explain.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q163a.png" width="400" height="405" alt="" title="" />
+</div>
+
+<p>This box will be found to serve excellently for the production of
+vortex rings. These rings, <span class='pagenum'>Pg 41<a name="Page_41" id="Page_41"></a></span>which were discussed by Von Helmholtz in
+1858, are most interesting, and the box (with the hole cut out) will
+produce them to perfection. Fill the box with tobacco smoke by blowing
+it gently through the hole. Now, if you hold it horizontally, and
+softly tap the side that is opposite to the hole, an immense number of
+perfect rings can be produced from one mouthful of smoke. It is best
+that there should be no currents of air in the room. People often do
+not realise that these rings are formed in the air when no smoke is
+used. The smoke only makes them visible. Now, one of these rings, if
+properly directed on its course, will travel across the room and put
+out the flame of a candle, and this feat is much more striking if you
+can manage to do it without the smoke. Of course, with a little
+practice, the rings may be blown from the mouth, but the box produces
+them in much greater perfection, and no skill whatever is required.
+Lord Kelvin propounded the theory that matter may consist of vortex
+rings in a fluid that fills all space, and by a development of the
+hypothesis he was able to explain chemical combination.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q163b.png" width="400" height="127" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_164_THE_POTATO_PUZZLE" id="X_164_THE_POTATO_PUZZLE"></a><a href="#X_164_THE_POTATO_PUZZLEa"><b>164.&mdash;THE POTATO PUZZLE.</b></a></p>
+
+<p>Take a circular slice of potato, place it on the table, and see into
+how large a number of pieces you can divide it with six cuts of a
+knife. Of course you must not readjust the pieces or pile them after a
+cut. What is the greatest number of pieces you can make?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q164.png" width="400" height="395" alt="" title="" />
+</div>
+
+<p>The illustration shows how to make sixteen pieces. This can, of
+course, be easily beaten.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_165_THE_SEVEN_PIGS" id="X_165_THE_SEVEN_PIGS"></a><a href="#X_165_THE_SEVEN_PIGSa"><b>165.&mdash;THE SEVEN PIGS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q165.png" width="400" height="410" alt="" title="" />
+</div>
+
+<p>Here is a little puzzle that was put to one of the sons of Erin the
+other day and perplexed him unduly, for it is really quite easy. It
+will be seen from the illustration that he was shown a sketch of a
+square pen containing seven pigs. He was asked how he would intersect
+the pen with three straight fences so as to enclose every pig in a
+separate sty. In other words, all you have to do is to take your
+pencil and, with three straight strokes across the square, enclose
+each pig separately. Nothing could be simpler.</p>
+
+<p>The Irishman complained that the pigs would not keep still while he
+was putting up the fences. He said that they would all flock together,
+or one obstinate beast would go into a corner and flock all by
+himself. It was pointed out to him that for the purposes of the puzzle
+the pigs were stationary. He answered that Irish pigs are not
+stationery&mdash;they are pork. Being persuaded to make the attempt, he
+drew three lines, one of which cut through a pig. When it was
+explained that this is not allowed, he protested <span class='pagenum'>Pg 42<a name="Page_42" id="Page_42"></a></span>that a pig was no
+use until you cut its throat. "Begorra, if it's bacon ye want without
+cutting your pig, it will be all gammon." We will not do the Irishman
+the injustice of suggesting that the miserable pun was intentional.
+However, he failed to solve the puzzle. Can you do it?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_166_THE_LANDOWNERS_FENCES" id="X_166_THE_LANDOWNERS_FENCES"></a><a href="#X_166_THE_LANDOWNERS_FENCESa"><b>166.&mdash;THE LANDOWNER'S FENCES.</b></a></p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q166.png" width="600" height="502" alt="" title="" />
+</div>
+
+<p>The landowner in the illustration is consulting with his bailiff over
+a rather puzzling little question. He has a large plan of one of his
+fields, in which there are eleven trees. Now, he wants to divide the
+field into just eleven enclosures by means of straight fences, so that
+every enclosure shall contain one tree as a shelter for his cattle.
+How is he to do it with as few fences as possible? Take your pencil
+and draw straight lines across the field until you have marked off the
+eleven enclosures (and no more), and then see how many fences you
+require. Of course the fences may cross one another.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_167_THE_WIZARDS_CATS" id="X_167_THE_WIZARDS_CATS"></a><a href="#X_167_THE_WIZARDS_CATSa"><b>167.&mdash;THE WIZARD'S CATS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q167.png" width="400" height="392" alt="" title="" />
+</div>
+
+<p>A wizard placed ten cats inside a magic circle as shown in our
+illustration, and hypnotized them so that they should remain
+stationary during his pleasure. He then proposed to draw three circles
+inside the large one, so that no cat could approach another cat
+without crossing a <span class='pagenum'>Pg 43<a name="Page_43" id="Page_43"></a></span>magic circle. Try to draw the three circles so
+that every cat has its own enclosure and cannot reach another cat
+without crossing a line.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_168_THE_CHRISTMAS_PUDDING" id="X_168_THE_CHRISTMAS_PUDDING"></a><a href="#X_168_THE_CHRISTMAS_PUDDINGa"><b>168.&mdash;THE CHRISTMAS PUDDING.</b></a></p>
+
+<p>"Speaking of Christmas puddings," said the host, as he glanced at the
+imposing delicacy at the other end of the table. "I am reminded of the
+fact that a friend gave me a new puzzle the other day respecting one.
+Here it is," he added, diving into his breast pocket.</p>
+
+<p>"'Problem: To find the contents,' I suppose," said the Eton boy.</p>
+
+<p>"No; the proof of that is in the eating. I will read you the
+conditions."</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q168.png" width="400" height="423" alt="" title="" />
+</div>
+
+<p>"'Cut the pudding into two parts, each of exactly the same size and
+shape, without touching any of the plums. The pudding is to be
+regarded as a flat disc, not as a sphere.'"</p>
+
+<p>"Why should you regard a Christmas pudding as a disc? And why should
+any reasonable person ever wish to make such an accurate division?"
+asked the cynic.</p>
+
+<p>"It is just a puzzle&mdash;a problem in dissection." All in turn had a look
+at the puzzle, but nobody succeeded in solving it. It is a little
+difficult unless you are acquainted with the principle involved in the
+making of such puddings, but easy enough when you know how it is done.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_169_A_TANGRAM_PARADOX" id="X_169_A_TANGRAM_PARADOX"></a><a href="#X_169_A_TANGRAM_PARADOXa"><b>169.&mdash;A TANGRAM PARADOX.</b></a></p>
+
+<p>Many pastimes of great antiquity, such as chess, have so developed and
+changed down the centuries that their original inventors would
+scarcely recognize them. This is not the case with Tangrams, a
+recreation that appears to be at least four thousand years old, that
+has apparently never been dormant, and that has not been altered or
+"improved upon" since the legendary Chinaman Tan first cut out the
+seven pieces shown in Diagram I. If you mark the point B, midway
+between A and C, on one side of a square of any size, and D, midway
+between C and E, on an adjoining side, the direction of the cuts is
+too obvious to need further explanation. Every design in this article
+is built up from the seven pieces of blackened cardboard. It will at
+once be understood that the possible combinations are infinite.</p>
+
+<div class="figcenter" style="width: 300px;">
+<img src="images/q169a.png" width="300" height="252" alt="" title="" />
+</div>
+
+<p>The late Mr. Sam Loyd, of New York, who published a small book of very
+ingenious designs, possessed the manuscripts of the late Mr.
+Challenor, who made a long and close study of Tangrams. This
+gentleman, it is said, records that there were originally seven books
+of Tangrams, compiled in China two thousand years before the Christian
+era. These books are so rare that, after forty years' residence in the
+country, he only succeeded in seeing perfect copies of the first and
+seventh volumes with fragments of the second. Portions of one of the
+books, printed in gold leaf upon parchment, were found in Peking by an
+English soldier and sold for three hundred pounds.</p>
+
+<p>A few years ago a little book came into my possession, from the
+library of the late Lewis Carroll, entitled <i>The Fashionable Chinese
+Puzzle</i>. It contains three hundred and twenty-three Tangram designs,
+mostly nondescript geometrical figures, to be constructed from the
+seven pieces. It was "Published by J. and E. Wallis, 42 Skinner
+Street, and J. Wallis, Jun., Marine Library, Sidmouth" (South Devon).
+There is no date, but the following note fixes the time of publication
+pretty closely: "This ingenious contrivance has for some time past
+been the favourite amusement of the ex-Emperor Napoleon, who, being
+now in a debilitated state and living very retired, passes many hours
+a day in thus exercising his patience and ingenuity." The reader will
+find, as did the great exile, that much amusement, not wholly
+uninstructive, may be derived from forming the designs of others. He
+will find many of the illustrations to this article quite easy to
+build up, and some rather difficult. Every picture may thus be
+regarded as a puzzle.</p>
+
+<p>But it is another pastime altogether to create new and original
+designs of a pictorial character, and it is surprising what
+extraordinary scope the Tangrams afford for producing pictures of real
+life&mdash;angular and often grotesque, it is true, but full of character.
+I give an example of a recumbent figure (2) that is particularly
+graceful, and only needs some slight reduction of its angularities to
+produce an entirely satisfactory outline.</p>
+
+<div class="figcenter" style="width: 300px;">
+<img src="images/q169b.png" width="300" height="233" alt="" title="" />
+</div>
+
+<p>As I have referred to the author of <i>Alice in Wonderland</i>, I give also
+my designs of the March <span class='pagenum'>Pg 44<a name="Page_44" id="Page_44"></a></span>Hare (3) and the Hatter (4). I also give an
+attempt at Napoleon (5), and a very excellent Red Indian with his
+Squaw by Mr. Loyd (6 and 7). A large number of other designs will be
+found in an article by me in <i>The Strand Magazine</i> for November, 1908.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q169b2.png" width="400" height="322" alt="" title="" />
+</div>
+
+<div class="figcenter" style="width: 150px;">
+<img src="images/q169c.png" width="150" height="250" alt="" title="" />
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q169d.png" width="400" height="370" alt="" title="" />
+</div>
+
+<p>On the appearance of this magazine article, the late Sir James Murray,
+the eminent philologist, tried, with that amazing industry that
+characterized all his work, to trace the word "tangram" to its source.
+At length he wrote as follows:&mdash;"One of my sons is a professor in the
+Anglo-Chinese college at Tientsin. Through him, his colleagues, and
+his students, I was able to make inquiries as to the alleged Tan among
+Chinese scholars. Our Chinese professor here (Oxford) also took an
+interest in the matter and obtained information from the secretary of
+the Chinese Legation in London, who is a very eminent representative
+of the Chinese literati."</p>
+
+<p>"The result has been to show that the man Tan, the god Tan, and the
+'Book of Tan' are entirely unknown to Chinese literature, history, or
+tradition. By most of the learned men the name, or allegation of the
+existence, of these had never been heard of. The puzzle is, of course,
+well known. It is called in Chinese <i>ch'i ch'iao t'u</i>; literally,
+'seven-ingenious-plan' or 'ingenious-puzzle figure of seven pieces.'
+No name approaching 'tangram,' or even 'tan,' occurs in Chinese, and
+the only suggestions for the latter were the Chinese <i>t'an</i>, 'to
+extend'; or <i>t'ang</i>, Cantonese dialect for 'Chinese.' It was suggested
+that probably some American or Englishman who knew a little Chinese or
+Cantonese, wanting a name for the puzzle, might concoct one out of one
+of these words and the European ending 'gram.' I should say the name
+'tangram' was probably invented by an American some little time before
+1864 and after 1847, but I cannot find it in print before the 1864
+edition of Webster. I have therefore had to deal very shortly with the
+word in the dictionary, telling what it is applied to and what
+conjectures or guesses have been made at the name, and giving a few
+quotations, one from your own article, which has enabled me to make
+more of the subject than I could otherwise have done."</p>
+
+<p>Several correspondents have informed me that they possess, or had
+possessed, specimens of the old Chinese books. An American gentleman
+writes to me as follows:&mdash;"I have in my possession a book made of
+tissue paper, printed in black (with a Chinese inscription on the
+front page), containing over three hundred designs, which belongs to
+the box of 'tangrams,' which I also own. The blocks are seven in
+number, made of mother-of-pearl, highly polished and finely engraved
+on either side. These are contained in a rosewood box 2<sup>1</sup>/<sub>8</sub>
+in. square. My great uncle, &mdash;&mdash;, was one of the first missionaries to
+visit China. This box and book, along with quite a collection of other
+relics, were sent to my grandfather and descended to myself."</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q169e.png" width="400" height="314" alt="" title="" />
+</div>
+
+<p>My correspondent kindly supplied me with rubbings of the Tangrams,
+from which it is clear that they are cut in the exact proportions that
+I have indicated. I reproduce the Chinese inscription (8) for this
+reason. The owner of the book informs me that he has submitted it to a
+number of Chinamen in the United States and offered as much as a
+dollar for a translation. But they all steadfastly refused to read the
+words, offering the lame excuse that the inscription is Japanese.
+Natives of Japan, however, insist that it is Chinese. Is there
+something occult and esoteric about Tangrams, that <span class='pagenum'>Pg 45<a name="Page_45" id="Page_45"></a></span>it is so difficult
+to lift the veil? Perhaps this page will come under the eye of some
+reader acquainted with the Chinese language, who will supply the
+required translation, which may, or may not, throw a little light on
+this curious question.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q169f.png" width="400" height="246" alt="" title="" />
+</div>
+
+<p>By using several sets of Tangrams at the same time we may construct
+more ambitious pictures. I was advised by a friend not to send my
+picture, "A Game of Billiards" (9), to the Academy. He assured me that
+it would not be accepted because the "judges are so hide-bound by
+convention." Perhaps he was right, and it will be more appreciated by
+Post-impressionists and Cubists. The players are considering a very
+delicate stroke at the top of the table. Of course, the two men, the
+table, and the clock are formed from four sets of Tangrams. My second
+picture is named "The Orchestra" (10), and it was designed for the
+decoration of a large hall of music. Here we have the conductor, the
+pianist, the fat little cornet-player, the left-handed player of the
+double-bass, whose attitude is life-like, though he does stand at an
+unusual distance from his instrument, and the drummer-boy, with his
+imposing music-stand. The dog at the back of the pianoforte is not
+howling: he is an appreciative listener.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q169g.png" width="600" height="233" alt="" title="" />
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q169h.png" width="400" height="292" alt="" title="" />
+</div>
+
+<p>One remarkable thing about these Tangram pictures is that they suggest
+to the imagination such a lot that is not really there. Who, for
+example, can look for a few minutes at Lady Belinda (11) and the Dutch
+girl (12) without soon feeling the haughty expression in the one case
+and the arch look in the other? Then look again at the stork (13), and
+see how it is suggested to the mind that the leg is actually much more
+slender than any one of the pieces employed. It is really an optical
+illusion. Again, notice in the case of the yacht (14) how, by leaving
+that little angular point at the top, a complete mast is suggested. If
+you place your Tangrams together on white paper so that they do not
+quite touch one another, in some cases the effect is improved by the
+white lines; in other cases it is almost destroyed.</p>
+
+<div class="figcenter" style="width: 499px;">
+<img src="images/q169i.png" width="499" height="292" alt="" title="" />
+</div>
+
+<p>Finally, I give an example from the many curious paradoxes that one
+happens upon in manipulating Tangrams. I show designs of <span class='pagenum'>Pg 46<a name="Page_46" id="Page_46"></a></span>two
+dignified individuals (15 and 16) who appear to be exactly alike,
+except for the fact that one has a foot and the other has not. Now,
+both of these figures are made from the same seven Tangrams. Where
+does the second man get his foot from?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q169j.png" width="400" height="361" alt="" title="" />
+</div>
+
+<hr style="width: 65%;" />
+<h2><a name="PATCHWORK_PUZZLES" id="PATCHWORK_PUZZLES"></a><a href="#CONTENTS">PATCHWORK PUZZLES.</a></h2>
+
+<p class='center'>"Of shreds and patches."&mdash;<i>Hamlet</i>, iii. 4.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_170_THE_CUSHION_COVERS" id="X_170_THE_CUSHION_COVERS"></a><a href="#X_170_THE_CUSHION_COVERSa"><b>170.&mdash;THE CUSHION COVERS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q170.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>The above represents a square of brocade. A lady wishes to cut it in
+four pieces so that two pieces will form one perfectly square cushion
+top, and the remaining two pieces another square cushion top. How is
+she to do it? Of course, she can only cut along the lines that divide
+the twenty-five squares, and the pattern must "match" properly without
+any irregularity whatever in the design of the material. There is only
+one way of doing it. Can you find it?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_171_THE_BANNER_PUZZLE" id="X_171_THE_BANNER_PUZZLE"></a><a href="#X_171_THE_BANNER_PUZZLEa"><b>171.&mdash;THE BANNER PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q171.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p>A Lady had a square piece of bunting with two lions on it, of which
+the illustration is an exactly reproduced reduction. She wished to cut
+the stuff into pieces that would fit together and form two square
+banners with a lion on each banner. She discovered that this could be
+done in as few as four pieces. How did she manage it? Of course, to
+cut the British Lion would be an unpardonable offence, so you must be
+careful that no cut passes through any portion of either of them.
+Ladies are informed that no allowance whatever has to be made for
+"turnings," and no part of the material may be wasted. It is quite a
+simple little dissection puzzle if rightly attacked. Remember that the
+banners have to be perfect squares, though they need not be both of
+the same size.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_172_MRS_SMILEYS_CHRISTMAS_PRESENT" id="X_172_MRS_SMILEYS_CHRISTMAS_PRESENT"></a><a href="#X_172_MRS_SMILEYS_CHRISTMAS_PRESENTa"><b>172.&mdash;MRS. SMILEY'S CHRISTMAS PRESENT.</b></a></p>
+
+<p>Mrs. Smiley's expression of pleasure was sincere when her six
+granddaughters sent to her, as a Christmas present, a very pretty
+patchwork quilt, which they had made with their own hands. It was
+constructed of square pieces of silk material, all of one size, and as
+they made a large quilt with fourteen of these little squares on each
+side, it is obvious that just 196 pieces had been stitched into it.
+Now, the six granddaughters each contributed a part of the work in the
+form of a perfect square (all six portions being different in size),
+but in order to join them up to form the square quilt it was necessary
+that the work of one girl should be unpicked into three separate
+pieces. Can you show how the joins might have been made? Of course, no
+portion can be turned over.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q172.png" width="600" height="465" alt="" title="" />
+</div>
+
+<p><span class='pagenum'>Pg 47<a name="Page_47" id="Page_47"></a></span></p>
+
+<hr style="width: 30%;" />
+<p><a name="X_173_MRS_PERKINSS_QUILT" id="X_173_MRS_PERKINSS_QUILT"></a><a href="#X_173_MRS_PERKINSS_QUILTa"><b>173.&mdash;MRS. PERKINS'S QUILT.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q173.png" width="400" height="461" alt="" title="" />
+</div>
+
+<p>It will be seen that in this case the square patchwork quilt is built
+up of 169 pieces. The puzzle is to find the smallest possible number
+of square portions of which the quilt could be composed and show how
+they might be joined together. Or, to put it the reverse way, divide
+the quilt into as few square portions as possible by merely cutting
+the stitches.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_174_THE_SQUARES_OF_BROCADE" id="X_174_THE_SQUARES_OF_BROCADE"></a><a href="#X_174_THE_SQUARES_OF_BROCADEa"><b>174.&mdash;THE SQUARES OF BROCADE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q174a.png" width="400" height="399" alt="" title="" />
+</div>
+
+<p>I happened to be paying a call at the house of a lady, when I took up
+from a table two lovely squares of brocade. They were beautiful
+specimens of Eastern workmanship&mdash;both of the same design, a delicate
+chequered pattern.</p>
+
+<p>"Are they not exquisite?" said my friend. "They were brought to me by
+a cousin who has just returned from India. Now, I want you <span class='pagenum'>Pg 48<a name="Page_48" id="Page_48"></a></span>to give me
+a little assistance. You see, I have decided to join them together so
+as to make one large square cushion-cover. How should I do this so as
+to mutilate the material as little as possible? Of course I propose to
+make my cuts only along the lines that divide the little chequers."</p>
+
+<div class="figcenter" style="width: 300px;">
+<img src="images/q174b.png" width="300" height="300" alt="" title="" />
+</div>
+
+<p>I cut the two squares in the manner desired into four pieces that
+would fit together and form another larger square, taking care that
+the pattern should match properly, and when I had finished I noticed
+that two of the pieces were of exactly the same area; that is, each of
+the two contained the same number of chequers. Can you show how the
+cuts were made in accordance with these conditions?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_175_ANOTHER_PATCHWORK_PUZZLE" id="X_175_ANOTHER_PATCHWORK_PUZZLE"></a><a href="#X_175_ANOTHER_PATCHWORK_PUZZLEa"><b>175.&mdash;ANOTHER PATCHWORK PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 550px;">
+<img src="images/q175.png" width="550" height="383" alt="" title="" />
+</div>
+
+<p>A lady was presented, by two of her girl friends, with the pretty
+pieces of silk patchwork shown in our illustration. It will be seen
+that both pieces are made up of squares all of the same size&mdash;one
+12x12 and the other 5x5. She proposes to join them together and make
+one square patchwork quilt, 13x13, but, of course, she will not cut
+any of the material&mdash;merely cut the stitches where necessary and join
+together again. What perplexes her is this. A friend assures her that
+there need be no more than four pieces in all to join up for the new
+quilt. Could you show her how this little needlework puzzle is to be
+solved in so few pieces?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_176_LINOLEUM_CUTTING" id="X_176_LINOLEUM_CUTTING"></a><a href="#X_176_LINOLEUM_CUTTINGa"><b>176.&mdash;LINOLEUM CUTTING.</b></a></p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/q176.png" width="500" height="288" alt="" title="" />
+</div>
+
+<p>The diagram herewith represents two separate pieces of linoleum. The
+chequered pattern is not repeated at the back, so that the pieces
+cannot be turned over. The puzzle is to cut the two squares into four
+pieces so that they shall fit together and form one perfect square
+10&times;10, so that the pattern shall properly <span class='pagenum'>Pg 49<a name="Page_49" id="Page_49"></a></span>match, and so that the
+larger piece shall have as small a portion as possible cut from it.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_177_ANOTHER_LINOLEUM_PUZZLE" id="X_177_ANOTHER_LINOLEUM_PUZZLE"></a><a href="#X_177_ANOTHER_LINOLEUM_PUZZLEa"><b>177.&mdash;ANOTHER LINOLEUM PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q177.png" width="400" height="200" alt="" title="" />
+</div>
+
+<p>Can you cut this piece of linoleum into four pieces that will fit
+together and form a perfect square? Of course the cuts may only be
+made along the lines.</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="VARIOUS_GEOMETRICAL_PUZZLES" id="VARIOUS_GEOMETRICAL_PUZZLES"></a><a href="#CONTENTS">VARIOUS GEOMETRICAL PUZZLES.</a></h2>
+
+<p class='center'>
+"So various are the tastes of men."<br />
+<span style="margin-left: 8em;">MARK AKENSIDE.<br /></span>
+</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_178_THE_CARDBOARD_BOX" id="X_178_THE_CARDBOARD_BOX"></a><a href="#X_178_THE_CARDBOARD_BOXa"><b>178.&mdash;THE CARDBOARD BOX.</b></a></p>
+
+<p>This puzzle is not difficult, but it will be found entertaining to
+discover the simple rule for its solution. I have a rectangular
+cardboard box. The top has an area of 120 square inches, the side 96
+square inches, and the end 80 square inches. What are the exact
+dimensions of the box?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_179_STEALING_THE_BELL-ROPES" id="X_179_STEALING_THE_BELL-ROPES"></a><a href="#X_179_STEALING_THE_BELL-ROPESa"><b>179.&mdash;STEALING THE BELL-ROPES.</b></a></p>
+
+<p>Two men broke into a church tower one night to steal the bell-ropes.
+The two ropes passed through holes in the wooden ceiling high above
+them, and they lost no time in climbing to the top. Then one man drew
+his knife and cut the rope above his head, in consequence of which he
+fell to the floor and was badly injured. His fellow-thief called out
+that it served him right for being such a fool. He said that he should
+have done as he was doing, upon which he cut the rope below the place
+at which he held on. Then, to his dismay, he found that he was in no
+better plight, for, after hanging on as long as his strength lasted,
+he was compelled to let go and fall beside his comrade. Here they were
+both found the next morning with their limbs broken. How far did they
+fall? One of the ropes when they found it was just touching the floor,
+and when you pulled the end to the wall, keeping the rope taut, it
+touched a point just three inches above the floor, and the wall was
+four feet from the rope when it hung at rest. How long was the rope
+from floor to ceiling?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_180_THE_FOUR_SONS" id="X_180_THE_FOUR_SONS"></a><a href="#X_180_THE_FOUR_SONSa"><b>180.&mdash;THE FOUR SONS.</b></a></p>
+
+<p>Readers will recognize the diagram as a familiar friend of their
+youth. A man possessed a square-shaped estate. He bequeathed to his
+widow the quarter of it that is shaded off. The remainder was to be
+divided equitably amongst his four sons, so that each should receive
+land of exactly the same area and exactly similar in shape. We are
+shown how this was done. But the remainder of the story is not so
+generally known. In the centre of the estate was a well, indicated by
+the dark spot, and Benjamin, Charles, and David complained that the
+division was not "equitable," since Alfred had access to this well,
+while they could not reach it without trespassing on somebody else's
+land. The puzzle is to show how the estate is to be apportioned so
+that each son shall have land of the same shape and area, and each
+have access to the well without going off his own land.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q180.png" width="400" height="392" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_181_THE_THREE_RAILWAY_STATIONS" id="X_181_THE_THREE_RAILWAY_STATIONS"></a><a href="#X_181_THE_THREE_RAILWAY_STATIONSa"><b>181.&mdash;THE THREE RAILWAY STATIONS.</b></a></p>
+
+<p>As I sat in a railway carriage I noticed at the other end of the
+compartment a worthy squire, whom I knew by sight, engaged in
+conversation with another passenger, who was evidently a friend of
+his.</p>
+
+<p>"How far have you to drive to your place from the railway station?"
+asked the stranger.</p>
+
+<p>"Well," replied the squire, "if I get out at Appleford, it is just the
+same distance as if I go to Bridgefield, another fifteen miles farther
+on; and if I changed at Appleford and went thirteen miles from there
+to Carterton, it would still be the same distance. You see, I am
+equidistant from the three stations, so I get a good choice of
+trains."</p>
+
+<p>Now I happened to know that Bridgefield is just fourteen miles from
+Carterton, so I amused myself in working out the exact distance that
+the squire had to drive home whichever station he got out at. What was
+the distance?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_182_THE_GARDEN_PUZZLE" id="X_182_THE_GARDEN_PUZZLE"></a><a href="#X_182_THE_GARDEN_PUZZLEa"><b>182.&mdash;THE GARDEN PUZZLE.</b></a></p>
+
+<p>Professor Rackbrain tells me that he was recently smoking a friendly
+pipe under a tree in the garden of a country acquaintance. The garden
+was enclosed by four straight walls, and his friend informed him that
+he had measured these and found the lengths to be 80, 45, 100, and 63
+yards respectively. "Then," said the professor, "we can calculate the
+exact area of the garden." "Impossible," his host replied, <span class='pagenum'>Pg 50<a name="Page_50" id="Page_50"></a></span>"because
+you can get an infinite number of different shapes with those four
+sides." "But you forget," Rackbrane said, with a twinkle in his eye,
+"that you told me once you had planted this tree equidistant from all
+the four corners of the garden." Can you work out the garden's area?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_183_DRAWING_A_SPIRAL" id="X_183_DRAWING_A_SPIRAL"></a><a href="#X_183_DRAWING_A_SPIRALa"><b>183.&mdash;DRAWING A SPIRAL.</b></a></p>
+
+<p>If you hold the page horizontally and give it a quick rotary motion
+while looking at the centre of the spiral, it will appear to revolve.
+Perhaps a good many readers are acquainted with this little optical
+illusion. But the puzzle is to show how I was able to draw this spiral
+with so much exactitude without using anything but a pair of compasses
+and the sheet of paper on which the diagram was made. How would you
+proceed in such circumstances?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q183.png" width="400" height="392" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_184_HOW_TO_DRAW_AN_OVAL" id="X_184_HOW_TO_DRAW_AN_OVAL"></a><a href="#X_184_HOW_TO_DRAW_AN_OVALa"><b>184.&mdash;HOW TO DRAW AN OVAL.</b></a></p>
+
+<p>Can you draw a perfect oval on a sheet of paper with one sweep of the
+compasses? It is one of the easiest things in the world when you know
+how.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_185_ST_GEORGES_BANNER" id="X_185_ST_GEORGES_BANNER"></a><a href="#X_185_ST_GEORGES_BANNERa"><b>185.&mdash;ST. GEORGE'S BANNER.</b></a></p>
+
+<p>At a celebration of the national festival of St. George's Day I was
+contemplating the familiar banner of the patron saint of our country.
+We all know the red cross on a white ground, shown in our
+illustration. This is the banner of St. George. The banner of St.
+Andrew (Scotland) is a white "St. Andrew's Cross" on a blue ground.
+That of St. Patrick (Ireland) is a similar cross in red on a white
+ground. These three are united in one to form our Union Jack.</p>
+
+<p>Now on looking at St. George's banner it occurred to me that the
+following question would make a simple but pretty little puzzle.
+Supposing the flag measures four feet by three feet, how wide must the
+arm of the cross be if it is required that there shall be used just
+the same quantity of red and of white bunting?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q185.png" width="400" height="420" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_186_THE_CLOTHES_LINE_PUZZLE" id="X_186_THE_CLOTHES_LINE_PUZZLE"></a><a href="#X_186_THE_CLOTHES_LINE_PUZZLEa"><b>186.&mdash;THE CLOTHES LINE PUZZLE.</b></a></p>
+
+<p>A boy tied a clothes line from the top of each of two poles to the
+base of the other. He then proposed to his father the following
+question. As one pole was exactly seven feet above the ground and the
+other exactly five feet, what was the height from the ground where the
+two cords crossed one another?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_187_THE_MILKMAID_PUZZLE" id="X_187_THE_MILKMAID_PUZZLE"></a><a href="#X_187_THE_MILKMAID_PUZZLEa"><b>187.&mdash;THE MILKMAID PUZZLE.</b></a></p>
+
+<p>Here is a little pastoral puzzle that the reader may, at first sight,
+be led into supposing is very profound, involving deep calculations.
+He may even say that it is quite impossible to give any answer unless
+we are told something definite as to the distances. And yet it is
+really quite "childlike and bland."</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q187.png" width="400" height="349" alt="" title="" />
+</div>
+
+<p><span class='pagenum'>Pg 51<a name="Page_51" id="Page_51"></a></span>In the corner of a field is seen a milkmaid milking a cow, and on the
+other side of the field is the dairy where the extract has to be
+deposited. But it has been noticed that the young woman always goes
+down to the river with her pail before returning to the dairy. Here
+the suspicious reader will perhaps ask why she pays these visits to
+the river. I can only reply that it is no business of ours. The
+alleged milk is entirely for local consumption.</p>
+
+<p>
+<span style="margin-left: 2em;">"Where are you going to, my pretty maid?"</span><br />
+<span style="margin-left: 2em;">"Down to the river, sir," she said.</span><br />
+<span style="margin-left: 2em;">"I'll <i>not</i> choose your dairy, my pretty maid."</span><br />
+<span style="margin-left: 2em;">"Nobody axed you, sir," she said.</span><br />
+</p>
+
+<p>If one had any curiosity in the matter, such an independent spirit
+would entirely disarm one. So we will pass from the point of
+commercial morality to the subject of the puzzle.</p>
+
+<p>Draw a line from the milking-stool down to the river and thence to the
+door of the dairy, which shall indicate the shortest possible route
+for the milkmaid. That is all. It is quite easy to indicate the exact
+spot on the bank of the river to which she should direct her steps if
+she wants as short a walk as possible. Can you find that spot?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_188_THE_BALL_PROBLEM" id="X_188_THE_BALL_PROBLEM"></a><a href="#X_188_THE_BALL_PROBLEMa"><b>188.&mdash;THE BALL PROBLEM.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q188.png" width="400" height="312" alt="" title="" />
+</div>
+
+<p>A stonemason was engaged the other day in cutting out a round ball for
+the purpose of some architectural decoration, when a smart schoolboy
+came upon the scene.</p>
+
+<p>"Look here," said the mason, "you seem to be a sharp youngster, can
+you tell me this? If I placed this ball on the level ground, how many
+other balls of the same size could I lay around it (also on the
+ground) so that every ball should touch this one?"</p>
+
+<p>The boy at once gave the correct answer, and then put this little
+question to the mason:&mdash;</p>
+
+<p>"If the surface of that ball contained just as many square feet as its
+volume contained cubic feet, what would be the length of its
+diameter?"</p>
+
+<p>The stonemason could not give an answer. Could you have replied
+correctly to the mason's and the boy's questions?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_189_THE_YORKSHIRE_ESTATES" id="X_189_THE_YORKSHIRE_ESTATES"></a><a href="#X_189_THE_YORKSHIRE_ESTATESa"><b>189.&mdash;THE YORKSHIRE ESTATES.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q189.png" width="400" height="278" alt="" title="" />
+</div>
+
+<p>I was on a visit to one of the large towns of Yorkshire. While walking
+to the railway station on the day of my departure a man thrust a
+hand-bill upon me, and I took this into the railway carriage and read
+it at my leisure. It informed me that three Yorkshire neighbouring
+estates were to be offered for sale. Each estate was square in shape,
+and they joined one another at their corners, just as shown in the
+diagram. Estate A contains exactly 370 acres, B contains 116 acres,
+and C 74 acres.</p>
+
+<p>Now, the little triangular bit of land enclosed by the three square
+estates was not offered for sale, and, for no reason in particular, I
+became curious as to the area of that piece. How many acres did it
+contain?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_190_FARMER_WURZELS_ESTATE" id="X_190_FARMER_WURZELS_ESTATE"></a><a href="#X_190_FARMER_WURZELS_ESTATEa"><b>190.&mdash;FARMER WURZEL'S ESTATE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q190.png" width="400" height="376" alt="" title="" />
+</div>
+
+<p>I will now present another land problem. The demonstration of the
+answer that I shall give will, I think, be found both interesting and
+easy of comprehension.</p>
+
+<p>Farmer Wurzel owned the three square fields shown in the annexed plan,
+containing respectively 18, 20, and 26 acres. In order to get a
+ring-fence round his property he bought the <span class='pagenum'>Pg 52<a name="Page_52" id="Page_52"></a></span>four intervening
+triangular fields. The puzzle is to discover what was then the whole
+area of his estate.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_191_THE_CRESCENT_PUZZLE" id="X_191_THE_CRESCENT_PUZZLE"></a><a href="#X_191_THE_CRESCENT_PUZZLEa"><b>191.&mdash;THE CRESCENT PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q191.png" width="400" height="381" alt="" title="" />
+</div>
+
+<p>Here is an easy geometrical puzzle. The crescent is formed by two
+circles, and C is the centre of the larger circle. The width of the
+crescent between B and D is 9 inches, and between E and F 5 inches.
+What are the diameters of the two circles?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_192_THE_PUZZLE_WALL" id="X_192_THE_PUZZLE_WALL"></a><a href="#X_192_THE_PUZZLE_WALLa"><b>192.&mdash;THE PUZZLE WALL.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q192.png" width="400" height="396" alt="" title="" />
+</div>
+
+<p>There was a small lake, around which four poor men built their
+cottages. Four rich men afterwards built their mansions, as shown in
+the illustration, and they wished to have the lake to themselves, so
+they instructed a builder to put up the shortest possible wall that
+would exclude the cottagers, but give themselves free access to the
+lake. How was the wall to be built?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_193_THE_SHEEP_FOLD" id="X_193_THE_SHEEP_FOLD"></a><a href="#X_193_THE_SHEEP_FOLDa"><b>193.&mdash;THE SHEEP-FOLD.</b></a></p>
+
+<p>It is a curious fact that the answers always given to some of the
+best-known puzzles that appear in every little book of fireside
+recreations that has been published for the last fifty or a hundred
+years are either quite unsatisfactory or clearly wrong. Yet nobody
+ever seems to detect their faults. Here is an example:&mdash;A farmer had a
+pen made of fifty hurdles, capable of holding a hundred sheep only.
+Supposing he wanted to make it sufficiently large to hold double that
+number, how many additional hurdles must he have?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_194_THE_GARDEN_WALLS" id="X_194_THE_GARDEN_WALLS"></a><a href="#X_194_THE_GARDEN_WALLSa"><b>194.&mdash;THE GARDEN WALLS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q194.png" width="400" height="409" alt="" title="" />
+</div>
+
+<p>A speculative country builder has a circular field, on which he has
+erected four cottages, as shown in the illustration. The field is
+surrounded by a brick wall, and the owner undertook to put up three
+other brick walls, so that the neighbours should not be overlooked by
+each other, but the four tenants insist that there shall be no
+favouritism, and that each shall have exactly the same length of wall
+space for his wall fruit trees. The puzzle is to show how the three
+walls may be built so that each tenant shall have the same area of
+ground, and precisely the same length of wall.</p>
+
+<p>Of course, each garden must be entirely enclosed by its walls, and it
+must be possible to prove that each garden has exactly the same length
+of wall. If the puzzle is properly solved no figures are necessary.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_195_LADY_BELINDAS_GARDEN" id="X_195_LADY_BELINDAS_GARDEN"></a><a href="#X_195_LADY_BELINDAS_GARDENa"><b>195.&mdash;LADY BELINDA'S GARDEN.</b></a></p>
+
+<p>Lady Belinda is an enthusiastic gardener. In the illustration she is
+depicted in the act of worrying out a pleasant little problem which I
+will relate. One of her gardens is oblong in shape, enclosed by a high
+holly hedge, and she is turning it into a rosary for the cultivation
+of <span class='pagenum'>Pg 53<a name="Page_53" id="Page_53"></a></span>some of her choicest roses. She wants to devote exactly half of
+the area of the garden to the flowers, in one large bed, and the other
+half to be a path going all round it of equal breadth throughout. Such
+a garden is shown in the diagram at the foot of the picture. How is
+she to mark out the garden under these simple conditions? She has only
+a tape, the length of the garden, to do it with, and, as the holly
+hedge is so thick and dense, she must make all her measurements
+inside. Lady Belinda did not know the exact dimensions of the garden,
+and, as it was not necessary for her to know, I also give no
+dimensions. It is quite a simple task no matter what the size or
+proportions of the garden may be. Yet how many lady gardeners would
+know just how to proceed? The tape may be quite plain&mdash;that is, it
+need not be a graduated measure.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q195.png" width="400" height="351" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_196_THE_TETHERED_GOAT" id="X_196_THE_TETHERED_GOAT"></a><a href="#X_196_THE_TETHERED_GOATa"><b>196.&mdash;THE TETHERED GOAT.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q196.png" width="400" height="356" alt="" title="" />
+</div>
+
+<p>Here is a little problem that everybody should know how to solve. The
+goat is placed in a half-acre meadow, that is in shape an equilateral
+triangle. It is tethered to a post at one corner of the field. What
+should be the length of the tether (to the nearest inch) in order that
+the goat shall be able to eat just half the grass in the field? It is
+assumed that the goat can feed to the end of the tether.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_197_THE_COMPASSES_PUZZLE" id="X_197_THE_COMPASSES_PUZZLE"></a><a href="#X_197_THE_COMPASSES_PUZZLEa"><b>197.&mdash;THE COMPASSES PUZZLE.</b></a></p>
+
+<p>It is curious how an added condition or restriction will sometimes
+convert an absurdly easy puzzle into an interesting and perhaps
+difficult one. I remember buying in the street many years ago a little
+mechanical puzzle that had a tremendous sale at the time. It consisted
+of a medal with holes in it, and the puzzle was to work a ring with a
+gap in it from hole to hole until it was finally detached. As I was
+walking along the street I very soon acquired the trick of taking off
+the ring with one hand while holding the puzzle in my pocket. A friend
+to whom I showed the little feat set about accomplishing it himself,
+and when I met him some days afterwards he exhibited his proficiency
+in the art. But he was a little taken aback when I then took the
+puzzle from him and, while simply holding the medal between the finger
+and thumb of one hand, by a series of little shakes and jerks caused
+the ring, without my even touching it, to fall off upon the floor. The
+following little poser will probably prove a rather tough nut for a
+great many readers, simply on account of the restricted conditions:&mdash;</p>
+
+<p>Show how to find exactly the middle of any straight line by means of
+the compasses only. You are not allowed to use any ruler, pencil, or
+other article&mdash;only the compasses; and no trick or dodge, such as
+folding the paper, will be permitted. You must simply use the
+compasses in the ordinary legitimate way.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_198_THE_EIGHT_STICKS" id="X_198_THE_EIGHT_STICKS"></a><a href="#X_198_THE_EIGHT_STICKSa"><b>198.&mdash;THE EIGHT STICKS.</b></a></p>
+
+<p>I have eight sticks, four of them being exactly half the length of the
+others. I lay every one of these on the table, so that they enclose
+three squares, all of the same size. How do I do it? There must be no
+loose ends hanging over.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_199_PAPAS_PUZZLE" id="X_199_PAPAS_PUZZLE"></a><a href="#X_199_PAPAS_PUZZLEa"><b>199.&mdash;PAPA'S PUZZLE.</b></a></p>
+
+<p>Here is a puzzle by Pappus, who lived at Alexandria about the end of
+the third century. It is the fifth proposition in the eighth book of
+his <i>Mathematical Collections</i>. I give it in the form that I presented
+it some years ago under the title "Papa's Puzzle," just to see how
+many readers would discover that it was by Pappus himself. "The little
+maid's papa has taken two different-sized rectangular pieces of
+cardboard, and has clipped off a triangular piece from one of them, so
+that when it is suspended by a thread from the point A it hangs with
+the long side perfectly horizontal, as shown in the illustration. He
+has perplexed the child by asking her to find the point A on the other
+card, so as to produce a similar result when cut and suspended by a
+thread." Of course, the point must not be <span class='pagenum'>Pg 54<a name="Page_54" id="Page_54"></a></span>found by trial clippings. A
+curious and pretty point is involved in this setting of the puzzle.
+Can the reader discover it?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q199.png" width="400" height="411" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_200_A_KITE-FLYING_PUZZLE" id="X_200_A_KITE-FLYING_PUZZLE"></a><a href="#X_200_A_KITE-FLYING_PUZZLEa"><b>200.&mdash;A KITE-FLYING PUZZLE.</b></a></p>
+
+<p>While accompanying my friend Professor Highflite during a scientific
+kite-flying competition on the South Downs of Sussex I was led into a
+little calculation that ought to interest my readers. The Professor
+was paying out the wire to which his kite was attached from a winch on
+which it had been rolled into a perfectly spherical form. This ball of
+wire was just two feet in diameter, and the wire had a diameter of
+one-hundredth of an inch. What was the length of the wire?</p>
+
+<p>Now, a simple little question like this that everybody can perfectly
+understand will puzzle many people to answer in any way. Let us see
+whether, without going into any profound mathematical calculations, we
+can get the answer roughly&mdash;say, within a mile of what is correct! We
+will assume that when the wire is all wound up the ball is perfectly
+solid throughout, and that no allowance has to be made for the axle
+that passes through it. With that simplification, I wonder how many
+readers can state within even a mile of the correct answer the length
+of that wire.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_201_HOW_TO_MAKE_CISTERNS" id="X_201_HOW_TO_MAKE_CISTERNS"></a><a href="#X_201_HOW_TO_MAKE_CISTERNSa"><b>201.&mdash;HOW TO MAKE CISTERNS.</b></a></p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q201.png" width="600" height="472" alt="" title="" />
+</div>
+
+<p>Our friend in the illustration has a large sheet of zinc, measuring
+(before cutting) eight feet by three feet, and he has cut out square
+pieces (all of the same size) from the four corners and now proposes
+to fold up the sides, solder the edges, and make a cistern. But the
+point that puzzles him is this: Has he cut out those square pieces of
+the correct size in order that the cistern may hold the greatest
+possible quantity of water? You see, if you cut them very small you
+get a <span class='pagenum'>Pg 55<a name="Page_55" id="Page_55"></a></span>very shallow cistern; if you cut them large you get a tall and
+slender one. It is all a question of finding a way of cutting put
+these four square pieces exactly the right size. How are we to avoid
+making them too small or too large?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_202_THE_CONE_PUZZLE" id="X_202_THE_CONE_PUZZLE"></a><a href="#X_202_THE_CONE_PUZZLEa"><b>202.&mdash;THE CONE PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q202.png" width="400" height="194" alt="" title="" />
+</div>
+
+<p>I have a wooden cone, as shown in Fig. 1. How am I to cut out of it
+the greatest possible cylinder? It will be seen that I can cut out one
+that is long and slender, like Fig. 2, or short and thick, like Fig.
+3. But neither is the largest possible. A child could tell you where
+to cut, if he knew the rule. Can you find this simple rule?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_203_CONCERNING_WHEELS" id="X_203_CONCERNING_WHEELS"></a><a href="#X_203_CONCERNING_WHEELSa"><b>203.&mdash;CONCERNING WHEELS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q203.png" width="400" height="537" alt="" title="" />
+</div>
+
+<p>There are some curious facts concerning the movements of wheels that
+are apt to perplex the novice. For example: when a railway train is
+travelling from London to Crewe certain parts of the train at any
+given moment are actually moving from Crewe towards London. Can you
+indicate those parts? It seems absurd that parts of the same train
+can at any time travel in opposite directions, but such is the case.</p>
+
+<p>In the accompanying illustration we have two wheels. The lower one is
+supposed to be fixed and the upper one running round it in the
+direction of the arrows. Now, how many times does the upper wheel turn
+on its own axis in making a complete revolution of the other wheel? Do
+not be in a hurry with your answer, or you are almost certain to be
+wrong. Experiment with two pennies on the table and the correct answer
+will surprise you, when you succeed in seeing it.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_204_A_NEW_MATCH_PUZZLE" id="X_204_A_NEW_MATCH_PUZZLE"></a><a href="#X_204_A_NEW_MATCH_PUZZLEa"><b>204.&mdash;A NEW MATCH PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q204.png" width="400" height="329" alt="" title="" />
+</div>
+
+<p>In the illustration eighteen matches are shown arranged so that they
+enclose two spaces, one just twice as large as the other. Can you
+rearrange them (1) so as to enclose two four-sided spaces, one exactly
+three times as large as the other, and (2) so as to enclose two
+five-sided spaces, one exactly three times as large as the other? All
+the eighteen matches must be fairly used in each case; the two spaces
+must be quite detached, and there must be no loose ends or duplicated
+matches.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_205_THE_SIX_SHEEP-PENS" id="X_205_THE_SIX_SHEEP-PENS"></a><a href="#X_205_THE_SIX_SHEEP-PENSa"><b>205.&mdash;THE SIX SHEEP-PENS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q205.png" width="400" height="154" alt="" title="" />
+</div>
+
+<p>Here is a new little puzzle with matches. It will be seen in the
+illustration that thirteen matches, representing a farmer's hurdles,
+have been so placed that they enclose six sheep-pens all of the same
+size. Now, one of these hurdles was stolen, and the farmer wanted
+still to enclose six pens of equal size with the remaining twelve. How
+was he to do it? All the twelve matches must be fairly used, and there
+must be no duplicated matches or loose ends.</p>
+
+
+
+<hr style="width: 65%;" /><p><span class='pagenum'>Pg 56<a name="Page_56" id="Page_56"></a></span></p>
+<h2><a name="POINTS_AND_LINES_PROBLEMS" id="POINTS_AND_LINES_PROBLEMS"></a><a href="#CONTENTS">POINTS AND LINES PROBLEMS.</a></h2>
+
+
+<p class='center'>"Line upon line, line upon line; here a little and there a
+little."&mdash;<i>Isa</i>. xxviii. 10.</p>
+
+<p>What are known as "Points and Lines" puzzles are found very
+interesting by many people. The most familiar example, here given, to
+plant nine trees so that they shall form ten straight rows with three
+trees in every row, is attributed to Sir Isaac Newton, but the
+earliest collection of such puzzles is, I believe, in a rare little
+book that I possess&mdash;published in 1821&mdash;<i>Rational Amusement for Winter
+Evenings</i>, by John Jackson. The author gives ten examples of "Trees
+planted in Rows."</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/points.png" width="400" height="208" alt="" title="" />
+</div>
+
+<p>These tree-planting puzzles have always been a matter of great
+perplexity. They are real "puzzles," in the truest sense of the word,
+because nobody has yet succeeded in finding a direct and certain way
+of solving them. They demand the exercise of sagacity, ingenuity, and
+patience, and what we call "luck" is also sometimes of service.
+Perhaps some day a genius will discover the key to the whole mystery.
+Remember that the trees must be regarded as mere points, for if we
+were allowed to make our trees big enough we might easily "fudge" our
+diagrams and get in a few extra straight rows that were more apparent
+than real.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_206_THE_KING_AND_THE_CASTLES" id="X_206_THE_KING_AND_THE_CASTLES"></a><a href="#X_206_THE_KING_AND_THE_CASTLESa"><b>206.&mdash;THE KING AND THE CASTLES.</b></a></p>
+
+<p>There was once, in ancient times, a powerful king, who had eccentric
+ideas on the subject of military architecture. He held that there was
+great strength and economy in symmetrical forms, and always cited the
+example of the bees, who construct their combs in perfect hexagonal
+cells, to prove that he had nature to support him. He resolved to
+build ten new castles in his country all to be connected by fortified
+walls, which should form five lines with four castles in every line.
+The royal architect presented his preliminary plan in the form I have
+shown. But the monarch pointed out that every castle could be
+approached from the outside, and commanded that the plan should be so
+modified that as many castles as possible should be free from attack
+from the outside, and could only be reached by crossing the fortified
+walls. The architect replied that he thought it impossible so to
+arrange them that even one castle, which the king proposed to use as a
+royal residence, could be so protected, but his majesty soon
+enlightened him by pointing out how it might be done. How would you
+have built the ten castles and fortifications so as best to fulfil the
+king's requirements? Remember that they must form five straight lines
+with four castles in every line.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q206.png" width="400" height="370" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_207_CHERRIES_AND_PLUMS" id="X_207_CHERRIES_AND_PLUMS"></a><a href="#X_207_CHERRIES_AND_PLUMSa"><b>207.&mdash;CHERRIES AND PLUMS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q207.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p>The illustration is a plan of a cottage as it stands surrounded by an
+orchard of fifty-five trees. Ten of these trees are cherries, ten are
+plums, and the remainder apples. The cherries are so planted as to
+form five straight lines, with four cherry trees in every line. The
+plum trees <span class='pagenum'>Pg 57<a name="Page_57" id="Page_57"></a></span>are also planted so as to form five straight lines with
+four plum trees in every line. The puzzle is to show which are the ten
+cherry trees and which are the ten plums. In order that the cherries
+and plums should have the most favourable aspect, as few as possible
+(under the conditions) are planted on the north and east sides of the
+orchard. Of course in picking out a group of ten trees (cherry or
+plum, as the case may be) you ignore all intervening trees. That is to
+say, four trees may be in a straight line irrespective of other trees
+(or the house) being in between. After the last puzzle this will be
+quite easy.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_208_A_PLANTATION_PUZZLE" id="X_208_A_PLANTATION_PUZZLE"></a><a href="#X_208_A_PLANTATION_PUZZLEa"><b>208.&mdash;A PLANTATION PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q208.png" width="400" height="405" alt="" title="" />
+</div>
+
+<p>A man had a square plantation of forty-nine trees, but, as will be
+seen by the omissions in the illustration, four trees were blown down
+and removed. He now wants to cut down all the remainder except ten
+trees, which are to be so left that they shall form five straight rows
+with four trees in every row. Which are the ten trees that he must
+leave?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_209_THE_TWENTY-ONE_TREES" id="X_209_THE_TWENTY-ONE_TREES"></a><a href="#X_209_THE_TWENTY-ONE_TREESa"><b>209.&mdash;THE TWENTY-ONE TREES.</b></a></p>
+
+<p>A gentleman wished to plant twenty-one trees in his park so that they
+should form twelve straight rows with five trees in every row. Could
+you have supplied him with a pretty symmetrical arrangement that would
+satisfy these conditions?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_210_THE_TEN_COINS" id="X_210_THE_TEN_COINS"></a><a href="#X_210_THE_TEN_COINSa"><b>210.&mdash;THE TEN COINS.</b></a></p>
+
+<p>Place ten pennies on a large sheet of paper or cardboard, as shown in
+the diagram, five on each edge. Now remove four of the coins, without
+disturbing the others, and replace them on the paper so that the ten
+shall form five straight lines with four coins in every line. This in
+itself is not difficult, but you should try to discover in how many
+different ways the puzzle may be solved, assuming that in every case
+the two rows at starting are exactly the same.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q210.png" width="400" height="394" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_211_THE_TWELVE_MINCE-PIES" id="X_211_THE_TWELVE_MINCE-PIES"></a><a href="#X_211_THE_TWELVE_MINCE-PIESa"><b>211.&mdash;THE TWELVE MINCE-PIES.</b></a></p>
+
+<p>It will be seen in our illustration how twelve mince-pies may be
+placed on the table so as to form six straight rows with four pies in
+every row. The puzzle is to remove only four of them to new positions
+so that there shall be <i>seven</i> straight rows with four in every row.
+Which four would you remove, and where would you replace them?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q211.png" width="400" height="509" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 58<a name="Page_58" id="Page_58"></a></span><a name="X_212_THE_BURMESE_PLANTATION" id="X_212_THE_BURMESE_PLANTATION"></a><a href="#X_212_THE_BURMESE_PLANTATIONa"><b>212.&mdash;THE BURMESE PLANTATION.</b></a></p>
+
+<p>A short time ago I received an interesting communication from the
+British chaplain at Meiktila, Upper Burma, in which my correspondent
+informed me that he had found some amusement on board ship on his way
+out in trying to solve this little poser.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q212.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>If he has a plantation of forty-nine trees, planted in the form of a
+square as shown in the accompanying illustration, he wishes to know
+how he may cut down twenty-seven of the trees so that the twenty-two
+left standing shall form as many rows as possible with four trees in
+every row.</p>
+
+<p>Of course there may not be more than four trees in any row.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_213_TURKS_AND_RUSSIANS" id="X_213_TURKS_AND_RUSSIANS"></a><a href="#X_213_TURKS_AND_RUSSIANSa"><b>213.&mdash;TURKS AND RUSSIANS.</b></a></p>
+
+<p>This puzzle is on the lines of the Afridi problem published by me in
+<i>Tit-Bits</i> some years ago.</p>
+
+<p>On an open level tract of country a party of Russian infantry, no two
+of whom were stationed at the same spot, were suddenly surprised by
+thirty-two Turks, who opened fire on the Russians from all directions.
+Each of the Turks simultaneously fired a bullet, and each bullet
+passed immediately over the heads of three Russian soldiers. As each
+of these bullets when fired killed a different man, the puzzle is to
+discover what is the smallest possible number of soldiers of which the
+Russian party could have consisted and what were the casualties on
+each side.</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="MOVING_COUNTER_PROBLEMS" id="MOVING_COUNTER_PROBLEMS"></a><a href="#CONTENTS">MOVING COUNTER PROBLEMS.</a></h2>
+
+
+<p class='center'>
+"I cannot do't without counters."<br />
+<span style="margin-left: 8em;"><i>Winter's Tale</i>, iv. 3.<br /></span>
+</p>
+
+<p>Puzzles of this class, except so far as they occur in connection with
+actual games, such as chess, seem to be a comparatively modern
+introduction. Mathematicians in recent times, notably Vandermonde and
+Reiss, have devoted some attention to them, but they do not appear to
+have been considered by the old writers. So far as games with counters
+are concerned, perhaps the most ancient and widely known in old times
+is "Nine Men's Morris" (known also, as I shall show, under a great
+many other names), unless the simpler game, distinctly mentioned in
+the works of Ovid (No. 110, "Ovid's Game," in <i>The Canterbury
+Puzzles</i>), from which "Noughts and Crosses" seems to be derived, is
+still more ancient.</p>
+
+<p>In France the game is called Marelle, in Poland Siegen Wulf Myll
+(She-goat Wolf Mill, or Fight), in Germany and Austria it is called
+Muhle (the Mill), in Iceland it goes by the name of Mylla, while the
+Bogas (or native bargees) of South America are said to play it, and on
+the Amazon it is called Trique, and held to be of Indian origin. In
+our own country it has different names in different districts, such as
+Meg Merrylegs, Peg Meryll, Nine Peg o'Merryal, Nine-Pin Miracle, Merry
+Peg, and Merry Hole. Shakespeare refers to it in "Midsummer Night's
+Dream" (Act ii., scene 1):&mdash;</p>
+
+<p>
+<span style="margin-left: 2em;">"The nine-men's morris is filled up with mud;</span><br />
+<span style="margin-left: 2.5em;">And the quaint mazes in the wanton green,</span><br />
+<span style="margin-left: 2.5em;">For lack of tread, are undistinguishable."</span><br />
+</p>
+
+<p>It was played by the shepherds with stones in holes cut in the turf.
+John Clare, the peasant poet of Northamptonshire, in "The Shepherd
+Boy" (1835) says:&mdash;"Oft we track his haunts .... By nine-peg-morris
+nicked upon the green." It is also mentioned by Drayton in his
+"Polyolbion."</p>
+
+<p>It was found on an old Roman tile discovered during the excavations at
+Silchester, and cut upon the steps of the Acropolis at Athens. When
+visiting the Christiania Museum a few years ago I was shown the great
+Viking ship that was discovered at Gokstad in 1880. On the oak planks
+forming the deck of the vessel were found boles and lines marking out
+the game, the holes being made to receive pegs. While inspecting the
+ancient oak furniture in the Rijks Museum at Amsterdam I became
+interested in an old catechumen's settle, and was surprised to find
+the game diagram cut in the centre of the seat&mdash;quite conveniently for
+surreptitious play. It has been discovered cut in the choir stalls of
+several of our English cathedrals. In the early eighties it was found
+scratched upon a stone built into a wall (probably about the date
+1200), during the restoration of Hargrave church in Northamptonshire.
+This stone is now in the Northampton Museum. A similar stone has since
+been found at Sempringham, Lincolnshire. It is to be seen on an
+ancient tombstone in the Isle of Man, and painted on old Dutch tiles.
+And in 1901 a stone was dug out of a gravel pit near Oswestry bearing
+an undoubted diagram of the game.</p>
+
+<p>The game has been played with different <span class='pagenum'>Pg 59<a name="Page_59" id="Page_59"></a></span>rules at different periods
+and places. I give a copy of the board. Sometimes the diagonal lines
+are omitted, but this evidently was not intended to affect the play:
+it simply meant that the angles alone were thought sufficient to
+indicate the points. This is how Strutt, in <i>Sports and Pastimes</i>,
+describes the game, and it agrees with the way I played it as a
+boy:&mdash;"Two persons, having each of them nine pieces, or men, lay them
+down alternately, one by one, upon the spots; and the business of
+either party is to prevent his antagonist from placing three of his
+pieces so as to form a row of three, without the intervention of an
+opponent piece. If a row be formed, he that made it is at liberty to
+take up one of his competitor's pieces from any part he thinks most to
+his advantage; excepting he has made a row, which must not be touched
+if he have another piece upon the board that is not a component part
+of that row. When all the pieces are laid down, they are played
+backwards and forwards, in any direction that the lines run, but only
+can move from one spot to another (next to it) at one time. He that
+takes off all his antagonist's pieces is the conqueror."</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/counter.png" width="400" height="393" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_214_THE_SIX_FROGS" id="X_214_THE_SIX_FROGS"></a><a href="#X_214_THE_SIX_FROGSa"><b>214.&mdash;THE SIX FROGS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q214.png" width="400" height="63" alt="" title="" />
+</div>
+
+<p>The six educated frogs in the illustration are trained to reverse
+their order, so that their numbers shall read 6, 5, 4, 3, 2, 1, with
+the blank square in its present position. They can jump to the next
+square (if vacant) or leap over one frog to the next square beyond (if
+vacant), just as we move in the game of draughts, and can go backwards
+or forwards at pleasure. Can you show how they perform their feat in
+the fewest possible moves? It is quite easy, so when you have done it
+add a seventh frog to the right and try again. Then add more frogs
+until you are able to give the shortest solution for any number. For
+it can always be done, with that single vacant square, no matter how
+many frogs there are.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_215_THE_GRASSHOPPER_PUZZLE" id="X_215_THE_GRASSHOPPER_PUZZLE"></a><a href="#X_215_THE_GRASSHOPPER_PUZZLEa"><b>215.&mdash;THE GRASSHOPPER PUZZLE.</b></a></p>
+
+<p>It has been suggested that this puzzle was a great favourite among the
+young apprentices of the City of London in the sixteenth and
+seventeenth centuries. Readers will have noticed the curious brass
+grasshopper on the Royal Exchange. This long-lived creature escaped
+the fires of 1666 and 1838. The grasshopper, after his kind, was the
+crest of Sir Thomas Gresham, merchant grocer, who died in 1579, and
+from this cause it has been used as a sign by grocers in general.
+Unfortunately for the legend as to its origin, the puzzle was only
+produced by myself so late as the year 1900. On twelve of the thirteen
+black discs are placed numbered counters or grasshoppers. The puzzle
+is to reverse their order, so that they shall read, 1, 2, 3, 4, etc.,
+in the opposite direction, with the vacant disc left in the same
+position as at present. Move one at a time in any order, either to the
+adjoining vacant disc or by jumping over one grasshopper, like the
+moves in draughts. The moves or leaps may be made in either direction
+that is at any time possible. What are the fewest possible moves in
+which it can be done?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q215.png" width="400" height="399" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_216_THE_EDUCATED_FROGS" id="X_216_THE_EDUCATED_FROGS"></a><a href="#X_216_THE_EDUCATED_FROGSa"><b>216.&mdash;THE EDUCATED FROGS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q216.png" width="400" height="108" alt="" title="" />
+</div>
+
+<p>Our six educated frogs have learnt a new and pretty feat. When placed
+on glass tumblers, as shown in the illustration, they change sides so
+that the three black ones are to the left and the white frogs to the
+right, with the unoccupied tumbler at the opposite end&mdash;No. 7. They
+can jump to the next tumbler (if unoccupied), or over one, or two,
+frogs to an unoccupied tumbler. The jumps can be made in either
+direction, and a frog may jump over his own or the opposite colour, or
+both colours. Four sue<span class='pagenum'>Pg 60<a name="Page_60" id="Page_60"></a></span>cessive specimen jumps will make everything
+quite plain: 4 to 1, 5 to 4, 3 to 5, 6 to 3. Can you show how they do
+it in ten jumps?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_217_THE_TWICKENHAM_PUZZLE" id="X_217_THE_TWICKENHAM_PUZZLE"></a><a href="#X_217_THE_TWICKENHAM_PUZZLEa"><b>217.&mdash;THE TWICKENHAM PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q217.png" width="400" height="406" alt="" title="" />
+</div>
+
+<p>In the illustration we have eleven discs in a circle. On five of the
+discs we place white counters with black letters&mdash;as shown&mdash;and on
+five other discs the black counters with white letters. The bottom
+disc is left vacant. Starting thus, it is required to get the counters
+into order so that they spell the word "Twickenham" in a clockwise
+direction, leaving the vacant disc in the original position. The black
+counters move in the direction that a clock-hand revolves, and the
+white counters go the opposite way. A counter may jump over one of the
+opposite colour if the vacant disc is next beyond. Thus, if your first
+move is with K, then C can jump over K. If then K moves towards E, you
+may next jump W over C, and so on. The puzzle may be solved in
+twenty-six moves. Remember a counter cannot jump over one of its own
+colour.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_218_THE_VICTORIA_CROSS_PUZZLE" id="X_218_THE_VICTORIA_CROSS_PUZZLE"></a><a href="#X_218_THE_VICTORIA_CROSS_PUZZLEa"><b>218.&mdash;THE VICTORIA CROSS PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q218.png" width="400" height="397" alt="" title="" />
+</div>
+
+<p>The puzzle-maker is peculiarly a "snapper-up of unconsidered trifles,"
+and his productions are often built up with the slenderest materials.
+Trivialities that might entirely escape the observation of others, or,
+if they were observed, would be regarded as of no possible moment,
+often supply the man who is in quest of posers with a pretty theme or
+an idea that he thinks possesses some "basal value."</p>
+
+<p>When seated opposite to a lady in a railway carriage at the time of
+Queen Victoria's Diamond Jubilee, my attention was attracted to a
+brooch that she was wearing. It was in the form of a Maltese or
+Victoria Cross, and bore the letters of the word VICTORIA. The number
+and arrangement of the letters immediately gave me the suggestion for
+the puzzle which I now present.</p>
+
+<p>The diagram, it will be seen, is composed of nine divisions. The
+puzzle is to place eight counters, bearing the letters of the word
+VICTORIA, exactly in the manner shown, and then slide one letter at a
+time from black to white and white to black alternately, until the
+word reads round in the same direction, only with the initial letter V
+on one of the black arms of the cross. At no time may two letters be
+in the same division. It is required to find the shortest method.</p>
+
+<p>Leaping moves are, of course, not permitted. The first move must
+obviously be made with A, I, T, or R. Supposing you move T to the
+centre, the next counter played will be O or C, since I or R cannot be
+moved. There is something a little remarkable in the solution of this
+puzzle which I will explain.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_219_THE_LETTER_BLOCK_PUZZLE" id="X_219_THE_LETTER_BLOCK_PUZZLE"></a><a href="#X_219_THE_LETTER_BLOCK_PUZZLEa"><b>219.&mdash;THE LETTER BLOCK PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q219.png" width="400" height="404" alt="" title="" />
+</div>
+
+<p>Here is a little reminiscence of our old friend the Fifteen Block
+Puzzle. Eight wooden blocks are lettered, and are placed in a box, as
+shown in the illustration. It will be seen that you can only move one
+block at a time to the place vacant for the time being, as no block
+may be lifted out of the box. The puzzle is to shift them about until
+you get them in the order&mdash;</p>
+
+<p><span class='pagenum'>Pg 61<a name="Page_61" id="Page_61"></a></span></p><p>
+<span style="margin-left: 2em;">A&nbsp;  B&nbsp;  C</span><br />
+<span style="margin-left: 2em;">D&nbsp;  E&nbsp;  F</span><br />
+<span style="margin-left: 2em;">G&nbsp;  H</span><br />
+</p>
+
+<p>This you will find by no means difficult if you are allowed as many
+moves as you like. But the puzzle is to do it in the fewest possible
+moves. I will not say what this smallest number of moves is, because
+the reader may like to discover it for himself. In writing down your
+moves you will find it necessary to record no more than the letters in
+the order that they are shifted. Thus, your first five moves might be
+C, H, G, E, F; and this notation can have no possible ambiguity. In
+practice you only need eight counters and a simple diagram on a sheet
+of paper.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_220_A_LODGING-HOUSE_DIFFICULTY" id="X_220_A_LODGING-HOUSE_DIFFICULTY"></a><a href="#X_220_A_LODGING-HOUSE_DIFFICULTYa"><b>220.&mdash;A LODGING-HOUSE DIFFICULTY.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q220.png" width="400" height="268" alt="" title="" />
+</div>
+
+<p>The Dobsons secured apartments at Slocomb-on-Sea. There were six rooms
+on the same floor, all communicating, as shown in the diagram. The
+rooms they took were numbers 4, 5, and 6, all facing the sea. But a
+little difficulty arose. Mr. Dobson insisted that the piano and the
+bookcase should change rooms. This was wily, for the Dobsons were not
+musical, but they wanted to prevent any one else playing the
+instrument. Now, the rooms were very small and the pieces of furniture
+indicated were very big, so that no two of these articles could be got
+into any room at the same time. How was the exchange to be made with
+the least possible labour? Suppose, for example, you first move the
+wardrobe into No. 2; then you can move the bookcase to No. 5 and the
+piano to No. 6, and so on. It is a fascinating puzzle, but the
+landlady had reasons for not appreciating it. Try to solve her
+difficulty in the fewest possible removals with counters on a sheet of
+paper.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_221_THE_EIGHT_ENGINES" id="X_221_THE_EIGHT_ENGINES"></a><a href="#X_221_THE_EIGHT_ENGINESa"><b>221.&mdash;THE EIGHT ENGINES.</b></a></p>
+
+<p>The diagram represents the engine-yard of a railway company under
+eccentric management. The engines are allowed to be stationary only at
+the nine points indicated, one of which is at present vacant. It is
+required to move the engines, one at a time, from point to point, in
+seventeen moves, so that their numbers shall be in numerical order
+round the circle, with the central point left vacant. But one of the
+engines has had its fire drawn, and therefore cannot move. How is the
+thing to be done? And which engine remains stationary throughout?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q221.png" width="400" height="425" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_222_A_RAILWAY_PUZZLE" id="X_222_A_RAILWAY_PUZZLE"></a><a href="#X_222_A_RAILWAY_PUZZLEa"><b>222.&mdash;A RAILWAY PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q222.png" width="400" height="482" alt="" title="" />
+</div>
+
+<p>Make a diagram, on a large sheet of paper, like the illustration, and
+have three counters marked A, three marked B, and three marked C. It
+will be seen that at the intersection of lines there are nine
+stopping-places, and a tenth stopping-<span class='pagenum'>Pg 62<a name="Page_62" id="Page_62"></a></span>place is attached to the outer
+circle like the tail of a Q. Place the three counters or engines
+marked A, the three marked B, and the three marked C at the places
+indicated. The puzzle is to move the engines, one at a time, along the
+lines, from stopping-place to stopping-place, until you succeed in
+getting an A, a B, and a C on each circle, and also A, B, and C on
+each straight line. You are required to do this in as few moves as
+possible. How many moves do you need?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_223_A_RAILWAY_MUDDLE" id="X_223_A_RAILWAY_MUDDLE"></a><a href="#X_223_A_RAILWAY_MUDDLEa"><b>223.&mdash;A RAILWAY MUDDLE.</b></a></p>
+
+<p>The plan represents a portion of the line of the London, Clodville,
+and Mudford Railway Company. It is a single line with a loop. There is
+only room for eight wagons, or seven wagons and an engine, between B
+and C on either the left line or the right line of the loop. It
+happened that two goods trains (each consisting of an engine and
+sixteen wagons) got into the position shown in the illustration. It
+looked like a hopeless deadlock, and each engine-driver wanted the
+other to go back to the next station and take off nine wagons. But an
+ingenious stoker undertook to pass the trains and send them on their
+respective journeys with their engines properly in front. He also
+contrived to reverse the engines the fewest times possible. Could you
+have performed the feat? And how many times would you require to
+reverse the engines? A "reversal" means a change of direction,
+backward or forward. No rope-shunting, fly-shunting, or other trick is
+allowed. All the work must be done legitimately by the two engines. It
+is a simple but interesting puzzle if attempted with counters.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q223.png" width="400" height="638" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_224_THE_MOTOR-GARAGE_PUZZLE" id="X_224_THE_MOTOR-GARAGE_PUZZLE"></a><a href="#X_224_THE_MOTOR-GARAGE_PUZZLEa"><b>224.&mdash;THE MOTOR-GARAGE PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q224.png" width="400" height="491" alt="" title="" />
+</div>
+
+<p>The difficulties of the proprietor of a motor garage are converted
+into a little pastime of a kind that has a peculiar fascination. All
+you need is to make a simple plan or diagram on a sheet of paper or
+cardboard and number eight counters, 1 to 8. Then a whole family can
+enter into an amusing competition to find the best possible solution
+of the difficulty.</p>
+
+<p>The illustration represents the plan of a motor garage, with
+accommodation for twelve cars. But the premises are so inconveniently
+restricted that the proprietor is often caused considerable
+perplexity. Suppose, for example, that the eight cars numbered 1 to 8
+are in the positions shown, how are they to be shifted in the quickest
+possible way so that 1, 2, 3, and 4 shall change places with 5, 6, 7,
+and 8&mdash;that is, with the numbers still running from left to right, as
+at present, but the top row exchanged with the bottom row? What are
+the fewest possible moves?</p>
+
+<p>One car moves at a time, and any distance counts as one move. To
+prevent misunderstanding, the stopping-places are marked in squares,
+and only one car can be in a square at the same time.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_225_THE_TEN_PRISONERS" id="X_225_THE_TEN_PRISONERS"></a><a href="#X_225_THE_TEN_PRISONERSa"><b>225.&mdash;THE TEN PRISONERS.</b></a></p>
+
+<p>If prisons had no other use, they might still be preserved for the
+special benefit of puzzle-makers. They appear to be an inexhaustible
+mine of perplexing ideas. Here is a little poser that will perhaps
+interest the reader for a short period. We have in the illustration a
+prison of sixteen cells. The locations of the ten prisoners will be
+seen. The jailer has queer superstitions about odd and even numbers,
+and he <span class='pagenum'>Pg 63<a name="Page_63" id="Page_63"></a></span>wants to rearrange the ten prisoners so that there shall be as
+many even rows of men, vertically, horizontally, and diagonally, as
+possible. At present it will be seen, as indicated by the arrows, that
+there are only twelve such rows of 2 and 4. I will state at once that
+the greatest number of such rows that is possible is sixteen. But the
+jailer only allows four men to be removed to other cells, and informs
+me that, as the man who is seated in the bottom right-hand corner is
+infirm, he must not be moved. Now, how are we to get those sixteen
+rows of even numbers under such conditions?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q225.png" width="400" height="392" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_226_ROUND_THE_COAST" id="X_226_ROUND_THE_COAST"></a><a href="#X_226_ROUND_THE_COASTa"><b>226.&mdash;ROUND THE COAST.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q226.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>Here is a puzzle that will, I think, be found as amusing as
+instructive. We are given a ring of eight circles. Leaving circle 8
+blank, we are required to write in the name of a seven-lettered port
+in the United Kingdom in this manner. Touch a blank circle with your
+pencil, then jump over two circles in either direction round the ring,
+and write down the first letter. Then touch another vacant circle,
+jump over two circles, and write down your second letter. Proceed
+similarly with the other letters in their proper order until you have
+completed the word. Thus, suppose we select "Glasgow," and proceed as
+follows: 6&mdash;1, 7&mdash;2, 8&mdash;3, 7&mdash;4, 8&mdash;5, which means that we touch 6,
+jump over 7 and and write down "G" on 1; then touch 7, jump over 8 and
+1, and write down "l" on 2; and so on. It will be found that after we
+have written down the first five letters&mdash;"Glasg"&mdash;as above, we cannot
+go any further. Either there is something wrong with "Glasgow," or we
+have not managed our jumps properly. Can you get to the bottom of the
+mystery?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_227_CENTRAL_SOLITAIRE" id="X_227_CENTRAL_SOLITAIRE"></a><a href="#X_227_CENTRAL_SOLITAIREa"><b>227.&mdash;CENTRAL SOLITAIRE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q227.png" width="400" height="397" alt="" title="" />
+</div>
+
+<p>This ancient puzzle was a great favourite with our grandmothers, and
+most of us, I imagine, have on occasions come across a "Solitaire"
+board&mdash;a round polished board with holes cut in it in a geometrical
+pattern, and a glass marble in every hole. Sometimes I have noticed
+one on a side table in a suburban front parlour, or found one on a
+shelf in a country cottage, or had one brought under my notice at a
+wayside inn. Sometimes they are of the form shown above, but it is
+equally common for the board to have four more holes, at the points
+indicated by dots. I select the simpler form.</p>
+
+<p>Though "Solitaire" boards are still sold at the toy shops, it will be
+sufficient if the reader will make an enlarged copy of the above on a
+sheet of cardboard or paper, number the "holes," and provide himself
+with 33 counters, buttons, or beans. Now place a counter in every hole
+except the central one, No. 17, and the puzzle is to take off all the
+counters in a series of jumps, except the last counter, which must be
+left in that central hole. You are <span class='pagenum'>Pg 64<a name="Page_64" id="Page_64"></a></span>allowed to jump one counter over
+the next one to a vacant hole beyond, just as in the game of draughts,
+and the counter jumped over is immediately taken off the board. Only
+remember every move must be a jump; consequently you will take off a
+counter at each move, and thirty-one single jumps will of course
+remove all the thirty-one counters. But compound moves are allowed (as
+in draughts, again), for so long as one counter continues to jump, the
+jumps all count as one move.</p>
+
+<p>Here is the beginning of an imaginary solution which will serve to
+make the manner of moving perfectly plain, and show how the solver
+should write out his attempts: 5-17, 12-10, 26-12, 24-26 (13-11,
+11-25), 9-11 (26-24, 24-10, 10-12), etc., etc. The jumps contained
+within brackets count as one move, because they are made with the same
+counter. Find the fewest possible moves. Of course, no diagonal jumps
+are permitted; you can only jump in the direction of the lines.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_228_THE_TEN_APPLES" id="X_228_THE_TEN_APPLES"></a><a href="#X_228_THE_TEN_APPLESa"><b>228.&mdash;THE TEN APPLES.</b></a></p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q228.png" width="600" height="425" alt="" title="" />
+</div>
+
+<p>The family represented in the illustration are amusing themselves with
+this little puzzle, which is not very difficult but quite interesting.
+They have, it will be seen, placed sixteen plates on the table in the
+form of a square, and put an apple in each of ten plates. They want to
+find a way of removing all the apples except one by jumping over one
+at a time to the next vacant square, as in draughts; or, better, as in
+solitaire, for you are not allowed to make any diagonal moves&mdash;only
+moves parallel to the sides of the square. It is obvious that as the
+apples stand no move can be made, but you are permitted to transfer
+any single apple you like to a vacant plate before starting. Then the
+moves must be all leaps, taking off the apples leaped over.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_229_THE_NINE_ALMONDS" id="X_229_THE_NINE_ALMONDS"></a><a href="#X_229_THE_NINE_ALMONDSa"><b>229.&mdash;THE NINE ALMONDS.</b></a></p>
+
+<p>"Here is a little puzzle," said a Parson, "that I have found
+peculiarly fascinating. It is so simple, and yet it keeps you
+interested indefinitely."</p>
+
+<p>The reverend gentleman took a sheet of paper and divided it off into
+twenty-five squares, like a square portion of a chessboard. Then he
+placed nine almonds on the central squares, as shown in the
+illustration, where we have represented numbered counters for
+convenience in giving the solution.</p>
+
+<p>"Now, the puzzle is," continued the Parson, "to remove eight of the
+almonds and leave the ninth in the central square. You make the
+removals by jumping one almond over another to the vacant square
+beyond and taking off the one jumped over&mdash;just as in draughts, only
+here you can jump in any direction, and not diagonally only. The point
+is to do the thing in the fewest possible moves."</p>
+
+<p>The following specimen attempt will make everything clear. Jump 4 over
+1, 5 over 9, 3 over 6, 5 over 3, 7 over 5 and 2, 4 over 7, 8 over 4.
+But 8 is not left in the central square, as it <span class='pagenum'>Pg 65<a name="Page_65" id="Page_65"></a></span>should be. Remember to
+remove those you jump over. Any number of jumps in succession with the
+same almond count as one move.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q229.png" width="400" height="399" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_230_THE_TWELVE_PENNIES" id="X_230_THE_TWELVE_PENNIES"></a><a href="#X_230_THE_TWELVE_PENNIESa"><b>230.&mdash;THE TWELVE PENNIES.</b></a></p>
+
+<p>Here is a pretty little puzzle that only requires twelve pennies or
+counters. Arrange them in a circle, as shown in the illustration. Now
+take up one penny at a time and, passing it over two pennies, place it
+on the third penny. Then take up another single penny and do the same
+thing, and so on, until, in six such moves, you have the coins in six
+pairs in the positions 1, 2, 3, 4, 5, 6. You can move in either
+direction round the circle at every play, and it does not matter
+whether the two jumped over are separate or a pair. This is quite easy
+if you use just a little thought.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q230.png" width="400" height="401" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_231_PLATES_AND_COINS" id="X_231_PLATES_AND_COINS"></a><a href="#X_231_PLATES_AND_COINSa"><b>231.&mdash;PLATES AND COINS.</b></a></p>
+
+<p>Place twelve plates, as shown, on a round table, with a penny or
+orange in every plate. Start from any plate you like and, always going
+in one direction round the table, take up one penny, pass it over two
+other pennies, and place it in the next plate. Go on again; take up
+another penny and, having passed it over two pennies, place it in a
+plate; and so continue your journey. Six coins only are to be removed,
+and when these have been placed there should be two coins in each of
+six plates and six plates empty. An important point of the puzzle is
+to go round the table as few times as possible. It does not matter
+whether the two coins passed over are in one or two plates, nor how
+many empty plates you pass a coin over. But you must always go in one
+direction round the table and end at the point from which you set out.
+Your hand, that is to say, goes steadily forward in one direction,
+without ever moving backwards.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q231.png" width="400" height="288" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_232_CATCHING_THE_MICE" id="X_232_CATCHING_THE_MICE"></a><a href="#X_232_CATCHING_THE_MICEa"><b>232.&mdash;CATCHING THE MICE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q232.png" width="400" height="408" alt="" title="" />
+</div>
+
+<p>"Play fair!" said the mice. "You know the rules of the game."</p>
+
+<p>"Yes, I know the rules," said the cat. "I've got to go round and round
+the circle, in the <span class='pagenum'>Pg 66<a name="Page_66" id="Page_66"></a></span>direction that you are looking, and eat every
+thirteenth mouse, but I must keep the white mouse for a tit-bit at the
+finish. Thirteen is an unlucky number, but I will do my best to oblige
+you."</p>
+
+<p>"Hurry up, then!" shouted the mice.</p>
+
+<p>"Give a fellow time to think," said the cat. "I don't know which of
+you to start at. I must figure it out."</p>
+
+<p>While the cat was working out the puzzle he fell asleep, and, the
+spell being thus broken, the mice returned home in safety. At which
+mouse should the cat have started the count in order that the white
+mouse should be the last eaten?</p>
+
+<p>When the reader has solved that little puzzle, here is a second one
+for him. What is the smallest number that the cat can count round and
+round the circle, if he must start at the white mouse (calling that
+"one" in the count) and still eat the white mouse last of all?</p>
+
+<p>And as a third puzzle try to discover what is the smallest number that
+the cat can count round and round if she must start at the white mouse
+(calling that "one") and make the white mouse the third eaten.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_233_THE_ECCENTRIC_CHEESEMONGER" id="X_233_THE_ECCENTRIC_CHEESEMONGER"></a><a href="#X_233_THE_ECCENTRIC_CHEESEMONGERa"><b>233.&mdash;THE ECCENTRIC CHEESEMONGER.</b></a></p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q233.png" width="600" height="516" alt="" title="" />
+</div>
+
+<p>The cheesemonger depicted in the illustration is an inveterate puzzle
+lover. One of his favourite puzzles is the piling of cheeses in his
+warehouse, an amusement that he finds good exercise for the body as
+well as for the mind. He places sixteen cheeses on the floor in a
+straight row and then makes them into four piles, with four cheeses in
+every pile, by always passing a cheese over four others. If you use
+sixteen counters and number them in order from 1 to 16, then you may
+place 1 on 6, 11 on 1, 7 on 4, and so on, until there are four in
+every pile. It will be seen that it does not matter whether the four
+passed over are standing alone or piled; they count just the same, and
+you can always carry a cheese in either direction. There are a great
+many different ways of doing it in twelve moves, so it makes a good
+game of "patience" to try to solve it so that the four piles shall be
+left in different stipulated places. For example, try to leave the
+piles at the extreme ends of the row, on Nos. 1, 2, 15 and 16; this is
+quite easy. Then try to leave three piles together, on Nos. 13, 14,
+and 15. Then again play so that they shall be left on Nos. 3, 5, 12,
+and 14.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_234_THE_EXCHANGE_PUZZLE" id="X_234_THE_EXCHANGE_PUZZLE"></a><a href="#X_234_THE_EXCHANGE_PUZZLEa"><b>234.&mdash;THE EXCHANGE PUZZLE.</b></a></p>
+
+
+<p>Here is a rather entertaining little puzzle with moving counters. You
+only need twelve counters&mdash;six of one colour, marked A, C, E, G, I,
+and K, and the other six marked B, D, F, H, J, and L. You first place
+them on the diagram, as shown in the illustration, and the puzzle is
+to get them into regular alphabetical order, as follows:&mdash;<span class='pagenum'>Pg 67<a name="Page_67" id="Page_67"></a></span></p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>A</td><td align='center'>B</td><td align='center'>C</td><td align='center'>D</td></tr>
+<tr><td align='center'>E</td><td align='center'>F</td><td align='center'>G</td><td align='center'>H</td></tr>
+<tr><td align='center'>I</td><td align='center'>J</td><td align='center'>K</td><td align='center'>L</td></tr>
+</table></div>
+
+<p>The moves are made by exchanges of opposite colours standing on the
+same line. Thus, G and J may exchange places, or F and A, but you
+cannot exchange G and C, or F and D, because in one case they are both
+white and in the other case both black. Can you bring about the
+required arrangement in seventeen exchanges?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q234.png" width="400" height="282" alt="" title="" />
+</div>
+
+<p>It cannot be done in fewer moves. The puzzle is really much easier
+than it looks, if properly attacked.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_235_TORPEDO_PRACTICE" id="X_235_TORPEDO_PRACTICE"></a><a href="#X_235_TORPEDO_PRACTICEa"><b>235.&mdash;TORPEDO PRACTICE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q235.png" width="400" height="405" alt="" title="" />
+</div>
+
+<p>If a fleet of sixteen men-of-war were lying at anchor and surrounded
+by the enemy, how many ships might be sunk if every torpedo, projected
+in a straight line, passed under three vessels and sank the fourth? In
+the diagram we have arranged the fleet in square formation, where it
+will be seen that as many as seven ships may be sunk (those in the top
+row and first column) by firing the torpedoes indicated by arrows.
+Anchoring the fleet as we like, to what extent can we increase this
+number? Remember that each successive ship is sunk before another
+torpedo is launched, and that every torpedo proceeds in a different
+direction; otherwise, by placing the ships in a straight line, we
+might sink as many as thirteen! It is an interesting little study in
+naval warfare, and eminently practical&mdash;provided the enemy will allow
+you to arrange his fleet for your convenience and promise to lie still
+and do nothing!</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_236_THE_HAT_PUZZLE" id="X_236_THE_HAT_PUZZLE"></a><a href="#X_236_THE_HAT_PUZZLEa"><b>236.&mdash;THE HAT PUZZLE.</b></a></p>
+
+<p>Ten hats were hung on pegs as shown in the illustration&mdash;five silk
+hats and five felt "bowlers," alternately silk and felt. The two pegs
+at the end of the row were empty.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q236.png" width="600" height="61" alt="" title="" />
+</div>
+
+<p>The puzzle is to remove two contiguous hats to the vacant pegs, then
+two other adjoining hats to the pegs now unoccupied, and so on until
+five pairs have been moved and the hats again hang in an unbroken row,
+but with all the silk ones together and all the felt hats together.</p>
+
+<p>Remember, the two hats removed must always be contiguous ones, and you
+must take one in each hand and place them on their new pegs without
+reversing their relative position. You are not allowed to cross your
+hands, nor to hang up one at a time.</p>
+
+<p>Can you solve this old puzzle, which I give as introductory to the
+next? Try it with counters of two colours or with coins, and remember
+that the two empty pegs must be left at one end of the row.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_237_BOYS_AND_GIRLS" id="X_237_BOYS_AND_GIRLS"></a><a href="#X_237_BOYS_AND_GIRLSa"><b>237.&mdash;BOYS AND GIRLS.</b></a></p>
+
+<p>If you mark off ten divisions on a sheet of paper to represent the
+chairs, and use eight numbered counters for the children, you will
+have a fascinating pastime. Let the odd numbers represent boys and
+even numbers girls, or you can use counters of two colours, or coins.</p>
+
+<p>The puzzle is to remove two children who are occupying adjoining
+chairs and place them in two empty chairs, <i>making them first change
+sides</i>; then remove a second pair of children from adjoining chairs
+and place them in the two now vacant, making them change sides; and so
+on, until all the boys are together and all the girls together, with
+the two vacant chairs at one end as at present. To solve the puzzle
+you must do this in five moves. The two children must always be taken
+from chairs that are next to one another; and remember the important
+point of making the two children change sides, <span class='pagenum'>Pg 68<a name="Page_68" id="Page_68"></a></span>as this latter is the
+distinctive feature of the puzzle. By "change sides" I simply mean
+that if, for example, you first move 1 and 2 to the vacant chairs,
+then the first (the outside) chair will be occupied by 2 and the
+second one by 1.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q237.png" width="600" height="148" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_238_ARRANGING_THE_JAMPOTS" id="X_238_ARRANGING_THE_JAMPOTS"></a><a href="#X_238_ARRANGING_THE_JAMPOTSa"><b>238.&mdash;ARRANGING THE JAMPOTS.</b></a></p>
+
+<p>I happened to see a little girl sorting out some jam in a cupboard for
+her mother. She was putting each different kind of preserve apart on
+the shelves. I noticed that she took a pot of damson in one hand and a
+pot of gooseberry in the other and made them change places; then she
+changed a strawberry with a raspberry, and so on. It was interesting
+to observe what a lot of unnecessary trouble she gave herself by
+making more interchanges than there was any need for, and I thought it
+would work into a good puzzle.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q238.png" width="600" height="500" alt="" title="" />
+</div>
+
+<p>It will be seen in the illustration that little Dorothy has to
+manipulate twenty-four large jampots in as many pigeon-holes. She
+wants to get them in correct numerical order&mdash;that is, 1, 2, 3, 4, 5,
+6 on the top shelf, 7, 8, 9, 10, 11, 12 on the next shelf, and so on.
+Now, if she always takes one pot in the right hand and another in the
+left and makes them change places, how many of these interchanges will
+be necessary to get all the jampots in proper order? She would
+naturally first change the 1 and the 3, then the 2 and the 3, when she
+would have the first three pots in their places. How would you advise
+her to go on then? Place some numbered counters on a sheet of paper
+divided into squares for the pigeon-holes, and you will find it an
+amusing puzzle.</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="UNICURSAL_AND_ROUTE_PROBLEMS" id="UNICURSAL_AND_ROUTE_PROBLEMS"></a><a href="#CONTENTS">UNICURSAL AND ROUTE PROBLEMS.</a></h2>
+
+<p class='center'>
+"I see them on their winding way."<br />
+<span style="margin-left: 8em;">REGINALD HEBER.</span><br />
+</p>
+
+<p>It is reasonable to suppose that from the earliest ages one man has
+asked another such questions as these: "Which is the nearest way
+home?" "Which is the easiest or pleasantest way?" "How can we find a
+way that will enable us to dodge the mastodon and the plesiosaurus?"
+"How can we get there without ever crossing the track of the enemy?"
+All these are elementary route problems, and they can be turned into
+good puzzles by the introduction of some conditions that complicate
+matters. A variety of such complications will be found in the
+following examples. I have also included some enumerations of more or
+less difficulty. These afford excellent practice for the reasoning
+faculties, and enable one to generalize in the case of symmetrical
+forms in a manner that is most instructive.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_239_A_JUVENILE_PUZZLE" id="X_239_A_JUVENILE_PUZZLE"></a><a href="#X_239_A_JUVENILE_PUZZLEa"><b>239.&mdash;A JUVENILE PUZZLE.</b></a></p>
+
+<p>For years I have been perpetually consulted by my juvenile friends
+about this little puzzle. Most children seem to know it, and yet,
+curiously enough, they are invariably unacquainted with the answer.
+The question they always ask is, "Do, please, tell me whether it is
+really possible." I believe Houdin the conjurer used to be very fond
+of giving it to his child friends, but I cannot say whether he
+invented the little puzzle or not. No doubt a large number of my
+readers will be glad to have the mystery of the solution cleared up,
+so I make no apology for introducing this old "teaser."</p>
+
+<p>The puzzle is to draw with three strokes of the pencil the diagram
+that the little girl is exhibiting in the illustration. Of course, you
+must not remove your pencil from the paper during a stroke or go over
+the same line a second time. You will find that you can get <span class='pagenum'>Pg 69<a name="Page_69" id="Page_69"></a></span>in a good
+deal of the figure with one continuous stroke, but it will always
+appear as if four strokes are necessary.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q239.png" width="400" height="380" alt="" title="" />
+</div>
+
+<p>Another form of the puzzle is to draw the diagram on a slate and then
+rub it out in three rubs.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_240_THE_UNION_JACK" id="X_240_THE_UNION_JACK"></a><a href="#X_240_THE_UNION_JACKa"><b>240.&mdash;THE UNION JACK.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q240.png" width="400" height="287" alt="" title="" />
+</div>
+
+<p>The illustration is a rough sketch somewhat resembling the British
+flag, the Union Jack. It is not possible to draw the whole of it
+without lifting the pencil from the paper or going over the same line
+twice. The puzzle is to find out just <i>how much</i> of the drawing it is
+possible to make without lifting your pencil or going twice over the
+same line. Take your pencil and see what is the best you can do.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_241_THE_DISSECTED_CIRCLE" id="X_241_THE_DISSECTED_CIRCLE"></a><a href="#X_241_THE_DISSECTED_CIRCLEa"><b>241.&mdash;THE DISSECTED CIRCLE.</b></a></p>
+
+<p>How many continuous strokes, without lifting your pencil from the
+paper, do you require to draw the design shown in our illustration?
+Directly you change the direction of your pencil it begins a new
+stroke. You may go over the same line more than once if you like. It
+requires just a little care, or you may find yourself beaten by one
+stroke.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q241.png" width="400" height="392" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_242_THE_TUBE_INSPECTORS_PUZZLE" id="X_242_THE_TUBE_INSPECTORS_PUZZLE"></a><a href="#X_242_THE_TUBE_INSPECTORS_PUZZLEa"><b>242.&mdash;THE TUBE INSPECTOR'S PUZZLE.</b></a></p>
+
+<p>The man in our illustration is in a little dilemma. He has just been
+appointed inspector of a certain system of tube railways, and it is
+his duty to inspect regularly, within a stated period, all the
+company's seventeen lines connecting twelve stations, as shown on the
+big poster plan that he is contemplating. Now he wants to arrange his
+route so that it shall take him over all the lines with as little
+travelling as possible. He may begin where he likes and end where he
+likes. What is his shortest route?</p>
+
+<p>Could anything be simpler? But the reader will soon find that, however
+he decides to proceed, the inspector must go over some of the lines
+more than once. In other words, if we say that the stations are a mile
+apart, he will have to travel more than seventeen miles to inspect
+every line. There is the little difficulty. How far is he compelled to
+travel, and which route do you recommend?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q242.png" width="400" height="461" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 70<a name="Page_70" id="Page_70"></a></span><a name="X_243_VISITING_THE_TOWNS" id="X_243_VISITING_THE_TOWNS"></a><a href="#X_243_VISITING_THE_TOWNSa"><b>243.&mdash;VISITING THE TOWNS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q243.png" width="400" height="251" alt="" title="" />
+</div>
+
+<p>A traveller, starting from town No. 1, wishes to visit every one of
+the towns once, and once only, going only by roads indicated by
+straight lines. How many different routes are there from which he can
+select? Of course, he must end his journey at No. 1, from which he
+started, and must take no notice of cross roads, but go straight from
+town to town. This is an absurdly easy puzzle, if you go the right way
+to work.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_244_THE_FIFTEEN_TURNINGS" id="X_244_THE_FIFTEEN_TURNINGS"></a><a href="#X_244_THE_FIFTEEN_TURNINGSa"><b>244.&mdash;THE FIFTEEN TURNINGS.</b></a></p>
+
+<p>Here is another queer travelling puzzle, the solution of which calls
+for ingenuity. In this case the traveller starts from the black town
+and wishes to go as far as possible while making only fifteen turnings
+and never going along the same road twice. The towns are supposed to
+be a mile apart. Supposing, for example, that he went straight to A,
+then straight to B, then to C, D, E, and F, you will then find that he
+has travelled thirty-seven miles in five turnings. Now, how far can he
+go in fifteen turnings?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q244.png" width="400" height="384" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_245_THE_FLY_ON_THE_OCTAHEDRON" id="X_245_THE_FLY_ON_THE_OCTAHEDRON"></a><a href="#X_245_THE_FLY_ON_THE_OCTAHEDRONa"><b>245.&mdash;THE FLY ON THE OCTAHEDRON.</b></a></p>
+
+<p>"Look here," said the professor to his colleague, "I have been
+watching that fly on the octahedron, and it confines its walks
+entirely to the edges. What can be its reason for avoiding the sides?"</p>
+
+<p>"Perhaps it is trying to solve some route problem," suggested the
+other. "Supposing it to start from the top point, how many different
+routes are there by which it may walk over all the edges, without ever
+going twice along the same edge in any route?"</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q245.png" width="400" height="528" alt="" title="" />
+</div>
+
+<p>The problem was a harder one than they expected, and after working at
+it during leisure moments for several days their results did not
+agree&mdash;in fact, they were both wrong. If the reader is surprised at
+their failure, let him attempt the little puzzle himself. I will just
+explain that the octahedron is one of the five regular, or Platonic,
+bodies, and is contained under eight equal and equilateral triangles.
+If you cut out the two pieces of cardboard of the shape shown in the
+margin of the illustration, cut half through along the dotted lines
+and then bend them and put them together, you will have a perfect
+octahedron. In any route over all the edges it will be found that the
+fly must end at the point of departure at the top.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_246_THE_ICOSAHEDRON_PUZZLE" id="X_246_THE_ICOSAHEDRON_PUZZLE"></a><a href="#X_246_THE_ICOSAHEDRON_PUZZLEa"><b>246.&mdash;THE ICOSAHEDRON PUZZLE.</b></a></p>
+
+<p>The icosahedron is another of the five regular, or Platonic, bodies
+having all their sides, angles, and planes similar and equal. It is
+bounded by twenty similar equilateral triangles. If you cut out a
+piece of cardboard of the form shown in the smaller diagram, and cut
+half through along the dotted lines, it will fold up and form a
+perfect icosahedron.</p>
+
+<p>Now, a Platonic body does not mean a <span class='pagenum'>Pg 71<a name="Page_71" id="Page_71"></a></span>heavenly body; but it will suit
+the purpose of our puzzle if we suppose there to be a habitable planet
+of this shape. We will also suppose that, owing to a superfluity of
+water, the only dry land is along the edges, and that the inhabitants
+have no knowledge of navigation. If every one of those edges is 10,000
+miles long and a solitary traveller is placed at the North Pole (the
+highest point shown), how far will he have to travel before he will
+have visited every habitable part of the planet&mdash;that is, have
+traversed every one of the edges?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q246.png" width="400" height="652" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_247_INSPECTING_A_MINE" id="X_247_INSPECTING_A_MINE"></a><a href="#X_247_INSPECTING_A_MINEa"><b>247.&mdash;INSPECTING A MINE.</b></a></p>
+
+<p>The diagram is supposed to represent the passages or galleries in a
+mine. We will assume that every passage, A to B, B to C, C to H, H to
+I, and so on, is one furlong in length. It will be seen that there are
+thirty-one of these passages. Now, an official has to inspect all of
+them, and he descends by the shaft to the point A. How far must he
+travel, and what route do you recommend? The reader may at first say,
+"As there are thirty-one passages, each a furlong in length, he will
+have to travel just thirty-one furlongs." But this is assuming that he
+need never go along a passage more than once, which is not the case.
+Take your pencil and try to find the shortest route. You will soon
+discover that there is room for considerable judgment. In fact, it is
+a perplexing puzzle.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q247.png" width="400" height="310" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_248_THE_CYCLISTS_TOUR" id="X_248_THE_CYCLISTS_TOUR"></a><a href="#X_248_THE_CYCLISTS_TOURa"><b>248.&mdash;THE CYCLISTS' TOUR.</b></a></p>
+
+<p>Two cyclists were consulting a road map in preparation for a little
+tour together. The circles represent towns, and all the good roads are
+represented by lines. They are starting from the town with a star, and
+must complete their tour at E. But before arriving there they want to
+visit every other town once, and only once. That is the difficulty.
+Mr. Spicer said, "I am certain we can find a way of doing it;" but Mr.
+Maggs replied, "No way, I'm sure." Now, which of them was correct?
+Take your pencil and see if you can find any way of doing it. Of
+course you must keep to the roads indicated.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q248.png" width="400" height="318" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_249_THE_SAILORS_PUZZLE" id="X_249_THE_SAILORS_PUZZLE"></a><a href="#X_249_THE_SAILORS_PUZZLEa"><b>249.&mdash;THE SAILOR'S PUZZLE.</b></a></p>
+
+<p>The sailor depicted in the illustration stated that he had since his
+boyhood been engaged in trading with a small vessel among some twenty
+little islands in the Pacific. He supplied the rough chart of which I
+have given a copy, and explained that the lines from island to island
+represented the only routes that he ever adopted. He always started
+from island A at the beginning of the season, and then visited every
+island <span class='pagenum'>Pg 72<a name="Page_72" id="Page_72"></a></span>once, and once only, finishing up his tour at the
+starting-point A. But he always put off his visit to C as long as
+possible, for trade reasons that I need not enter into. The puzzle is
+to discover his exact route, and this can be done with certainty. Take
+your pencil and, starting at A, try to trace it out. If you write down
+the islands in the order in which you visit them&mdash;thus, for example,
+A, I, O, L, G, etc.&mdash;you can at once see if you have visited an island
+twice or omitted any. Of course, the crossings of the lines must be
+ignored&mdash;that is, you must continue your route direct, and you are not
+allowed to switch off at a crossing and proceed in another direction.
+There is no trick of this kind in the puzzle. The sailor knew the best
+route. Can you find it?</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q249.png" width="600" height="777" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_250_THE_GRAND_TOUR" id="X_250_THE_GRAND_TOUR"></a><a href="#X_250_THE_GRAND_TOURa"><b>250.&mdash;THE GRAND TOUR.</b></a></p>
+
+<p>One of the everyday puzzles of life is the working out of routes. If
+you are taking a holiday on your bicycle, or a motor tour, there
+always arises the question of how you are to make the best of your
+time and other resources. You have determined to get as far as some
+particular place, to include visits to such-and-such a town, to try to
+see something of special interest elsewhere, and perhaps to try to
+look up an old friend at a spot that will not take you much out of
+your way. Then you have to plan your route so as to avoid bad roads,
+uninteresting country, and, if possible, the necessity of a return by
+the same way that you went. With a map before you, the interesting
+puzzle is attacked and solved. I will present a little poser based on
+these lines.</p>
+
+<p>I give a rough map of a country&mdash;it is not necessary to say what
+particular country&mdash;the circles representing towns and the dotted
+lines the railways connecting them. Now there lived in the town marked
+A a man who was born there, and during the whole of his life had never
+once left his native place. From his youth upwards he had been very
+industrious, sticking incessantly to his trade, and had no desire
+whatever to roam abroad. However, on attaining his fiftieth birthday
+he decided to see something of his country, and especially to pay a
+visit to a very old friend living at the town marked Z. <span class='pagenum'>Pg 73<a name="Page_73" id="Page_73"></a></span>What he
+proposed was this: that he would start from his home, enter every town
+once and only once, and finish his journey at Z. As he made up his
+mind to perform this grand tour by rail only, he found it rather a
+puzzle to work out his route, but he at length succeeded in doing so.
+How did he manage it? Do not forget that every town has to be visited
+once, and not more than once.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q250.png" width="400" height="345" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_251_WATER_GAS_AND_ELECTRICITY" id="X_251_WATER_GAS_AND_ELECTRICITY"></a><a href="#X_251_WATER_GAS_AND_ELECTRICITYa"><b>251.&mdash;WATER, GAS, AND ELECTRICITY.</b></a></p>
+
+<p>There are some half-dozen puzzles, as old as the hills, that are
+perpetually cropping up, and there is hardly a month in the year that
+does not bring inquiries as to their solution. Occasionally one of
+these, that one had thought was an extinct volcano, bursts into
+eruption in a surprising manner. I have received an extraordinary
+number of letters respecting the ancient puzzle that I have called
+"Water, Gas, and Electricity." It is much older than electric
+lighting, or even gas, but the new dress brings it up to date. The
+puzzle is to lay on water, gas, and electricity, from W, G, and E, to
+each of the three houses, A, B, and C, without any pipe crossing
+another. Take your pencil and draw lines showing how this should be
+done. You will soon find yourself landed in difficulties.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q251.png" width="400" height="233" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_252_A_PUZZLE_FOR_MOTORISTS" id="X_252_A_PUZZLE_FOR_MOTORISTS"></a><a href="#X_252_A_PUZZLE_FOR_MOTORISTSa"><b>252.&mdash;A PUZZLE FOR MOTORISTS.</b></a></p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q252.png" width="600" height="343" alt="" title="" />
+</div>
+
+<p>Eight motorists drove to church one morning. Their respective houses
+and churches, together with the only roads available (the dotted
+lines), are shown. One went from his house A to his church A, another
+from his house B to his church B, another from C to C, and so on, but
+it was afterwards found that no driver ever crossed the track of
+another car. Take your pencil and try to trace out their various
+routes.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_253_A_BANK_HOLIDAY_PUZZLE" id="X_253_A_BANK_HOLIDAY_PUZZLE"></a><a href="#X_253_A_BANK_HOLIDAY_PUZZLEa"><b>253.&mdash;A BANK HOLIDAY PUZZLE.</b></a></p>
+
+<p>Two friends were spending their bank holiday on a cycling trip.
+Stopping for a rest at a village inn, they consulted a route map,
+which is represented in our illustration in an exceedingly simplified
+form, for the puzzle is interesting enough without all the original
+complexities. They started from the town in the top left-hand corner
+marked A. It will be seen that there are one hundred and twenty such
+towns, all connected by straight roads. Now they discovered that there
+are exactly 1,365 different routes by which they may reach their
+destina<span class='pagenum'>Pg 74<a name="Page_74" id="Page_74"></a></span>tion, always travelling either due south or due east. The
+puzzle is to discover which town is their destination.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q253.png" width="400" height="343" alt="" title="" />
+</div>
+
+<p>Of course, if you find that there are more than 1,365 different routes
+to a town it cannot be the right one.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_254_THE_MOTOR-CAR_TOUR" id="X_254_THE_MOTOR-CAR_TOUR"></a><a href="#X_254_THE_MOTOR-CAR_TOURa"><b>254.&mdash;THE MOTOR-CAR TOUR.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q254.png" width="400" height="377" alt="" title="" />
+</div>
+
+<p>In the above diagram the circles represent towns and the lines good
+roads. In just how many different ways can a motorist, starting from
+London (marked with an L), make a tour of all these towns, visiting
+every town once, and only once, on a tour, and always coming back to
+London on the last ride? The exact reverse of any route is not
+counted as different.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_255_THE_LEVEL_PUZZLE" id="X_255_THE_LEVEL_PUZZLE"></a><a href="#X_255_THE_LEVEL_PUZZLEa"><b>255.&mdash;THE LEVEL PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q255.png" width="400" height="396" alt="" title="" />
+</div>
+
+<p>This is a simple counting puzzle. In how many different ways can you
+spell out the word LEVEL by placing the point of your pencil on an L
+and then passing along the lines from letter to letter. You may go in
+any direction, backwards or forwards. Of course you are not allowed to
+miss letters&mdash;that is to say, if you come to a letter you must use it.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_256_THE_DIAMOND_PUZZLE" id="X_256_THE_DIAMOND_PUZZLE"></a><a href="#X_256_THE_DIAMOND_PUZZLEa"><b>256.&mdash;THE DIAMOND PUZZLE.</b></a></p>
+
+<p>IN how many different ways may the word DIAMOND be read in the
+arrangement shown? You may start wherever you like at a D and go up or
+down, backwards or forwards, in and out, in any direction you like, so
+long as you always pass from one letter to another that adjoins it.
+How many ways are there?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q256.png" width="400" height="405" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_257_THE_DEIFIED_PUZZLE" id="X_257_THE_DEIFIED_PUZZLE"></a><a href="#X_257_THE_DEIFIED_PUZZLEa"><b>257.&mdash;THE DEIFIED PUZZLE.</b></a></p>
+
+<p>In how many different ways may the word DEIFIED be read in this
+arrangement under <span class='pagenum'>Pg 75<a name="Page_75" id="Page_75"></a></span>the same conditions as in the last puzzle, with the
+addition that you can use any letters twice in the same reading?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q257.png" width="400" height="400" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_258_THE_VOTERS_PUZZLE" id="X_258_THE_VOTERS_PUZZLE"></a><a href="#X_258_THE_VOTERS_PUZZLEa"><b>258.&mdash;THE VOTERS' PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q258.png" width="400" height="393" alt="" title="" />
+</div>
+
+<p>Here we have, perhaps, the most interesting form of the puzzle. In how
+many different ways can you read the political injunction, "RISE TO
+VOTE, SIR," under the same conditions as before? In this case every
+reading of the palindrome requires the use of the central V as the
+middle letter.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_259_HANNAHS_PUZZLE" id="X_259_HANNAHS_PUZZLE"></a><a href="#X_259_HANNAHS_PUZZLEa"><b>259.&mdash;HANNAH'S PUZZLE.</b></a></p>
+
+<p>A man was in love with a young lady whose Christian name was Hannah.
+When he asked her to be his wife she wrote down the letters of her
+name in this manner:&mdash;</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q259.png" width="400" height="210" alt="" title="" />
+</div>
+
+<p>and promised that she would be his if he could tell her correctly in
+how many different ways it was possible to spell out her name, always
+passing from one letter to another that was adjacent. Diagonal steps
+are here allowed. Whether she did this merely to tease him or to test
+his cleverness is not recorded, but it is satisfactory to know that he
+succeeded. Would you have been equally successful? Take your pencil
+and try. You may start from any of the H's and go backwards or
+forwards and in any direction, so long as all the letters in a
+spelling are adjoining one another. How many ways are there, no two
+exactly alike?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_260_THE_HONEYCOMB_PUZZLE" id="X_260_THE_HONEYCOMB_PUZZLE"></a><a href="#X_260_THE_HONEYCOMB_PUZZLEa"><b>260.&mdash;THE HONEYCOMB PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q260.png" width="400" height="353" alt="" title="" />
+</div>
+
+<p>Here is a little puzzle with the simplest possible conditions. Place
+the point of your pencil on a letter in one of the cells of the
+honeycomb, and trace out a very familiar proverb by passing always
+from a cell to one that is contiguous to it. If you take the right
+route you will have visited every cell once, and only once. The puzzle
+is much easier than it looks.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_261_THE_MONK_AND_THE_BRIDGES" id="X_261_THE_MONK_AND_THE_BRIDGES"></a><a href="#X_261_THE_MONK_AND_THE_BRIDGESa"><b>261.&mdash;THE MONK AND THE BRIDGES.</b></a></p>
+
+<p>In this case I give a rough plan of a river with an island and five
+bridges. On one side of the river is a monastery, and on the other
+side is seen a monk in the foreground. Now, the monk has decided that
+he will cross every bridge once, and only once, on his return to the
+monastery. This is, of course, quite easy to do, but on the way he
+thought to himself, "I wonder how many different routes there are from
+which I might have selected." Could you have told him? That is the
+puzzle. Take your pencil and trace out a route that will take you once
+<span class='pagenum'>Pg 76<a name="Page_76" id="Page_76"></a></span>over all the five bridges. Then trace out a second route, then a
+third, and see if you can count all the variations. You will find that
+the difficulty is twofold: you have to avoid dropping routes on the
+one hand and counting the same routes more than once on the other.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q261.png" width="600" height="400" alt="" title="" />
+</div>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="COMBINATION_AND_GROUP_PROBLEMS" id="COMBINATION_AND_GROUP_PROBLEMS"></a><a href="#CONTENTS">COMBINATION AND GROUP PROBLEMS.</a></h2>
+
+<p class='center'>
+"A combination and a form indeed."<br />
+<span style="margin-left: 8em;"><i>Hamlet</i>, iii. 4.</span><br />
+</p>
+
+<p>Various puzzles in this class might be termed problems in the
+"geometry of situation," but their solution really depends on the
+theory of combinations which, in its turn, is derived directly from
+the theory of permutations. It has seemed convenient to include here
+certain group puzzles and enumerations that might, perhaps, with equal
+reason have been placed elsewhere; but readers are again asked not to
+be too critical about the classification, which is very difficult and
+arbitrary. As I have included my problem of "The Round Table" (No.
+273), perhaps a few remarks on another well-known problem of the same
+class, known by the French as La Probl&ecirc;me des M&eacute;nages, may be
+interesting. If <i>n</i> married ladies are seated at a round table in any
+determined order, in how many different ways may their <i>n</i> husbands be
+placed so that every man is between two ladies but never next to his
+own wife?</p>
+
+<p>This difficult problem was first solved by Laisant, and the method
+shown in the following table is due to Moreau:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="2" summary="">
+<tr><td align='right'>4</td><td align='right'>0</td><td align='right'>2</td></tr>
+<tr><td align='right'>5</td><td align='right'>3</td><td align='right'>13</td></tr>
+<tr><td align='right'>6</td><td align='right'>13</td><td align='right'>80</td></tr>
+<tr><td align='right'>7</td><td align='right'>83</td><td align='right'>579</td></tr>
+<tr><td align='right'>8</td><td align='right'>592</td><td align='right'>4738</td></tr>
+<tr><td align='right'>9</td><td align='right'>4821</td><td align='right'>43387</td></tr>
+<tr><td align='right'>10</td><td align='right'>43979</td><td align='right'>439792</td></tr>
+</table></div>
+
+<p>The first column shows the number of married couples. The numbers in
+the second column are obtained in this way: 5&nbsp;&times;&nbsp;3&nbsp;+&nbsp;0&nbsp;-&nbsp;2&nbsp;=&nbsp;13; 6&nbsp;&times;&nbsp;13 +
+3&nbsp;+&nbsp;2&nbsp;=&nbsp;83; 7&nbsp;&times;&nbsp;83&nbsp;+&nbsp;13&nbsp;-&nbsp;2&nbsp;=&nbsp;592; 8&nbsp;&times;&nbsp;592&nbsp;+&nbsp;83&nbsp;+&nbsp;2&nbsp;=&nbsp;4821; and so on.
+Find all the numbers, except 2, in the table, and the method will be
+evident. It will be noted that the 2 is subtracted when the first
+number (the number of couples) is odd, and added when that number is
+even. The numbers in the third column are obtained thus: 13&nbsp;-&nbsp;0&nbsp;=&nbsp;13; 83
+- 3&nbsp;=&nbsp;80; 592&nbsp;-&nbsp;13&nbsp;=&nbsp;579; 4821&nbsp;-&nbsp;83&nbsp;=&nbsp;4738; and so on. The numbers in
+this last column give the required solutions. Thus, four husbands may
+be seated in two ways, five husbands may be placed in thirteen ways,
+and six husbands in eighty ways.</p>
+
+<p>The following method, by Lucas, will show the remarkable way in which
+chessboard analysis may be applied to the solution of a circular
+problem of this kind. Divide a square into thirty-six cells, six by
+six, and strike out all the cells in the long diagonal from the bottom
+left-hand corner to the top right-hand corner, also the five cells in
+the diagonal next above it and the cell in the bottom right-hand
+corner. The answer for six couples will be the same as the number of
+ways in which you can place six rooks (not using the cancelled cells)
+so that no rook shall ever attack another rook. It will be found that
+the six rooks may be placed in eighty different ways, which agrees
+with the above table.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 77<a name="Page_77" id="Page_77"></a></span><a name="X_262_THOSE_FIFTEEN_SHEEP" id="X_262_THOSE_FIFTEEN_SHEEP"></a><a href="#X_262_THOSE_FIFTEEN_SHEEPa"><b>262.&mdash;THOSE FIFTEEN SHEEP.</b></a></p>
+
+<p>A certain cyclop&aelig;dia has the following curious problem, I am told:
+"Place fifteen sheep in four pens so that there shall be the same
+number of sheep in each pen." No answer whatever is vouchsafed, so I
+thought I would investigate the matter. I saw that in dealing with
+apples or bricks the thing would appear to be quite impossible, since
+four times any number must be an even number, while fifteen is an odd
+number. I thought, therefore, that there must be some quality peculiar
+to the sheep that was not generally known. So I decided to interview
+some farmers on the subject. The first one pointed out that if we put
+one pen inside another, like the rings of a target, and placed all
+sheep in the smallest pen, it would be all right. But I objected to
+this, because you admittedly place all the sheep in one pen, not in
+four pens. The second man said that if I placed four sheep in each of
+three pens and three sheep in the last pen (that is fifteen sheep in
+all), and one of the ewes in the last pen had a lamb during the night,
+there would be the same number in each pen in the morning. This also
+failed to satisfy me.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q262.png" width="600" height="384" alt="" title="" />
+</div>
+
+<p>The third farmer said, "I've got four hurdle pens down in one of my
+fields, and a small flock of wethers, so if you will just step down
+with me I will show you how it is done." The illustration depicts my
+friend as he is about to demonstrate the matter to me. His lucid
+explanation was evidently that which was in the mind of the writer of
+the article in the cyclop&aelig;dia. What was it? Can you place those
+fifteen sheep?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_263_KING_ARTHURS_KNIGHTS" id="X_263_KING_ARTHURS_KNIGHTS"></a><a href="#X_263_KING_ARTHURS_KNIGHTSa"><b>263.&mdash;KING ARTHUR'S KNIGHTS.</b></a></p>
+
+<p>King Arthur sat at the Round Table on three successive evenings with
+his knights&mdash;Beleobus, Caradoc, Driam, Eric, Floll, and Galahad&mdash;but
+on no occasion did any person have as his neighbour one who had before
+sat next to him. On the first evening they sat in alphabetical order
+round the table. But afterwards King Arthur arranged the two next
+sittings so that he might have Beleobus as near to him as possible and
+Galahad as far away from him as could be managed. How did he seat the
+knights to the best advantage, remembering that rule that no knight
+may have the same neighbour twice?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_264_THE_CITY_LUNCHEONS" id="X_264_THE_CITY_LUNCHEONS"></a><a href="#X_264_THE_CITY_LUNCHEONSa"><b>264.&mdash;THE CITY LUNCHEONS.</b></a></p>
+
+<p>Twelve men connected with a large firm in the City of London sit down
+to luncheon together every day in the same room. The tables are small
+ones that only accommodate two persons at the same time. Can you show
+how these twelve men may lunch together on eleven days in pairs, so
+that no two of them shall ever sit twice together? We will represent
+the men by the first twelve letters of the alphabet, and suppose the
+first day's pairing to be as follows&mdash;</p>
+
+<p>
+<span style="margin-left: 2em;">(A B) (C D) (E F) (G H) (I J) (K L).</span><br />
+</p>
+
+<p>Then give any pairing you like for the next day, say&mdash;</p>
+
+<p>
+<span style="margin-left: 2em;">(A C) (B D) (E G) (F H) (I K) (J L),</span><br />
+</p>
+
+<p>and so on, until you have completed your eleven lines, with no pair
+ever occurring twice. There are a good many different arrangements
+possible. Try to find one of them.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 78<a name="Page_78" id="Page_78"></a></span><a name="X_265_A_PUZZLE_FOR_CARD-PLAYERS" id="X_265_A_PUZZLE_FOR_CARD-PLAYERS"></a><a href="#X_265_A_PUZZLE_FOR_CARD-PLAYERSa"><b>265.&mdash;A PUZZLE FOR CARD-PLAYERS.</b></a></p>
+
+<p>Twelve members of a club arranged to play bridge together on eleven
+evenings, but no player was ever to have the same partner more than
+once, or the same opponent more than twice. Can you draw up a scheme
+showing how they may all sit down at three tables every evening? Call
+the twelve players by the first twelve letters of the alphabet and try
+to group them.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_266_A_TENNIS_TOURNAMENT" id="X_266_A_TENNIS_TOURNAMENT"></a><a href="#X_266_A_TENNIS_TOURNAMENTa"><b>266.&mdash;A TENNIS TOURNAMENT.</b></a></p>
+
+<p>Four married couples played a "mixed double" tennis tournament, a man
+and a lady always playing against a man and a lady. But no person ever
+played with or against any other person more than once. Can you show
+how they all could have played together in the two courts on three
+successive days? This is a little puzzle of a quite practical kind,
+and it is just perplexing enough to be interesting.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_267_THE_WRONG_HATS" id="X_267_THE_WRONG_HATS"></a><a href="#X_267_THE_WRONG_HATSa"><b>267.&mdash;THE WRONG HATS.</b></a></p>
+
+<p>"One of the most perplexing things I have come across lately," said
+Mr. Wilson, "is this. Eight men had been dining not wisely but too
+well at a certain London restaurant. They were the last to leave, but
+not one man was in a condition to identify his own hat. Now,
+considering that they took their hats at random, what are the chances
+that every man took a hat that did not belong to him?"</p>
+
+<p>"The first thing," said Mr. Waterson, "is to see in how many different
+ways the eight hats could be taken."</p>
+
+<p>"That is quite easy," Mr. Stubbs explained. "Multiply together the
+numbers, 1, 2, 3, 4, 5, 6, 7, and 8. Let me see&mdash;half a minute&mdash;yes;
+there are 40,320 different ways."</p>
+
+<p>"Now all you've got to do is to see in how many of these cases no man
+has his own hat," said Mr. Waterson.</p>
+
+<p>"Thank you, I'm not taking any," said Mr. Packhurst. "I don't envy the
+man who attempts the task of writing out all those forty-thousand-odd
+cases and then picking out the ones he wants."</p>
+
+<p>They all agreed that life is not long enough for that sort of
+amusement; and as nobody saw any other way of getting at the answer,
+the matter was postponed indefinitely. Can you solve the puzzle?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_268_THE_PEAL_OF_BELLS" id="X_268_THE_PEAL_OF_BELLS"></a><a href="#X_268_THE_PEAL_OF_BELLSa"><b>268.&mdash;THE PEAL OF BELLS.</b></a></p>
+
+<p>A correspondent, who is apparently much interested in campanology,
+asks me how he is to construct what he calls a "true and correct" peal
+for four bells. He says that every possible permutation of the four
+bells must be rung once, and once only. He adds that no bell must move
+more than one place at a time, that no bell must make more than two
+successive strokes in either the first or the last place, and that the
+last change must be able to pass into the first. These fantastic
+conditions will be found to be observed in the little peal for three
+bells, as follows:&mdash;</p>
+
+
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>1</td><td align='right'>2</td><td align='right'>3</td></tr>
+<tr><td align='right'>2</td><td align='right'>1</td><td align='right'>3</td></tr>
+<tr><td align='right'>2</td><td align='right'>3</td><td align='right'>1</td></tr>
+<tr><td align='right'>3</td><td align='right'>2</td><td align='right'>1</td></tr>
+<tr><td align='right'>3</td><td align='right'>1</td><td align='right'>2</td></tr>
+<tr><td align='right'>1</td><td align='right'>3</td><td align='right'>2</td></tr>
+</table></div>
+
+
+<p>How are we to give him a correct solution for his four bells?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_269_THREE_MEN_IN_A_BOAT" id="X_269_THREE_MEN_IN_A_BOAT"></a><a href="#X_269_THREE_MEN_IN_A_BOATa"><b>269.&mdash;THREE MEN IN A BOAT.</b></a></p>
+
+<p>A certain generous London manufacturer gives his workmen every year a
+week's holiday at the seaside at his own expense. One year fifteen of
+his men paid a visit to Herne Bay. On the morning of their departure
+from London they were addressed by their employer, who expressed the
+hope that they would have a very pleasant time.</p>
+
+<p>"I have been given to understand," he added, "that some of you fellows
+are very fond of rowing, so I propose on this occasion to provide you
+with this recreation, and at the same time give you an amusing little
+puzzle to solve. During the seven days that you are at Herne Bay every
+one of you will go out every day at the same time for a row, but there
+must always be three men in a boat and no more. No two men may ever go
+out in a boat together more than once, and no man is allowed to go out
+twice in the same boat. If you can manage to do this, and use as few
+different boats as possible, you may charge the firm with the
+expense."</p>
+
+<p>One of the men tells me that the experience he has gained in such
+matters soon enabled him to work out the answer to the entire
+satisfaction of themselves and their employer. But the amusing part of
+the thing is that they never really solved the little mystery. I find
+their method to have been quite incorrect, and I think it will amuse
+my readers to discover how the men should have been placed in the
+boats. As their names happen to have been Andrews, Baker, Carter,
+Danby, Edwards, Frith, Gay, Hart, Isaacs, Jackson, Kent, Lang, Mason,
+Napper, and Onslow, we can call them by their initials and write out
+the five groups for each of the seven days in the following simple
+way:</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'></td><td align='center'>1</td><td align='center'>2</td><td align='center'>3</td><td align='center'>4</td><td align='center'>5</td></tr>
+<tr><td align='center'>First Day:</td><td align='center'>(ABC)</td><td align='center'>(DEF)</td><td align='center'>(GHI)</td><td align='center'>(JKL)</td><td align='center'>(MNO).</td></tr>
+</table></div>
+
+<p>The men within each pair of brackets are here seen to be in the same
+boat, and therefore A can never go out with B or with C again, and C
+can never go out again with B. The same applies to the other four
+boats. The figures show the number on the boat, so that A, B, or C,
+for example, can never go out in boat No. 1 again.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_270_THE_GLASS_BALLS" id="X_270_THE_GLASS_BALLS"></a><a href="#X_270_THE_GLASS_BALLSa"><b>270.&mdash;THE GLASS BALLS.</b></a></p>
+
+<p>A number of clever marksmen were staying at a country house, and the
+host, to provide a little amusement, suspended strings of glass balls,
+as shown in the illustration, to be fired <span class='pagenum'>Pg 79<a name="Page_79" id="Page_79"></a></span>at. After they had all put
+their skill to a sufficient test, somebody asked the following
+question: "What is the total number of different ways in which these
+sixteen balls may be broken, if we must always break the lowest ball
+that remains on any string?" Thus, one way would be to break all the
+four balls on each string in succession, taking the strings from left
+to right. Another would be to break all the fourth balls on the four
+strings first, then break the three remaining on the first string,
+then take the balls on the three other strings alternately from right
+to left, and so on. There is such a vast number of different ways
+(since every little variation of order makes a different way) that one
+is apt to be at first impressed by the great difficulty of the
+problem. Yet it is really quite simple when once you have hit on the
+proper method of attacking it. How many different ways are there?</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q270.png" width="600" height="488" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_271_FIFTEEN_LETTER_PUZZLE" id="X_271_FIFTEEN_LETTER_PUZZLE"></a><a href="#X_271_FIFTEEN_LETTER_PUZZLEa"><b>271.&mdash;FIFTEEN LETTER PUZZLE.</b></a></p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>ALE</td><td align='center'>FOE</td><td align='center'>HOD</td><td align='center'>BGN</td></tr>
+<tr><td align='center'>CAB</td><td align='center'>HEN</td><td align='center'>JOG</td><td align='center'>KFM</td></tr>
+<tr><td align='center'>HAG</td><td align='center'>GEM</td><td align='center'>MOB</td><td align='center'>BFH</td></tr>
+<tr><td align='center'>FAN</td><td align='center'>KIN</td><td align='center'>JEK</td><td align='center'>DFL</td></tr>
+<tr><td align='center'>JAM</td><td align='center'>HIM</td><td align='center'>GCL</td><td align='center'>LJH</td></tr>
+<tr><td align='center'>AID</td><td align='center'>JIB</td><td align='center'>FCJ</td><td align='center'>NJD</td></tr>
+<tr><td align='center'>OAK</td><td align='center'>FIG</td><td align='center'>HCK</td><td align='center'>MLN</td></tr>
+<tr><td align='center'>BED</td><td align='center'>OIL</td><td align='center'>MCD</td><td align='center'>BLK</td></tr>
+<tr><td align='center'>ICE</td><td align='center'>CON</td><td align='center'>DGK</td></tr>
+</table></div>
+
+<p>The above is the solution of a puzzle I gave in <i>Tit-bits</i> in the
+summer of 1896. It was required to take the letters, A, B, C, D, E, F,
+G, H, I, J, K, L, M, N, and O, and with them form thirty-five groups
+of three letters so that the combinations should include the greatest
+number possible of common English words. No two letters may appear
+together in a group more than once. Thus, A and L having been together
+in ALE, must never be found together again; nor may A appear again in
+a group with E, nor L with E. These conditions will be found complied
+with in the above solution, and the number of words formed is
+twenty-one. Many persons have since tried hard to beat this number,
+but so far have not succeeded.</p>
+
+<p>More than thirty-five combinations of the fifteen letters cannot be
+formed within the conditions. Theoretically, there cannot possibly be
+more than twenty-three words formed, because only this number of
+combinations is possible with a vowel or vowels in each. And as no
+English word can be formed from three of the given vowels (A, E, I,
+and O), we must reduce the number of possible words to twenty-two.
+This is correct theoretically, but practically that twenty-second word
+cannot be got in. If JEK, shown above, were a word it would be all
+right; but it is not, and no amount of juggling with the other letters
+has resulted in a better answer than the one shown. I should, say that
+proper nouns and abbreviations, such as Joe, Jim, Alf, Hal, Flo, Ike,
+etc., are disallowed.</p>
+
+<p>Now, the present puzzle is a variation of the above. It is simply
+this: Instead of using the fifteen letters given, the reader is
+allowed to select any fifteen different letters of the alphabet that
+he may prefer. Then construct thirty-five <span class='pagenum'>Pg 80<a name="Page_80" id="Page_80"></a></span>groups in accordance with
+the conditions, and show as many good English words as possible.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_272_THE_NINE_SCHOOLBOYS" id="X_272_THE_NINE_SCHOOLBOYS"></a><a href="#X_272_THE_NINE_SCHOOLBOYSa"><b>272.&mdash;THE NINE SCHOOLBOYS.</b></a></p>
+
+<p>This is a new and interesting companion puzzle to the "Fifteen
+Schoolgirls" (see solution of No. 269), and even in the simplest
+possible form in which I present it there are unquestionable
+difficulties. Nine schoolboys walk out in triplets on the six week
+days so that no boy ever walks <i>side by side</i> with any other boy more
+than once. How would you arrange them?</p>
+
+<p>If we represent them by the first nine letters of the alphabet, they
+might be grouped on the first day as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>A</td><td align='center'>B</td><td align='center'>C</td></tr>
+<tr><td align='center'>D</td><td align='center'>E</td><td align='center'>F</td></tr>
+<tr><td align='center'>G</td><td align='center'>H</td><td align='center'>I</td></tr>
+</table></div>
+
+<p>Then A can never walk again side by side with B, or B with C, or D
+with E, and so on. But A can, of course, walk side by side with C. It
+is here not a question of being together in the same triplet, but of
+walking side by side in a triplet. Under these conditions they can
+walk out on six days; under the "Schoolgirls" conditions they can only
+walk on four days.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_273_THE_ROUND_TABLE" id="X_273_THE_ROUND_TABLE"></a><a href="#X_273_THE_ROUND_TABLEa"><b>273.&mdash;THE ROUND TABLE.</b></a></p>
+
+<p>Seat the same <i>n</i> persons at a round table on</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' class='bb'>(n&nbsp;-&nbsp;1)(n&nbsp;-&nbsp;2)</td></tr>
+<tr><td align='center'>2</td></tr>
+</table></div>
+
+<p>occasions so that no person shall ever have the same two neighbours
+twice. This is, of course, equivalent to saying that every person must
+sit once, and once only, between every possible pair.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_274_THE_MOUSE-TRAP_PUZZLE" id="X_274_THE_MOUSE-TRAP_PUZZLE"></a><a href="#X_274_THE_MOUSE-TRAP_PUZZLEa"><b>274.&mdash;THE MOUSE-TRAP PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q274.png" width="400" height="397" alt="" title="" />
+</div>
+
+<p>This is a modern version, with a difference, of an old puzzle of the
+same name. Number twenty-one cards, 1, 2, 3, etc., up to 21, and place
+them in a circle in the particular order shown in the illustration.
+These cards represent mice. You start from any card, calling that card
+"one," and count, "one, two, three," etc., in a clockwise direction,
+and when your count agrees with the number on the card, you have made
+a "catch," and you remove the card. Then start at the next card,
+calling that "one," and try again to make another "catch." And so on.
+Supposing you start at 18, calling that card "one," your first "catch"
+will be 19. Remove 19 and your next "catch" is 10. Remove 10 and your
+next "catch" is 1. Remove the 1, and if you count up to 21 (you must
+never go beyond), you cannot make another "catch." Now, the ideal is
+to "catch" all the twenty-one mice, but this is not here possible, and
+if it were it would merely require twenty-one different trials, at the
+most, to succeed. But the reader may make any two cards change places
+before he begins. Thus, you can change the 6 with the 2, or the 7 with
+the 11, or any other pair. This can be done in several ways so as to
+enable you to "catch" all the twenty-one mice, if you then start at
+the right place. You may never pass over a "catch"; you must always
+remove the card and start afresh.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_275_THE_SIXTEEN_SHEEP" id="X_275_THE_SIXTEEN_SHEEP"></a><a href="#X_275_THE_SIXTEEN_SHEEPa"><b>275.&mdash;THE SIXTEEN SHEEP.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q275.png" width="400" height="402" alt="" title="" />
+</div>
+
+<p>Here is a new puzzle with matches and counters or coins. In the
+illustration the matches represent hurdles and the counters sheep. The
+sixteen hurdles on the outside, and the sheep, must be regarded as
+immovable; the puzzle has to do entirely with the nine hurdles on the
+inside. It will be seen that at present these nine hurdles enclose
+four groups of 8, 3, 3, and 2 sheep. The farmer requires to readjust
+some of the hurdles so as to enclose 6, 6, and 4 sheep. Can you do it
+by only replacing two hurdles? When you have succeeded, then try to
+do it by replacing three hurdles; then four, five, six, and seven in
+succession. Of course, the hurdles must be legitimately laid on the
+dotted lines, and no such tricks are allowed as leaving unconnected
+ends of hurdles, or two hurdles placed side by side, or merely making
+hurdles change places. In fact, the conditions are so simple that any
+farm labourer will understand it directly.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_276_THE_EIGHT_VILLAS" id="X_276_THE_EIGHT_VILLAS"></a><a href="#X_276_THE_EIGHT_VILLASa"><b>276.&mdash;THE EIGHT VILLAS.</b></a></p>
+
+<p>In one of the outlying suburbs of London a man had a square plot of
+ground on which he <span class='pagenum'>Pg 81<a name="Page_81" id="Page_81"></a></span>decided to build eight villas, as shown in the
+illustration, with a common recreation ground in the middle. After the
+houses were completed, and all or some of them let, he discovered that
+the number of occupants in the three houses forming a side of the
+square was in every case nine. He did not state how the occupants were
+distributed, but I have shown by the numbers on the sides of the
+houses one way in which it might have happened. The puzzle is to
+discover the total number of ways in which all or any of the houses
+might be occupied, so that there should be nine persons on each side.
+In order that there may be no misunderstanding, I will explain that
+although B is what we call a reflection of A, these would count as two
+different arrangements, while C, if it is turned round, will give four
+arrangements; and if turned round in front of a mirror, four other
+arrangements. All eight must be counted.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q276.png" width="400" height="533" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_277_COUNTER_CROSSES" id="X_277_COUNTER_CROSSES"></a><a href="#X_277_COUNTER_CROSSESa"><b>277.&mdash;COUNTER CROSSES.</b></a></p>
+
+<p>All that we need for this puzzle is nine counters, numbered 1, 2, 3,
+4, 5, 6, 7, 8, and 9. It will be seen that in the illustration A these
+are arranged so as to form a Greek cross, while in the case of B they
+form a Latin cross. In both cases the reader will find that the sum of
+the numbers in the upright of the cross is the same as the sum of the
+numbers in the horizontal arm. It is quite easy to hit on such an
+arrangement by trial, but the problem is to discover in exactly how
+many different ways it may be done in each case. Remember that
+reversals and reflections do not count as different. That is to say,
+if you turn this page round you get four arrangements of the Greek
+cross, and if you turn it round again in front of a mirror you will
+get four more. But these eight are all regarded as one and the same.
+Now, how many different ways are there in each case?</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q277.png" width="600" height="295" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_278_A_DORMITORY_PUZZLE" id="X_278_A_DORMITORY_PUZZLE"></a><a href="#X_278_A_DORMITORY_PUZZLEa"><b>278.&mdash;A DORMITORY PUZZLE.</b></a></p>
+
+<p>In a certain convent there were eight large dormitories on one floor,
+approached by a spiral staircase in the centre, as shown in our plan.
+On an inspection one Monday by the abbess it was found that the south
+aspect was so much preferred that six times as many nuns slept on the
+south side as on each of the other three sides. She objected to this
+overcrowding, and ordered that it should be reduced. On Tuesday she
+found that five times as many slept on the south side as on each of
+the other sides. Again she complained. On Wednesday she found four
+times as many on the south side, on Thursday three times as many, and
+on Friday twice as many. Urging the nuns to further efforts, she was
+pleased to find on Saturday that an equal number slept on each of the
+four <span class='pagenum'>Pg 82<a name="Page_82" id="Page_82"></a></span>sides of the house. What is the smallest number of nuns there
+could have been, and how might they have arranged themselves on each
+of the six nights? No room may ever be unoccupied.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q278.png" width="400" height="393" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_279_THE_BARRELS_OF_BALSAM" id="X_279_THE_BARRELS_OF_BALSAM"></a><a href="#X_279_THE_BARRELS_OF_BALSAMa"><b>279.&mdash;THE BARRELS OF BALSAM.</b></a></p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q279.png" width="600" height="446" alt="" title="" />
+</div>
+
+<p>A merchant of Bagdad had ten barrels of precious balsam for sale.
+They were numbered, and were arranged in two rows, one on top of the
+other, as shown in the picture. The smaller the number on the barrel,
+the greater was its value. So that the best quality was numbered "1" and
+the worst numbered "10," and all the other numbers of graduating values.
+Now, the rule of Ahmed Assan, the merchant, was that he never put a
+barrel either beneath or to the right of one of less value. The
+arrangement shown is, of course, the simplest way of complying with this
+condition. But there are many other ways&mdash;such, for example, as
+this:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>1</td><td align='right'>2</td><td align='right'>5</td><td align='right'>7</td><td align='right'>8</td></tr>
+<tr><td align='right'>3</td><td align='right'>4</td><td align='right'>6</td><td align='right'>9</td><td align='right'>10</td></tr>
+</table></div>
+
+<p>Here, again, no barrel has a smaller number than itself on its right
+or beneath it. The puzzle is to discover in how many different ways
+the merchant of Bagdad might have arranged his barrels in the two rows
+without breaking his rule. Can you count the number of ways?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_280_BUILDING_THE_TETRAHEDRON" id="X_280_BUILDING_THE_TETRAHEDRON"></a><a href="#X_280_BUILDING_THE_TETRAHEDRONa"><b>280.&mdash;BUILDING THE TETRAHEDRON.</b></a></p>
+
+<p>I possess a tetrahedron, or triangular pyramid, formed of six sticks
+glued together, as shown in the illustration. Can you count correctly
+the number of different ways in which these six sticks might have been
+stuck together so as to form the pyramid?</p>
+
+<p>Some friends worked at it together one evening, each person providing
+himself with six lucifer matches to aid his thoughts; but it was found
+that no two results were the same. You see, if we remove one of the
+sticks and turn it round the other way, that will be a different
+pyramid. If we make two of the sticks change <span class='pagenum'>Pg 83<a name="Page_83" id="Page_83"></a></span>places the result will
+again be different. But remember that every pyramid may be made to
+stand on either of its four sides without being a different one. How
+many ways are there altogether?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q280.png" width="400" height="347" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_281_PAINTING_A_PYRAMID" id="X_281_PAINTING_A_PYRAMID"></a><a href="#X_281_PAINTING_A_PYRAMIDa"><b>281.&mdash;PAINTING A PYRAMID.</b></a></p>
+
+<p>This puzzle concerns the painting of the four sides of a tetrahedron,
+or triangular pyramid. If you cut out a piece of cardboard of the
+triangular shape shown in Fig. 1, and then cut half through along the
+dotted lines, it will fold up and form a perfect triangular pyramid.
+And I would first remind my readers that the primary colours of the
+solar spectrum are seven&mdash;violet, indigo, blue, green, yellow, orange,
+and red. When I was a child I was taught to remember these by the
+ungainly word formed by the initials of the colours, "Vibgyor."</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q281.png" width="600" height="246" alt="" title="" />
+</div>
+
+<p>In how many different ways may the triangular pyramid be coloured,
+using in every case one, two, three, or four colours of the solar
+spectrum? Of course a side can only receive a single colour, and no
+side can be left uncoloured. But there is one point that I must make
+quite clear. The four sides are not to be regarded as individually
+distinct. That is to say, if you paint your pyramid as shown in Fig. 2
+(where the bottom side is green and the other side that is out of view
+is yellow), and then paint another in the order shown in Fig. 3, these
+are really both the same and count as one way. For if you tilt over
+No. 2 to the right it will so fall as to represent No. 3. The
+avoidance of repetitions of this kind is the real puzzle of the thing.
+If a coloured pyramid cannot be placed so that it exactly resembles in
+its colours and their relative order another pyramid, then they are
+different. Remember that one way would be to colour all the four sides
+red, another to colour two sides green, and the remaining sides yellow
+and blue; and so on.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_282_THE_ANTIQUARYS_CHAIN" id="X_282_THE_ANTIQUARYS_CHAIN"></a><a href="#X_282_THE_ANTIQUARYS_CHAINa"><b>282.&mdash;THE ANTIQUARY'S CHAIN.</b></a></p>
+
+<p>An antiquary possessed a number of curious old links, which he took to
+a blacksmith, and told him to join together to form one straight piece
+of chain, with the sole condition that the two circular links were not
+to be together. The following illustration shows the appearance of the
+chain and the form of each link. Now, supposing the owner should
+separate the links again, and then take them to another smith and
+repeat his former instructions exactly, what are the chances against
+the links being put together exactly as they were by the first man?
+Remember that every successive link can be joined on to another in one
+of two ways, just as you can put a ring on your finger in two ways, or
+link your forefingers and thumbs in two ways.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q282.png" width="400" height="191" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_283_THE_FIFTEEN_DOMINOES" id="X_283_THE_FIFTEEN_DOMINOES"></a><a href="#X_283_THE_FIFTEEN_DOMINOESa"><b>283.&mdash;THE FIFTEEN DOMINOES.</b></a></p>
+
+<p>In this case we do not use the complete set of twenty-eight dominoes
+to be found in the ordinary box. We dispense with all those dominoes
+<span class='pagenum'>Pg 84<a name="Page_84" id="Page_84"></a></span>that have a five or a six on them and limit ourselves to the fifteen
+that remain, where the double-four is the highest.</p>
+
+<p>In how many different ways may the fifteen dominoes be arranged in a
+straight line in accordance with the simple rule of the game that a
+number must always be placed against a similar number&mdash;that is, a four
+against a four, a blank against a blank, and so on? Left to right and
+right to left of the same arrangement are to be counted as two
+different ways.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_284_THE_CROSS_TARGET" id="X_284_THE_CROSS_TARGET"></a><a href="#X_284_THE_CROSS_TARGETa"><b>284.&mdash;THE CROSS TARGET.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q284.png" width="400" height="314" alt="" title="" />
+</div>
+
+<p>In the illustration we have a somewhat curious target designed by an
+eccentric sharpshooter. His idea was that in order to score you must
+hit four circles in as many shots so that those four shots shall form
+a square. It will be seen by the results recorded on the target that
+two attempts have been successful. The first man hit the four circles
+at the top of the cross, and thus formed his square. The second man
+intended to hit the four in the bottom arm, but his second shot, on
+the left, went too high. This compelled him to complete his four in a
+different way than he intended. It will thus be seen that though it is
+immaterial which circle you hit at the first shot, the second shot may
+commit you to a definite procedure if you are to get your square. Now,
+the puzzle is to say in just how many different ways it is possible to
+form a square on the target with four shots.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_285_THE_FOUR_POSTAGE_STAMPS" id="X_285_THE_FOUR_POSTAGE_STAMPS"></a><a href="#X_285_THE_FOUR_POSTAGE_STAMPSa"><b>285.&mdash;THE FOUR POSTAGE STAMPS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q285.png" width="400" height="300" alt="" title="" />
+</div>
+
+<p>"It is as easy as counting," is an expression one sometimes hears. But
+mere counting may be puzzling at times. Take the following simple
+example. Suppose you have just bought twelve postage stamps, in this
+form&mdash;three by four&mdash;and a friend asks you to oblige him with four
+stamps, all joined together&mdash;no stamp hanging on by a mere corner. In
+how many different ways is it possible for you to tear off those four
+stamps? You see, you can give him 1, 2, 3, 4, or 2, 3, 6, 7, or 1, 2,
+3, 6, or 1, 2, 3, 7, or 2, 3, 4, 8, and so on. Can you count the
+number of different ways in which those four stamps might be
+delivered? There are not many more than fifty ways, so it is not a big
+count. Can you get the exact number?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_286_PAINTING_THE_DIE" id="X_286_PAINTING_THE_DIE"></a><a href="#X_286_PAINTING_THE_DIEa"><b>286.&mdash;PAINTING THE DIE.</b></a></p>
+
+<p>In how many different ways may the numbers on a single die be marked,
+with the only condition that the 1 and 6, the 2 and 5, and the 3 and 4
+must be on opposite sides? It is a simple enough question, and yet it
+will puzzle a good many people.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_287_AN_ACROSTIC_PUZZLE" id="X_287_AN_ACROSTIC_PUZZLE"></a><a href="#X_287_AN_ACROSTIC_PUZZLEa"><b>287.&mdash;AN ACROSTIC PUZZLE.</b></a></p>
+
+<p>In the making or solving of double acrostics, has it ever occurred to
+you to consider the variety and limitation of the pair of initial and
+final letters available for cross words? You may have to find a word
+beginning with A and ending with B, or A and C, or A and D, and so on.
+Some combinations are obviously impossible&mdash;such, for example, as
+those with Q at the end. But let us assume that a good English word
+can be found for every case. Then how many possible pairs of letters
+are available?</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="CHESSBOARD_PROBLEMS" id="CHESSBOARD_PROBLEMS"></a><a href="#CONTENTS">CHESSBOARD PROBLEMS.</a></h2>
+
+<p class='center'>
+"You and I will goe to the chesse."<br />
+<span style="margin-left: 8em;">GREENE'S <i>Groatsworth of Wit.</i><br /></span>
+</p>
+
+<p>During a heavy gale a chimney-pot was hurled through the air, and
+crashed upon the pavement just in front of a pedestrian. He quite
+calmly said, "I have no use for it: I do not smoke." Some readers,
+when they happen to see a puzzle represented on a chessboard with
+chess pieces, are apt to make the equally inconsequent remark, "I have
+no use for it: I do not play chess." This is largely a result of the
+common, but erroneous, notion that the ordinary chess puzzle with
+which we are familiar in the press (dignified, for some reason, with
+the name "problem") has a vital connection with the game of chess
+itself. But there is no condition in the game that you shall checkmate
+your opponent in two moves, in three moves, or in four moves, while
+the majority of the positions given in these puzzles are such that one
+player would have so great a superiority in pieces that <span class='pagenum'>Pg 85<a name="Page_85" id="Page_85"></a></span>the other
+would have resigned before the situations were reached. And the
+solving of them helps you but little, and that quite indirectly, in
+playing the game, it being well known that, as a rule, the best "chess
+problemists" are indifferent players, and <i>vice versa</i>. Occasionally a
+man will be found strong on both subjects, but he is the exception to
+the rule.</p>
+
+<p>Yet the simple chequered board and the characteristic moves of the
+pieces lend themselves in a very remarkable manner to the devising of
+the most entertaining puzzles. There is room for such infinite variety
+that the true puzzle lover cannot afford to neglect them. It was with
+a view to securing the interest of readers who are frightened off by
+the mere presentation of a chessboard that so many puzzles of this
+class were originally published by me in various fanciful dresses.
+Some of these posers I still retain in their disguised form; others I
+have translated into terms of the chessboard. In the majority of cases
+the reader will not need any knowledge whatever of chess, but I have
+thought it best to assume throughout that he is acquainted with the
+terminology, the moves, and the notation of the game.</p>
+
+<p>I first deal with a few questions affecting the chessboard itself;
+then with certain statical puzzles relating to the Rook, the Bishop,
+the Queen, and the Knight in turn; then dynamical puzzles with the
+pieces in the same order; and, finally, with some miscellaneous
+puzzles on the chessboard. It is hoped that the formul&aelig; and tables
+given at the end of the statical puzzles will be of interest, as they
+are, for the most part, published for the first time.</p>
+
+
+
+<h2><a name="THE_CHESSBOARD" id="THE_CHESSBOARD"></a><a href="#CONTENTS">THE CHESSBOARD.</a></h2>
+
+<p class='center'>
+"Good company's a chessboard."<br />
+<span style="margin-left: 8em;">BYRON'S <i>Don Juan</i>, xiii. 89.</span><br />
+</p>
+
+<p>A chessboard is essentially a square plane divided into sixty-four
+smaller squares by straight lines at right angles. Originally it was
+not chequered (that is, made with its rows and columns alternately
+black and white, or of any other two colours), and this improvement
+was introduced merely to help the eye in actual play. The utility of
+the chequers is unquestionable. For example, it facilitates the
+operation of the bishops, enabling us to see at the merest glance that
+our king or pawns on black squares are not open to attack from an
+opponent's bishop running on the white diagonals. Yet the chequering
+of the board is not essential to the game of chess. Also, when we are
+propounding puzzles on the chessboard, it is often well to remember
+that additional interest may result from "generalizing" for boards
+containing any number of squares, or from limiting ourselves to some
+particular chequered arrangement, not necessarily a square. We will
+give a few puzzles dealing with chequered boards in this general way.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_288_CHEQUERED_BOARD_DIVISIONS" id="X_288_CHEQUERED_BOARD_DIVISIONS"></a><a href="#X_288_CHEQUERED_BOARD_DIVISIONSa"><b>288.&mdash;CHEQUERED BOARD DIVISIONS.</b></a></p>
+
+<p>I recently asked myself the question: In how many different ways may a
+chessboard be divided into two parts of the same size and shape by
+cuts along the lines dividing the squares? The problem soon proved to
+be both fascinating and bristling with difficulties. I present it in a
+simplified form, taking a board of smaller dimensions.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q288.png" width="400" height="679" alt="" title="" />
+</div>
+
+<p>It is obvious that a board of four squares can only be so divided in
+one way&mdash;by a straight cut down the centre&mdash;because we shall not count
+reversals and reflections as different. In the case of a board of
+sixteen squares&mdash;four by four&mdash;there are just six different ways. I
+have given all these in the diagram, and the reader will not find any
+others. Now, take the larger board of thirty-six squares, and try to
+discover in how many ways it may be cut into two parts of the same
+size and shape.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_289_LIONS_AND_CROWNS" id="X_289_LIONS_AND_CROWNS"></a><a href="#X_289_LIONS_AND_CROWNSa"><b>289.&mdash;LIONS AND CROWNS.</b></a></p>
+
+<p>The young lady in the illustration is confronted with a little
+cutting-out difficulty in which the reader may be glad to assist her.
+She wishes, for some reason that she has not communi<span class='pagenum'>Pg 86<a name="Page_86" id="Page_86"></a></span>cated to me, to
+cut that square piece of valuable material into four parts, all of
+exactly the same size and shape, but it is important that every piece
+shall contain a lion and a crown. As she insists that the cuts can
+only be made along the lines dividing the squares, she is considerably
+perplexed to find out how it is to be done. Can you show her the way?
+There is only one possible method of cutting the stuff.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q289.png" width="400" height="707" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_290_BOARDS_WITH_AN_ODD_NUMBER_OF_SQUARES" id="X_290_BOARDS_WITH_AN_ODD_NUMBER_OF_SQUARES"></a><a href="#X_290_BOARDS_WITH_AN_ODD_NUMBER_OF_SQUARESa"><b>290.&mdash;BOARDS WITH AN ODD NUMBER OF SQUARES.</b></a></p>
+
+<p>We will here consider the question of those boards that contain an odd
+number of squares. We will suppose that the central square is first
+cut out, so as to leave an even number of squares for division. Now,
+it is obvious that a square three by three can only be divided in one
+way, as shown in Fig. 1. It will be seen that the pieces A and B are
+of the same size and shape, and that any other way of cutting would
+only produce the same shaped pieces, so remember that these variations
+are not counted as different ways. The puzzle I propose is to cut the
+board five by five (Fig. 2) into two pieces of the same size and shape
+in as many different ways as possible. I have shown in the
+illustration one way of doing it. How many different ways are there
+altogether? A piece which when turned over resembles another piece is
+not considered to be of a different shape.</p>
+
+<div class="figcenter" style="width: 130px;">
+<img src="images/q290fig1.png" width="130" height="134" alt="" title="" />
+</div>
+
+<div class="figcenter" style="width: 250px;">
+<img src="images/q290fig2.png" width="250" height="250" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_291_THE_GRAND_LAMAS_PROBLEM" id="X_291_THE_GRAND_LAMAS_PROBLEM"></a><a href="#X_291_THE_GRAND_LAMAS_PROBLEMa"><b>291.&mdash;THE GRAND LAMA'S PROBLEM.</b></a></p>
+
+<p>Once upon a time there was a Grand Lama who had a chessboard made of
+pure gold, magnificently engraved, and, of course, of great value.
+Every year a tournament was held at Lhassa among the priests, and
+whenever any one beat the Grand Lama it was considered a great honour,
+and his name was inscribed on the back of the board, and a costly
+jewel set in the particular square on which the checkmate had been
+given. After this sovereign pontiff had been defeated on four
+occasions he died&mdash;possibly of chagrin.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q291.png" width="400" height="394" alt="" title="" />
+</div>
+
+<p>Now the new Grand Lama was an inferior chess-player, and preferred
+other forms of innocent amusement, such as cutting off people's heads.
+So he discouraged chess as a degrading game, that did not improve
+either the mind or the morals, and abolished the tournament summarily.
+Then he sent for the four priests who had had the effrontery to play
+better than a Grand Lama, and addressed them as follows: <span class='pagenum'>Pg 87<a name="Page_87" id="Page_87"></a></span>"Miserable
+and heathenish men, calling yourselves priests! Know ye not that to
+lay claim to a capacity to do anything better than my predecessor is a
+capital offence? Take that chessboard and, before day dawns upon the
+torture chamber, cut it into four equal parts of the same shape, each
+containing sixteen perfect squares, with one of the gems in each part!
+If in this you fail, then shall other sports be devised for your
+special delectation. Go!" The four priests succeeded in their
+apparently hopeless task. Can you show how the board may be divided
+into four equal parts, each of exactly the same shape, by cuts along
+the lines dividing the squares, each part to contain one of the gems?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_292_THE_ABBOTS_WINDOW" id="X_292_THE_ABBOTS_WINDOW"></a><a href="#X_292_THE_ABBOTS_WINDOWa"><b>292.&mdash;THE ABBOT'S WINDOW.</b></a></p>
+
+<p>Once upon a time the Lord Abbot of St. Edmondsbury, in consequence of
+"devotions too strong for his head," fell sick and was unable to leave
+his bed. As he lay awake, tossing his head restlessly from side to
+side, the attentive monks noticed that something was disturbing his
+mind; but nobody dared ask what it might be, for the abbot was of a
+stern disposition, and never would brook inquisitiveness. Suddenly he
+called for Father John, and that venerable monk was soon at the
+bedside.</p>
+
+<p>"Father John," said the Abbot, "dost thou know that I came into this
+wicked world on a Christmas Even?"</p>
+
+<p>The monk nodded assent.</p>
+
+<p>"And have I not often told thee that, having been born on Christmas
+Even, I have no love for the things that are odd? Look there!"</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q292.png" width="400" height="382" alt="" title="" />
+</div>
+
+<p>The Abbot pointed to the large dormitory window, of which I give a
+sketch. The monk looked, and was perplexed.</p>
+
+<p>"Dost thou not see that the sixty-four lights add up an even number
+vertically and horizontally, but that all the <i>diagonal</i> lines, except
+fourteen are of a number that is odd? Why is this?"</p>
+
+<p>"Of a truth, my Lord Abbot, it is of the very nature of things, and
+cannot be changed."</p>
+
+<p>"Nay, but it <i>shall</i> be changed. I command thee that certain of the
+lights be closed this day, so that every line shall have an even
+number of lights. See thou that this be done without delay, lest the
+cellars be locked up for a month and other grievous troubles befall
+thee."</p>
+
+<p>Father John was at his wits' end, but after consultation with one who
+was learned in strange mysteries, a way was found to satisfy the whim
+of the Lord Abbot. Which lights were blocked up, so that those which
+remained added up an even number in every line horizontally,
+vertically, and diagonally, while the least possible obstruction of
+light was caused?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_293_THE_CHINESE_CHESSBOARD" id="X_293_THE_CHINESE_CHESSBOARD"></a><a href="#X_293_THE_CHINESE_CHESSBOARDa"><b>293.&mdash;THE CHINESE CHESSBOARD.</b></a></p>
+
+<p>Into how large a number of different pieces may the chessboard be cut
+(by cuts along the lines only), no two pieces being exactly alike?
+Remember that the arrangement of black and white constitutes a
+difference. Thus, a single black square will be different from a
+single white square, a row of three containing two white squares will
+differ from a row of three containing two black, and so on. If two
+pieces cannot be placed on the table so as to be exactly alike, they
+count as different. And as the back of the board is plain, the pieces
+cannot be turned over.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_294_THE_CHESSBOARD_SENTENCE" id="X_294_THE_CHESSBOARD_SENTENCE"></a><a href="#X_294_THE_CHESSBOARD_SENTENCEa"><b>294.&mdash;THE CHESSBOARD SENTENCE.</b></a></p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/q294.png" width="500" height="71" alt="" title="" />
+</div>
+
+<p>I once set myself the amusing task of so dissecting an ordinary
+chessboard into letters of the alphabet that they would form a
+complete sentence. It will be seen from the illustration that the
+pieces assembled give the sentence, "CUT THY LIFE," with the stops
+between. The ideal sentence would, of course, have only one full stop,
+but that I did not succeed in obtaining.</p>
+
+<p>The sentence is an appeal to the transgressor to cut himself adrift
+from the evil life he is living. Can you fit these pieces together to
+form a perfect chessboard?</p>
+
+
+
+<hr style="width: 65%;" /><p><span class='pagenum'>Pg 88<a name="Page_88" id="Page_88"></a></span></p>
+<h2><a name="STATICAL_CHESS_PUZZLES" id="STATICAL_CHESS_PUZZLES"></a><a href="#CONTENTS">STATICAL CHESS PUZZLES.</a></h2>
+
+<p class='center'>
+"They also serve who only stand and wait."<br />
+<span style="margin-left: 8em;">MILTON.</span><br />
+</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_295_THE_EIGHT_ROOKS" id="X_295_THE_EIGHT_ROOKS"></a><a href="#X_295_THE_EIGHT_ROOKSa"><b>295.&mdash;THE EIGHT ROOKS.</b></a></p>
+
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q295fig1.png" width="400" height="394" alt="" title="" />
+</div>
+
+<p>It will be seen in the first diagram that every square on the board is
+either occupied or attacked by a rook, and that every rook is
+"guarded" (if they were alternately black and white rooks we should
+say "attacked") by another rook. Placing the eight rooks on any row or
+file obviously will have the same effect. In diagram 2 every square is
+again either occupied or attacked, but in this case every rook is
+unguarded. Now, in how many different ways can you so place the eight
+rooks on the board that every square shall be occupied or attacked and
+no rook ever guarded by another? I do not wish to go into the question
+of reversals and reflections on this occasion, so that placing the
+rooks on the other diagonal will count as different, and similarly
+with other repetitions obtained by turning the board round.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q295fig2.png" width="400" height="391" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_296_THE_FOUR_LIONS" id="X_296_THE_FOUR_LIONS"></a><a href="#X_296_THE_FOUR_LIONSa"><b>296.&mdash;THE FOUR LIONS.</b></a></p>
+
+<p>The puzzle is to find in how many different ways the four lions may be
+placed so that there shall never be more than one lion in any row or
+column. Mere reversals and reflections will not count as different.
+Thus, regarding the example given, if we place the lions in the other
+diagonal, it will be considered the same arrangement. For if you hold
+the second arrangement in front of a mirror or give it a quarter turn,
+you merely get the first arrangement. It is a simple little puzzle,
+but requires a certain amount of careful consideration.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q296.png" width="400" height="400" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_297_BISHOPSmdashUNGUARDED" id="X_297_BISHOPSmdashUNGUARDED"></a><a href="#X_297_BISHOPSmdashUNGUARDEDa"><b>297.&mdash;BISHOPS&mdash;UNGUARDED.</b></a></p>
+
+<p>Place as few bishops as possible on an ordinary chessboard so that
+every square of the board shall be either occupied or attacked. It
+will be seen that the rook has more scope than the bishop: for
+wherever you place the former, it will always attack fourteen other
+squares; whereas the latter will attack seven, nine, eleven, or
+thirteen squares, according to the position of the diagonal on which
+it is placed. And it is well here to state that when we speak of
+"diagonals" in connection with the chessboard, we do not limit
+ourselves to the two long diagonals from corner to corner, but include
+all the shorter lines that are parallel to these. To prevent
+misunderstanding on future occasions, it will be well for the reader
+to note carefully this fact.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_298_BISHOPSmdashGUARDED" id="X_298_BISHOPSmdashGUARDED"></a><a href="#X_298_BISHOPSmdashGUARDEDa"><b>298.&mdash;BISHOPS&mdash;GUARDED.</b></a></p>
+
+<p>Now, how many bishops are necessary in order that every square shall
+be either occupied or attacked, and every bishop guarded by another
+bishop? And how may they be placed?</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 89<a name="Page_89" id="Page_89"></a></span><a name="X_299_BISHOPS_IN_CONVOCATION" id="X_299_BISHOPS_IN_CONVOCATION"></a><a href="#X_299_BISHOPS_IN_CONVOCATIONa"><b>299.&mdash;BISHOPS IN CONVOCATION.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q299.png" width="400" height="391" alt="" title="" />
+</div>
+
+<p>The greatest number of bishops that can be placed at the same time on
+the chessboard, without any bishop attacking another, is fourteen. I
+show, in diagram, the simplest way of doing this. In fact, on a square
+chequered board of any number of squares the greatest number of
+bishops that can be placed without attack is always two less than
+twice the number of squares on the side. It is an interesting puzzle
+to discover in just how many different ways the fourteen bishops may
+be so placed without mutual attack. I shall give an exceedingly simple
+rule for determining the number of ways for a square chequered board
+of any number of squares.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_300_THE_EIGHT_QUEENS" id="X_300_THE_EIGHT_QUEENS"></a><a href="#X_300_THE_EIGHT_QUEENSa"><b>300.&mdash;THE EIGHT QUEENS.</b></a></p>
+
+<p>The queen is by far the strongest piece on the chessboard. If you
+place her on one of the four squares in the centre of the board, she
+attacks no fewer than twenty-seven other squares; and if you try to
+hide her in a corner, she still attacks twenty-one squares. Eight
+queens may be placed on the board so that no queen attacks another,
+and it is an old puzzle (first proposed by Nauck in 1850, and it has
+quite a little literature of its own) to discover in just how many
+different ways this may be done. I show one way in the diagram, and
+there are in all twelve of these fundamentally different ways. These
+twelve produce ninety-two ways if we regard reversals and reflections
+as different. The diagram is in a way a symmetrical arrangement. If
+you turn the page upside down, it will reproduce itself exactly; but
+if you look at it with one of the other sides at the bottom, you get
+another way that is not identical. Then if you reflect these two ways
+in a mirror you get two more ways. Now, all the other eleven solutions
+are non-symmetrical, and therefore each of them may be presented in
+eight ways by these reversals and reflections. It will thus be seen
+why the twelve fundamentally different solutions produce only
+ninety-two arrangements, as I have said, and not ninety-six, as would
+happen if all twelve were non-symmetrical. It is well to have a clear
+understanding on the matter of reversals and reflections when dealing
+with puzzles on the chessboard.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q300.png" width="400" height="391" alt="" title="" />
+</div>
+
+<p>Can the reader place the eight queens on the board so that no queen
+shall attack another and so that no three queens shall be in a
+straight line in any oblique direction? Another glance at the diagram
+will show that this arrangement will not answer the conditions, for in
+the two directions indicated by the dotted lines there are three
+queens in a straight line. There is only one of the twelve fundamental
+ways that will solve the puzzle. Can you find it?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_301_THE_EIGHT_STARS" id="X_301_THE_EIGHT_STARS"></a><a href="#X_301_THE_EIGHT_STARSa"><b>301.&mdash;THE EIGHT STARS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q301.png" width="400" height="393" alt="" title="" />
+</div>
+
+<p>The puzzle in this case is to place eight stars in the diagram so that
+no star shall be in line with another star horizontally, vertically,
+or diagonally. One star is already placed, and that must not be moved,
+so there are only seven for the reader now to place. But you must not
+place a star on any one of the shaded squares. There is only one way
+of solving this little puzzle.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 90<a name="Page_90" id="Page_90"></a></span><a name="X_302_A_PROBLEM_IN_MOSAICS" id="X_302_A_PROBLEM_IN_MOSAICS"></a><a href="#X_302_A_PROBLEM_IN_MOSAICSa"><b>302.&mdash;A PROBLEM IN MOSAICS.</b></a></p>
+
+<p>The art of producing pictures or designs by means of joining together
+pieces of hard substances, either naturally or artificially coloured,
+is of very great antiquity. It was certainly known in the time of the
+Pharaohs, and we find a reference in the Book of Esther to "a pavement
+of red, and blue, and white, and black marble." Some of this ancient
+work that has come down to us, especially some of the Roman mosaics,
+would seem to show clearly, even where design is not at first evident,
+that much thought was bestowed upon apparently disorderly
+arrangements. Where, for example, the work has been produced with a
+very limited number of colours, there are evidences of great ingenuity
+in preventing the same tints coming in close proximity. Lady readers
+who are familiar with the construction of patchwork quilts will know
+how desirable it is sometimes, when they are limited in the choice of
+material, to prevent pieces of the same stuff coming too near
+together. Now, this puzzle will apply equally to patchwork quilts or
+tesselated pavements.</p>
+
+<p>It will be seen from the diagram how a square piece of flooring may be
+paved with sixty-two square tiles of the eight colours violet, red,
+yellow, green, orange, purple, white, and blue (indicated by the
+initial letters), so that no tile is in line with a similarly coloured
+tile, vertically, horizontally, or diagonally. Sixty-four such tiles
+could not possibly be placed under these conditions, but the two
+shaded squares happen to be occupied by iron ventilators.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q302.png" width="400" height="394" alt="" title="" />
+</div>
+
+<p>The puzzle is this. These two ventilators have to be removed to the
+positions indicated by the darkly bordered tiles, and two tiles placed
+in those bottom corner squares. Can you readjust the thirty-two tiles
+so that no two of the same colour shall still be in line?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_303_UNDER_THE_VEIL" id="X_303_UNDER_THE_VEIL"></a><a href="#X_303_UNDER_THE_VEILa"><b>303.&mdash;UNDER THE VEIL.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q303.png" width="400" height="394" alt="" title="" />
+</div>
+
+<p>If the reader will examine the above diagram, he will see that I have
+so placed eight V's, eight E's, eight I's, and eight L's in the
+diagram that no letter is in line with a similar one horizontally,
+vertically, or diagonally. Thus, no V is in line with another V, no E
+with another E, and so on. There are a great many different ways of
+arranging the letters under this condition. The puzzle is to find an
+arrangement that produces the greatest possible number of four-letter
+words, reading upwards and downwards, backwards and forwards, or
+diagonally. All repetitions count as different words, and the five
+variations that may be used are: VEIL, VILE, LEVI, LIVE, and EVIL.</p>
+
+<p>This will be made perfectly clear when I say that the above
+arrangement scores eight, because the top and bottom row both give
+VEIL; the second and seventh columns both give VEIL; and the two
+diagonals, starting from the L in the 5th row and E in the 8th row,
+both give LIVE and EVIL. There are therefore eight different readings
+of the words in all.</p>
+
+<p>This difficult word puzzle is given as an example of the use of
+chessboard analysis in solving such things. Only a person who is
+familiar with the "Eight Queens" problem could hope to solve it.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_304_BACHETS_SQUARE" id="X_304_BACHETS_SQUARE"></a><a href="#X_304_BACHETS_SQUAREa"><b>304.&mdash;BACHET'S SQUARE.</b></a></p>
+
+<p>One of the oldest card puzzles is by Claude Caspar Bachet de M&eacute;ziriac,
+first published, I believe, in the 1624 edition of his work. Rearrange
+the sixteen court cards (including the aces) in a square so that in no
+row of four cards, horizontal, vertical, or diagonal, shall be found
+two cards of the same suit or the same value. This in itself is easy
+enough, but a point of the puzzle is to find in how many different
+ways this may be done. The eminent French mathematician A. Labosne, in
+his modern edition of Bachet, gives the answer incorrectly. And yet
+the puzzle is really quite easy. Any arrangement produces seven more
+by turning the square round and reflecting it in a mirror. These are
+counted as different by Bachet.</p>
+
+<p><span class='pagenum'>Pg 91<a name="Page_91" id="Page_91"></a></span>Note "row of four cards," so that the only diagonals we have here to
+consider are the two long ones.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_305_THE_THIRTY_SIX_LETTER_BLOCKS" id="X_305_THE_THIRTY_SIX_LETTER_BLOCKS"></a><a href="#X_305_THE_THIRTY_SIX_LETTER_BLOCKSa"><b>305.&mdash;THE THIRTY-SIX LETTER BLOCKS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q305.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p>The illustration represents a box containing thirty-six letter-blocks.
+The puzzle is to rearrange these blocks so that no A shall be in a
+line vertically, horizontally, or diagonally with another A, no B with
+another B, no C with another C, and so on. You will find it impossible
+to get all the letters into the box under these conditions, but the
+point is to place as many as possible. Of course no letters other than
+those shown may be used.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_306_THE_CROWDED_CHESSBOARD" id="X_306_THE_CROWDED_CHESSBOARD"></a><a href="#X_306_THE_CROWDED_CHESSBOARDa"><b>306.&mdash;THE CROWDED CHESSBOARD.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q306.png" width="400" height="395" alt="" title="" />
+</div>
+
+<p>The puzzle is to rearrange the fifty-one pieces on the chessboard so
+that no queen shall attack another queen, no rook attack another rook,
+no bishop attack another bishop, and no knight attack another knight.
+No notice is to be taken of the intervention of pieces of another type
+from that under consideration&mdash;that is, two queens will be considered
+to attack one another although there may be, say, a rook, a bishop,
+and a knight between them. And so with the rooks and bishops. It is
+not difficult to dispose of each type of piece separately; the
+difficulty comes in when you have to find room for all the
+arrangements on the board simultaneously.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_307_THE_COLOURED_COUNTERS" id="X_307_THE_COLOURED_COUNTERS"></a><a href="#X_307_THE_COLOURED_COUNTERSa"><b>307.&mdash;THE COLOURED COUNTERS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q307.png" width="400" height="404" alt="" title="" />
+</div>
+
+<p>The diagram represents twenty-five coloured counters, Red, Blue,
+Yellow, Orange, and Green (indicated by their initials), and there are
+five of each colour, numbered 1, 2, 3, 4, and 5. The problem is so to
+place them in a square that neither colour nor number shall be found
+repeated in any one of the five rows, five columns, and two diagonals.
+Can you so rearrange them?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_308_THE_GENTLE_ART_OF_STAMP-LICKING" id="X_308_THE_GENTLE_ART_OF_STAMP-LICKING"></a><a href="#X_308_THE_GENTLE_ART_OF_STAMP-LICKINGa"><b>308.&mdash;THE GENTLE ART OF STAMP-LICKING.</b></a></p>
+
+<p>The Insurance Act is a most prolific source of entertaining puzzles,
+particularly entertaining if you happen to be among the exempt. One's
+initiation into the gentle art of stamp-licking suggests the following
+little poser: If you have a card divided into sixteen spaces (4&nbsp;&times;&nbsp;4),
+and are provided with plenty of stamps of the values 1<i>d</i>., 2<i>d</i>., 3<i>d</i>.,
+4<i>d</i>., and 5<i>d</i>., what is the greatest value that you can stick on the
+card if the Chancellor of the Exchequer forbids you to place any stamp
+in a straight line (that is, horizontally, vertically, or diagonally)
+with another stamp of similar value? Of course, only one stamp can be
+affixed in a space. The reader will probably find, when he sees the
+solution, that, like the stamps themselves, he is licked <span class='pagenum'>Pg 92<a name="Page_92" id="Page_92"></a></span>He will most
+likely be twopence short of the maximum. A friend asked the Post
+Office how it was to be done; but they sent him to the Customs and
+Excise officer, who sent him to the Insurance Commissioners, who sent
+him to an approved society, who profanely sent him&mdash;but no matter.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_309_THE_FORTY-NINE_COUNTERS" id="X_309_THE_FORTY-NINE_COUNTERS"></a><a href="#X_309_THE_FORTY-NINE_COUNTERSa"><b>309.&mdash;THE FORTY-NINE COUNTERS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q309.png" width="400" height="401" alt="" title="" />
+</div>
+
+<p>Can you rearrange the above forty-nine counters in a square so that no
+letter, and also no number, shall be in line with a similar one,
+vertically, horizontally, or diagonally? Here I, of course, mean in
+the lines parallel with the diagonals, in the chessboard sense.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_310_THE_THREE_SHEEP" id="X_310_THE_THREE_SHEEP"></a><a href="#X_310_THE_THREE_SHEEPa"><b>310.&mdash;THE THREE SHEEP.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q310.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>A farmer had three sheep and an arrangement of sixteen pens, divided
+off by hurdles in the manner indicated in the illustration. In how
+many different ways could he place those sheep, each in a separate
+pen, so that every pen should be either occupied or in line
+(horizontally, vertically, or diagonally) with at least one sheep? I
+have given one arrangement that fulfils the conditions. How many
+others can you find? Mere reversals and reflections must not be
+counted as different. The reader may regard the sheep as queens. The
+problem is then to place the three queens so that every square shall
+be either occupied or attacked by at least one queen&mdash;in the maximum
+number of different ways.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_311_THE_FIVE_DOGS_PUZZLE" id="X_311_THE_FIVE_DOGS_PUZZLE"></a><a href="#X_311_THE_FIVE_DOGS_PUZZLEa"><b>311.&mdash;THE FIVE DOGS PUZZLE.</b></a></p>
+
+<p>In 1863, C.F. de Jaenisch first discussed the "Five Queens Puzzle"&mdash;to
+place five queens on the chessboard so that every square shall be
+attacked or occupied&mdash;which was propounded by his friend, a "Mr. de
+R." Jaenisch showed that if no queen may attack another there are
+ninety-one different ways of placing the five queens, reversals and
+reflections not counting as different. If the queens may attack one
+another, I have recorded hundreds of ways, but it is not practicable
+to enumerate them exactly.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q311.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>The illustration is supposed to represent an arrangement of sixty-four
+kennels. It will be seen that five kennels each contain a dog, and on
+further examination it will be seen that every one of the sixty-four
+kennels is in a straight line with at least one dog&mdash;either
+horizontally, vertically, or diagonally. Take any kennel you like, and
+you will find that you can draw a straight line to a dog in one or
+other of the three ways mentioned. The puzzle is to replace the five
+dogs and discover in just how many different ways they may be placed
+in five kennels <i>in a straight row</i>, so that every kennel shall always
+be in line with at least one dog. Reversals and reflections are here
+counted as different.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_312_THE_FIVE_CRESCENTS_OF_BYZANTIUM" id="X_312_THE_FIVE_CRESCENTS_OF_BYZANTIUM"></a><a href="#X_312_THE_FIVE_CRESCENTS_OF_BYZANTIUMa"><b>312.&mdash;THE FIVE CRESCENTS OF BYZANTIUM.</b></a></p>
+
+<p>When Philip of Macedon, the father of Alexander the Great, found
+himself confronted with great difficulties in the siege of Byzantium,
+he <span class='pagenum'>Pg 93<a name="Page_93" id="Page_93"></a></span>set his men to undermine the walls. His desires, however,
+miscarried, for no sooner had the operations been begun than a
+crescent moon suddenly appeared in the heavens and discovered his
+plans to his adversaries. The Byzantines were naturally elated, and in
+order to show their gratitude they erected a statue to Diana, and the
+crescent became thenceforward a symbol of the state. In the temple
+that contained the statue was a square pavement composed of sixty-four
+large and costly tiles. These were all plain, with the exception of
+five, which bore the symbol of the crescent. These five were for
+occult reasons so placed that every tile should be watched over by
+(that is, in a straight line, vertically, horizontally, or diagonally
+with) at least one of the crescents. The arrangement adopted by the
+Byzantine architect was as follows:&mdash;</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q312.png" width="400" height="402" alt="" title="" />
+</div>
+
+<p>Now, to cover up one of these five crescents was a capital offence,
+the death being something very painful and lingering. But on a certain
+occasion of festivity it was necessary to lay down on this pavement a
+square carpet of the largest dimensions possible, and I have shown in
+the illustration by dark shading the largest dimensions that would be
+available.</p>
+
+<p>The puzzle is to show how the architect, if he had foreseen this
+question of the carpet, might have so arranged his five crescent tiles
+in accordance with the required conditions, and yet have allowed for
+the largest possible square carpet to be laid down without any one of
+the five crescent tiles being covered, or any portion of them.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_313_QUEENS_AND_BISHOP_PUZZLE" id="X_313_QUEENS_AND_BISHOP_PUZZLE"></a><a href="#X_313_QUEENS_AND_BISHOP_PUZZLEa"><b>313.&mdash;QUEENS AND BISHOP PUZZLE.</b></a></p>
+
+<p>It will be seen that every square of the board is either occupied or
+attacked. The puzzle is to substitute a bishop for the rook on the
+same square, and then place the four queens on other squares so that
+every square shall again be either occupied or attacked.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q313.png" width="400" height="412" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_314_THE_SOUTHERN_CROSS" id="X_314_THE_SOUTHERN_CROSS"></a><a href="#X_314_THE_SOUTHERN_CROSSa"><b>314.&mdash;THE SOUTHERN CROSS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q314.png" width="400" height="392" alt="" title="" />
+</div>
+
+<p>In the above illustration we have five Planets and eighty-one Fixed
+Stars, five of the latter being hidden by the Planets. It will be
+found that every Star, with the exception of the ten that have a black
+spot in their centres, is in a straight line, vertically,
+horizontally, or diagonally, with at least one of the Planets. The
+puzzle is so to rearrange the Planets that all the Stars shall be in
+line with one or more of them.</p>
+
+<p>In rearranging the Planets, each of the five may be moved once in a
+straight line, in either of the three directions mentioned. They will,
+of course, obscure five other Stars in place of those at present
+covered.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_315_THE_HAT-PEG_PUZZLE" id="X_315_THE_HAT-PEG_PUZZLE"></a><a href="#X_315_THE_HAT-PEG_PUZZLEa"><b>315.&mdash;THE HAT-PEG PUZZLE.</b></a></p>
+
+<p>Here is a five-queen puzzle that I gave in a fanciful dress in 1897.
+As the queens were <span class='pagenum'>Pg 94<a name="Page_94" id="Page_94"></a></span>there represented as hats on sixty-four pegs, I
+will keep to the title, "The Hat-Peg Puzzle." It will be seen that
+every square is occupied or attacked. The puzzle is to remove one
+queen</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q315.png" width="400" height="396" alt="" title="" />
+</div>
+
+<p>to a different square so that still every square is occupied or
+attacked, then move a second queen under a similar condition, then a
+third queen, and finally a fourth queen. After the fourth move every
+square must be attacked or occupied, but no queen must then attack
+another. Of course, the moves need not be "queen moves;" you can move
+a queen to any part of the board.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_316_THE_AMAZONS" id="X_316_THE_AMAZONS"></a><a href="#X_316_THE_AMAZONSa"><b>316.&mdash;THE AMAZONS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q316.png" width="400" height="402" alt="" title="" />
+</div>
+
+<p>This puzzle is based on one by Captain Turton. Remove three of the
+queens to other squares so that there shall be eleven squares on the
+board that are not attacked. The removal of the three queens need not
+be by "queen moves." You may take them up and place them anywhere.
+There is only one solution.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_317_A_PUZZLE_WITH_PAWNS" id="X_317_A_PUZZLE_WITH_PAWNS"></a><a href="#X_317_A_PUZZLE_WITH_PAWNSa"><b>317.&mdash;A PUZZLE WITH PAWNS.</b></a></p>
+
+<p>Place two pawns in the middle of the chessboard, one at Q 4 and the
+other at K 5. Now, place the remaining fourteen pawns (sixteen in all)
+so that no three shall be in a straight line in any possible
+direction.</p>
+
+<p>Note that I purposely do not say queens, because by the words "any
+possible direction" I go beyond attacks on diagonals. The pawns must
+be regarded as mere points in space&mdash;at the centres of the squares.
+See dotted lines in the case of No. <a href="#X_300_THE_EIGHT_QUEENS">300</a>, "<a href="#X_300_THE_EIGHT_QUEENS">The Eight Queens</a>."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_318_LION-HUNTING" id="X_318_LION-HUNTING"></a><a href="#X_318_LION-HUNTINGa"><b>318.&mdash;LION-HUNTING.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q318.png" width="400" height="402" alt="" title="" />
+</div>
+
+<p>My friend Captain Potham Hall, the renowned hunter of big game, says
+there is nothing more exhilarating than a brush with a herd&mdash;a
+pack&mdash;a team&mdash;a flock&mdash;a swarm (it has taken me a full quarter of an
+hour to recall the right word, but I have it at last)&mdash;a <i>pride</i> of
+lions. Why a number of lions are called a "pride," a number of whales
+a "school," and a number of foxes a "skulk" are mysteries of philology
+into which I will not enter.</p>
+
+<p>Well, the captain says that if a spirited lion crosses your path in
+the desert it becomes lively, for the lion has generally been looking
+for the man just as much as the man has sought the king of the forest.
+And yet when they meet they always quarrel and fight it out. A little
+contemplation of this unfortunate and long-standing feud between two
+estimable families has led me to figure out a few calculations as to
+the probability of the man and the lion crossing one another's path in
+the jungle. In all these cases one has to start on certain more <span class='pagenum'>Pg 95<a name="Page_95" id="Page_95"></a></span>or
+less arbitrary assumptions. That is why in the above illustration I
+have thought it necessary to represent the paths in the desert with
+such rigid regularity. Though the captain assures me that the tracks
+of the lions usually run much in this way, I have doubts.</p>
+
+<p>The puzzle is simply to find out in how many different ways the man
+and the lion may be placed on two different spots that are not on the
+same path. By "paths" it must be understood that I only refer to the
+ruled lines. Thus, with the exception of the four corner spots, each
+combatant is always on two paths and no more. It will be seen that
+there is a lot of scope for evading one another in the desert, which
+is just what one has always understood.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_319_THE_KNIGHT-GUARDS" id="X_319_THE_KNIGHT-GUARDS"></a><a href="#X_319_THE_KNIGHT-GUARDSa"><b>319.&mdash;THE KNIGHT-GUARDS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q319.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>The knight is the irresponsible low comedian of the chessboard. "He is
+a very uncertain, sneaking, and demoralizing rascal," says an American
+writer. "He can only move two squares, but makes up in the quality of
+his locomotion for its quantity, for he can spring one square sideways
+and one forward simultaneously, like a cat; can stand on one leg in
+the middle of the board and jump to any one of eight squares he
+chooses; can get on one side of a fence and blackguard three or four
+men on the other; has an objectionable way of inserting himself in
+safe places where he can scare the king and compel him to move, and
+then gobble a queen. For pure cussedness the knight has no equal, and
+when you chase him out of one hole he skips into another." Attempts
+have been made over and over again to obtain a short, simple, and
+exact definition of the move of the knight&mdash;without success. It really
+consists in moving one square like a rook, and then another square
+like a bishop&mdash;the two operations being done in one leap, so that it
+does not matter whether the first square passed over is occupied by
+another piece or not. It is, in fact, the only leaping move in chess.
+But difficult as it is to define, a child can learn it by inspection
+in a few minutes.</p>
+
+<p>I have shown in the diagram how twelve knights (the fewest possible
+that will perform the feat) may be placed on the chessboard so that
+every square is either occupied or attacked by a knight. Examine every
+square in turn, and you will find that this is so. Now, the puzzle in
+this case is to discover what is the smallest possible number of
+knights that is required in order that every square shall be either
+occupied or attacked, and every knight protected by another knight.
+And how would you arrange them? It will be found that of the twelve
+shown in the diagram only four are thus protected by being a knight's
+move from another knight.</p>
+
+
+<h2><a name="THE_GUARDED_CHESSBOARD" id="THE_GUARDED_CHESSBOARD"></a><a href="#CONTENTS">THE GUARDED CHESSBOARD.</a></h2>
+
+<p>On an ordinary chessboard, 8 by 8, every square can be guarded&mdash;that
+is, either occupied or attacked&mdash;by 5 queens, the fewest possible.
+There are exactly 91 fundamentally different arrangements in which no
+queen attacks another queen. If every queen must attack (or be
+protected by) another queen, there are at fewest 41 arrangements, and
+I have recorded some 150 ways in which some of the queens are attacked
+and some not, but this last case is very difficult to enumerate
+exactly.</p>
+
+<p>On an ordinary chessboard every square can be guarded by 8 rooks (the
+fewest possible) in 40,320 ways, if no rook may attack another rook,
+but it is not known how many of these are fundamentally different.
+(See solution to No. <a href="#X_295_THE_EIGHT_ROOKSa">295</a>, "<a href="#X_295_THE_EIGHT_ROOKSa">The Eight Rooks</a>.") I have not enumerated
+the ways in which every rook shall be protected by another rook.</p>
+
+<p>On an ordinary chessboard every square can be guarded by 8 bishops
+(the fewest possible), if no bishop may attack another bishop. Ten
+bishops are necessary if every bishop is to be protected. (See Nos.
+<a href="#X_297_BISHOPSmdashUNGUARDED">297</a> and <a href="#X_298_BISHOPSmdashGUARDED">298</a>, "<a href="#X_297_BISHOPSmdashUNGUARDED">Bishops unguarded</a>" and "<a href="#X_298_BISHOPSmdashGUARDED">Bishops guarded</a>.")</p>
+
+<p>On an ordinary chessboard every square can be guarded by 12 knights if
+all but 4 are unprotected. But if every knight must be protected, 14
+are necessary. (See No. <a href="#X_319_THE_KNIGHT-GUARDS">319</a>, "<a href="#X_319_THE_KNIGHT-GUARDS">The Knight-Guards</a>.")</p>
+
+<p>Dealing with the queen on <i>n</i><sup>2</sup> boards generally, where <i>n</i> is less
+than 8, the following results will be of interest:&mdash;</p>
+
+
+<p>1 queen guards 2<sup>2</sup> board in 1 fundamental way.</p>
+
+<p>1 queen guards 3<sup>2</sup> board in 1 fundamental way.</p>
+
+<p>2 queens guard 4<sup>2</sup> board in 3 fundamental ways (protected).</p>
+
+<p>3 queens guard 4<sup>2</sup> board in 2 fundamental ways (not protected).</p>
+
+<p>3 queens guard 5<sup>2</sup> board in 37 fundamental ways (protected).</p>
+
+<p>3 queens guard 5<sup>2</sup> board in 2 fundamental ways (not protected).</p>
+
+<p>3 queens guard 6<sup>2</sup> board in 1 fundamental way (protected).</p><p><span class='pagenum'>Pg 96<a name="Page_96" id="Page_96"></a></span></p>
+
+<p>4 queens guard 6<sup>2</sup> board in 17 fundamental ways (not protected).</p>
+
+<p>4 queens guard 7<sup>2</sup> board in 5 fundamental ways (protected).</p>
+
+<p>4 queens guard 7<sup>2</sup> board in 1 fundamental way (not protected).</p>
+
+
+<h3>NON-ATTACKING CHESSBOARD ARRANGEMENTS.</h3>
+
+<p>We know that n queens may always be placed on a square board of n<sup>2</sup>
+squares (if n be greater than 3) without any queen attacking another
+queen. But no general formula for enumerating the number of different
+ways in which it may be done has yet been discovered; probably it is
+undiscoverable. The known results are as follows:&mdash;</p>
+
+<p>Where n = 4 there is 1 fundamental solution and 2 in all.</p>
+
+<p>Where n = 5 there are 2 fundamental solutions and 10 in all.</p>
+
+<p>Where n = 6 there is 1 fundamental solution and 4 in all.</p>
+
+<p>Where n = 7 there are 6 fundamental solutions and 40 in all.</p>
+
+<p>Where n = 8 there are 12 fundamental solutions and 92 in all.</p>
+
+<p>Where n = 9 there are 46 fundamental solutions.</p>
+
+<p>Where n = 10 there are 92 fundamental solutions.</p>
+
+<p>Where n = 11 there are 341 fundamental solutions.</p>
+
+<p>Obviously n rooks may be placed without attack on an n<sup>2</sup> board in n!
+ways, but how many of these are fundamentally different I have only
+worked out in the four cases where n equals 2, 3, 4, and 5. The
+answers here are respectively 1, 2, 7, and 23. (See No. <a href="#X_296_THE_FOUR_LIONS">296</a>, "<a href="#X_296_THE_FOUR_LIONS">The Four
+Lions</a>.")</p>
+
+<p>We can place 2n-2 bishops on an n<sup>2</sup> board in 2<sup>n</sup> ways. (See No.
+<a href="#X_299_BISHOPS_IN_CONVOCATION">299</a>, "<a href="#X_299_BISHOPS_IN_CONVOCATION">Bishops in Convocation</a>.") For boards containing 2, 3, 4, 5, 6,
+7, 8 squares, on a side there are respectively 1, 2, 3, 6, 10, 20, 36
+fundamentally different arrangements. Where n is odd there are
+2<sup>&frac12;(n-1)</sup> such arrangements, each giving 4 by reversals and
+reflections, and 2<sup>n-3</sup>&nbsp;-&nbsp;2<sup>&frac12;(n-3)</sup> giving 8. Where n is even there
+are 2<sup>&frac12;(n-2)</sup>, each giving 4 by reversals and reflections, and
+2<sup>n-3</sup>&nbsp;-&nbsp;2<sup>&frac12;(n-4)</sup>, each giving 8.</p>
+
+<p>We can place &frac12;(n<sup>2</sup>+1) knights on an n<sup>2</sup> board without attack, when
+n is odd, in 1 fundamental way; and &frac12;n<sup>2</sup> knights on an n<sup>2</sup> board,
+when n is even, in 1 fundamental way. In the first case we place all
+the knights on the same colour as the central square; in the second
+case we place them all on black, or all on white, squares.</p>
+
+
+<h3>THE TWO PIECES PROBLEM.</h3>
+
+<p>On a board of n<sup>2</sup> squares, two queens, two rooks, two bishops, or
+two knights can always be placed, irrespective of attack or not, in
+&frac12;(n<sup>4</sup>&nbsp;-&nbsp;n<sup>2</sup>) ways. The following formul&aelig; will show in how many of
+these ways the two pieces may be placed with attack and without:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="1" cellspacing="3" summary="">
+<tr><td align='left'></td><td align='center'>With Attack.</td><td align='center'>Without Attack.</td></tr>
+<tr><td align='left' rowspan='2'>2 Queens</td><td align='center' class='bb'>5n3&nbsp;-&nbsp;6n2&nbsp;+&nbsp;n</td><td align='center' class='bb'>3n4&nbsp;-&nbsp;10n3&nbsp;+&nbsp;9n2&nbsp;-&nbsp;2n</td></tr>
+<tr><td align='center'>3</td><td align='center'>6</td></tr>
+<tr><td align='left' rowspan='2'>2 Rooks</td><td align='center' rowspan='2'>n3&nbsp;-&nbsp;n2</td><td align='center' class='bb'>n4&nbsp;-&nbsp;2n3&nbsp;+&nbsp;n2</td></tr>
+<tr><td align='center'>2</td></tr>
+<tr><td align='left' rowspan='2'>2 Bishops</td><td align='center' class='bb'>4n3&nbsp;-&nbsp;6n2&nbsp;+&nbsp;2n</td><td align='center' class='bb'>3n4&nbsp;-&nbsp;4n3&nbsp;+&nbsp;3n2&nbsp;-&nbsp;2n</td></tr>
+<tr><td align='center'>6</td><td align='center'>6</td></tr>
+<tr><td align='left' rowspan='2'>2 Knights</td><td align='center' rowspan='2'>4n2&nbsp;-&nbsp;12n&nbsp;+&nbsp;8</td><td align='center' class='bb'>n4&nbsp;-&nbsp;9n2&nbsp;+&nbsp;24n</td></tr>
+<tr><td align='center'>2</td></tr>
+</table></div>
+
+<p>(See No. <a href="#X_318_LION-HUNTING">318</a>, "<a href="#X_318_LION-HUNTING">Lion Hunting</a>.")</p>
+
+
+<hr style="width: 65%;" />
+<h2><a name="DYNAMICAL_CHESS_PUZZLES" id="DYNAMICAL_CHESS_PUZZLES"></a><a href="#CONTENTS">DYNAMICAL CHESS PUZZLES.</a></h2>
+
+<p class='center'>
+<span style="margin-left: 0em;">"Push on&mdash;keep moving."<br /></span>
+<span style="margin-left: 8em;">THOS. MORTON: <i>Cure for the Heartache</i>.</span><br />
+</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_320_THE_ROOKS_TOUR" id="X_320_THE_ROOKS_TOUR"></a><a href="#X_320_THE_ROOKS_TOURa"><b>320.&mdash;THE ROOK'S TOUR.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q320.png" width="400" height="397" alt="" title="" />
+</div>
+
+
+<p>The puzzle is to move the single rook over the whole board, so that it
+shall visit every square of the board once, and only once, and end its
+tour on the square from which it starts. You have to do this in as few
+moves as possible, and unless you are very careful you will take just
+one move too many. Of course, a square is regarded equally as
+"visited" whether you merely pass over it or make it a stopping-place,
+and we will not quibble over the point whether the original square is
+actually visited twice. We will assume that it is not.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_321_THE_ROOKS_JOURNEY" id="X_321_THE_ROOKS_JOURNEY"></a><a href="#X_321_THE_ROOKS_JOURNEYa"><b>321.&mdash;THE ROOK'S JOURNEY.</b></a></p>
+
+<p>This puzzle I call "The Rook's Journey," because the word "tour"
+(derived from a turner's wheel) implies that we return to the point
+from which we set out, and we do not do this in the present case. We
+should not be satisfied with <span class='pagenum'>Pg 97<a name="Page_97" id="Page_97"></a></span>a personally conducted holiday tour that
+ended by leaving us, say, in the middle of the Sahara. The rook here
+makes twenty-one moves, in the course of which journey it visits every
+square of the board once and only once, stopping at the square marked
+10 at the end of its tenth move, and ending at the square marked 21.
+Two consecutive moves cannot be made in the same direction&mdash;that is to
+say, you must make a turn after every move.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q321.png" width="400" height="392" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_322_THE_LANGUISHING_MAIDEN" id="X_322_THE_LANGUISHING_MAIDEN"></a><a href="#X_322_THE_LANGUISHING_MAIDENa"><b>322.&mdash;THE LANGUISHING MAIDEN.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q322.png" width="400" height="393" alt="" title="" />
+</div>
+
+<p>A wicked baron in the good old days imprisoned an innocent maiden in
+one of the deepest dungeons beneath the castle moat. It will be seen
+from our illustration that there were sixty-three cells in the
+dungeon, all connected by open doors, and the maiden was chained in
+the cell in which she is shown. Now, a valiant knight, who loved the
+damsel, succeeded in rescuing her from the enemy. Having gained an
+entrance to the dungeon at the point where he is seen, he succeeded in
+reaching the maiden after entering every cell once and only once. Take
+your pencil and try to trace out such a route. When you have
+succeeded, then try to discover a route in twenty-two straight paths
+through the cells. It can be done in this number without entering any
+cell a second time.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_323_A_DUNGEON_PUZZLE" id="X_323_A_DUNGEON_PUZZLE"></a><a href="#X_323_A_DUNGEON_PUZZLEa"><b>323.&mdash;A DUNGEON PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q323.png" width="400" height="394" alt="" title="" />
+</div>
+
+<p>A French prisoner, for his sins (or other people's), was confined in
+an underground dungeon containing sixty-four cells, all communicating
+with open doorways, as shown in our illustration. In order to reduce
+the tedium of his restricted life, he set himself various puzzles, and
+this is one of them. Starting from the cell in which he is shown, how
+could he visit every cell once, and only once, and make as many
+turnings as possible? His first attempt is shown by the dotted track.
+It will be found that there are as many as fifty-five straight lines
+in his path, but after many attempts he improved upon this. Can you
+get more than fifty-five? You may end your path in any cell you like.
+Try the puzzle with a pencil on chessboard diagrams, or you may regard
+them as rooks' moves on a board.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_324_THE_LION_AND_THE_MAN" id="X_324_THE_LION_AND_THE_MAN"></a><a href="#X_324_THE_LION_AND_THE_MANa"><b>324.&mdash;THE LION AND THE MAN.</b></a></p>
+
+<p>In a public place in Rome there once stood a prison divided into
+sixty-four cells, all open to the sky and all communicating with one
+another, as shown in the illustration. The sports that here took place
+were watched from a high tower. The favourite game was to place a
+Christian in one corner cell and a lion in the diagonally opposite
+corner and then leave them with all the inner doors open. The
+consequent effect was sometimes most laughable. On one occasion the
+man was given a sword. He was <span class='pagenum'>Pg 98<a name="Page_98" id="Page_98"></a></span>no coward, and was as anxious to find
+the lion as the lion undoubtedly was to find him.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q324.png" width="400" height="390" alt="" title="" />
+</div>
+
+<p>The man visited every cell once and only once in the fewest possible
+straight lines until he reached the lion's cell. The lion, curiously
+enough, also visited every cell once and only once in the fewest
+possible straight lines until he finally reached the man's cell. They
+started together and went at the same speed; yet, although they
+occasionally got glimpses of one another, they never once met. The
+puzzle is to show the route that each happened to take.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_325_AN_EPISCOPAL_VISITATION" id="X_325_AN_EPISCOPAL_VISITATION"></a><a href="#X_325_AN_EPISCOPAL_VISITATIONa"><b>325.&mdash;AN EPISCOPAL VISITATION.</b></a></p>
+
+<p>The white squares on the chessboard represent the parishes of a
+diocese. Place the bishop on any square you like, and so contrive that
+(using the ordinary bishop's move of chess) he shall visit every one
+of his parishes in the fewest possible moves. Of course, all the
+parishes passed through on any move are regarded as "visited." You can
+visit any squares more than once, but you are not allowed to move
+twice between the same two adjoining squares. What are the fewest
+possible moves? The bishop need not end his visitation at the parish
+from which he first set out.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_326_A_NEW_COUNTER_PUZZLE" id="X_326_A_NEW_COUNTER_PUZZLE"></a><a href="#X_326_A_NEW_COUNTER_PUZZLEa"><b>326.&mdash;A NEW COUNTER PUZZLE.</b></a></p>
+
+<p>Here is a new puzzle with moving counters, or coins, that at first
+glance looks as if it must be absurdly simple. But it will be found
+quite a little perplexity. I give it in this place for a reason that I
+will explain when we come to the next puzzle. Copy the simple diagram,
+enlarged, on a sheet of paper; then place two white counters on the
+points 1 and 2, and two red counters on 9 and 10, The puzzle is to
+make the red and white change places. You may move the counters one at
+a time in any order you like, along the lines from point to point,
+with the only restriction that a red and a white counter may never
+stand at once on the same straight line. Thus the first move can only
+be from 1 or 2 to 3, or from 9 or 10 to 7.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q326.png" width="400" height="307" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_327_A_NEW_BISHOPS_PUZZLE" id="X_327_A_NEW_BISHOPS_PUZZLE"></a><a href="#X_327_A_NEW_BISHOPS_PUZZLEa"><b>327.&mdash;A NEW BISHOP'S PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q327.png" width="400" height="493" alt="" title="" />
+</div>
+
+<p>This is quite a fascinating little puzzle. Place eight bishops (four
+black and four white) on the reduced chessboard, as shown in the
+illustration. The problem is to make the black bishops change places
+with the white ones, no bishop ever attacking another of the opposite
+colour. They must move alternately&mdash;first a white, then a black, then
+a white, and so on. When you have succeeded in doing it at all, try to
+find the fewest possible moves.</p>
+
+<p>If you leave out the bishops standing on black squares, and only play
+on the white squares, you will discover my last puzzle turned on its
+side.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_328_THE_QUEENS_TOUR" id="X_328_THE_QUEENS_TOUR"></a><a href="#X_328_THE_QUEENS_TOURa"><b>328.&mdash;THE QUEEN'S TOUR.</b></a></p>
+
+<p>The puzzle of making a complete tour of the chessboard with the queen
+in the fewest possible moves (in which squares may be visited more
+than once) was first given by the late Sam Loyd <span class='pagenum'>Pg 99<a name="Page_99" id="Page_99"></a></span>in his <i>Chess
+Strategy</i>. But the solution shown below is the one he gave in
+<i>American Chess-Nuts</i> in 1868. I have recorded at least six different
+solutions in the minimum number of moves&mdash;fourteen&mdash;but this one is
+the best of all, for reasons I will explain.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q328a.png" width="400" height="397" alt="" title="" />
+</div>
+
+<p>If you will look at the lettered square you will understand that there
+are only ten really differently placed squares on a chessboard&mdash;those
+enclosed by a dark line&mdash;all the others are mere reversals or
+reflections. For example, every A is a corner square, and every J a
+central square. Consequently, as the solution shown has a
+turning-point at the enclosed D square, we can obtain a solution
+starting from and ending at any square marked D&mdash;by just turning the
+board about. Now, this scheme will give you a tour starting from any
+A, B, C, D, E, F, or H, while no other route that I know can be
+adapted to more than five different starting-points. There is no
+Queen's Tour in fourteen moves (remember a tour must be re-entrant)
+that may start from a G, I, or J. But we can have a non-re-entrant
+path over the whole board in fourteen moves, starting from any given
+square. Hence the following puzzle:&mdash;</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q328b.png" width="400" height="397" alt="" title="" />
+</div>
+
+<p>Start from the J in the enclosed part of the lettered diagram and
+visit every square of the board in fourteen moves, ending wherever you
+like.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_329_THE_STAR_PUZZLE" id="X_329_THE_STAR_PUZZLE"></a><a href="#X_329_THE_STAR_PUZZLEa"><b>329.&mdash;THE STAR PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q329.png" width="400" height="401" alt="" title="" />
+</div>
+
+<p>Put the point of your pencil on one of the white stars and (without
+ever lifting your pencil from the paper) strike out all the stars in
+fourteen continuous straight strokes, ending at the second white star.
+Your straight strokes may be in any direction you like, only every
+turning must be made on a star. There is no objection to striking out
+any star more than once.</p>
+
+<p>In this case, where both your starting and ending squares are fixed
+inconveniently, you cannot obtain a solution by breaking a Queen's
+Tour, or in any other way by queen moves alone. But you are allowed to
+use oblique straight lines&mdash;such as from the upper white star direct
+to a corner star.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_330_THE_YACHT_RACE" id="X_330_THE_YACHT_RACE"></a><a href="#X_330_THE_YACHT_RACEa"><b>330.&mdash;THE YACHT RACE.</b></a></p>
+
+<p>Now then, ye land-lubbers, hoist your baby-jib-topsails, break out
+your spinnakers, ease off your balloon sheets, and get your head-sails
+set!</p>
+
+<p>Our race consists in starting from the point at which the yacht is
+lying in the illustration and touching every one of the sixty-four
+buoys in fourteen straight courses, returning in the final tack to the
+buoy from which we start. The seventh course must finish at the buoy
+from which a flag is flying.</p>
+
+<p>This puzzle will call for a lot of skilful seamanship on account of
+the sharp angles at which it will occasionally be necessary to tack.
+<span class='pagenum'>Pg 100<a name="Page_100" id="Page_100"></a></span>The point of a lead pencil and a good nautical eye are all the outfit
+that we require.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q330.png" width="400" height="342" alt="" title="" />
+</div>
+
+<p>This is difficult, because of the condition as to the flag-buoy, and
+because it is a re-entrant tour. But again we are allowed those
+oblique lines.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_331_THE_SCIENTIFIC_SKATER" id="X_331_THE_SCIENTIFIC_SKATER"></a><a href="#X_331_THE_SCIENTIFIC_SKATERa"><b>331.&mdash;THE SCIENTIFIC SKATER.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q331.png" width="400" height="467" alt="" title="" />
+</div>
+
+<p>It will be seen that this skater has marked on the ice sixty-four
+points or stars, and he proposes to start <i>from his present position</i>
+near the corner and enter every one of the points in fourteen straight
+lines. How will he do it? Of course there is no objection to his
+passing over any point more than once, but his last straight stroke
+must bring him back to the position from which he started.</p>
+
+<p>It is merely a matter of taking your pencil and starting from the spot
+on which the skater's foot is at present resting, and striking out all
+the stars in fourteen continuous straight lines, returning to the
+point from which you set out.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_332_THE_FORTY-NINE_STARS" id="X_332_THE_FORTY-NINE_STARS"></a><a href="#X_332_THE_FORTY-NINE_STARSa"><b>332.&mdash;THE FORTY-NINE STARS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q332.png" width="400" height="383" alt="" title="" />
+</div>
+
+<p>The puzzle in this case is simply to take your pencil and, starting
+from one black star, strike out all the stars in twelve straight
+strokes, ending at the other black star. It will be seen that the
+attempt shown in the illustration requires fifteen strokes. Can you do
+it in twelve? Every turning must be made on a star, and the lines must
+be parallel to the sides and diagonals of the square, as shown. In
+this case we are dealing with a chessboard of reduced dimensions, but
+only queen moves (without going outside the boundary as in the last
+case) are required.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_333_THE_QUEENS_JOURNEY" id="X_333_THE_QUEENS_JOURNEY"></a><a href="#X_333_THE_QUEENS_JOURNEYa"><b>333.&mdash;THE QUEEN'S JOURNEY.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q333.png" width="400" height="403" alt="" title="" />
+</div>
+
+<p>Place the queen on her own square, as shown in the illustration, and
+then try to discover the <span class='pagenum'>Pg 101<a name="Page_101" id="Page_101"></a></span>greatest distance that she can travel over
+the board in five queen's moves without passing over any square a
+second time. Mark the queen's path on the board, and note carefully
+also that she must never cross her own track. It seems simple enough,
+but the reader may find that he has tripped.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_334_ST_GEORGE_AND_THE_DRAGON" id="X_334_ST_GEORGE_AND_THE_DRAGON"></a><a href="#X_334_ST_GEORGE_AND_THE_DRAGONa"><b>334.&mdash;ST. GEORGE AND THE DRAGON.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q334.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p>Here is a little puzzle on a reduced chessboard of forty-nine squares.
+St. George wishes to kill the dragon. Killing dragons was a well-known
+pastime of his, and, being a knight, it was only natural that he
+should desire to perform the feat in a series of knight's moves. Can
+you show how, starting from that central square, he may visit once,
+and only once, every square of the board in a chain of chess knight's
+moves, and end by capturing the dragon on his last move? Of course a
+variety of different ways are open to him, so try to discover a route
+that forms some pretty design when you have marked each successive
+leap by a straight line from square to square.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_335_FARMER_LAWRENCES_CORNFIELDS" id="X_335_FARMER_LAWRENCES_CORNFIELDS"></a><a href="#X_335_FARMER_LAWRENCES_CORNFIELDSa"><b>335.&mdash;FARMER LAWRENCE'S CORNFIELDS.</b></a></p>
+
+<p>One of the most beautiful districts within easy distance of London for
+a summer ramble is that part of Buckinghamshire known as the Valley of
+the Chess&mdash;at least, it was a few years ago, before it was discovered
+by the speculative builder. At the beginning of the present century
+there lived, not far from Latimers, a worthy but eccentric farmer
+named Lawrence. One of his queer notions was that every person who
+lived near the banks of the river Chess ought to be in some way
+acquainted with the noble game of the same name, and in order to
+impress this fact on his men and his neighbours he adopted at times
+strange terminology. For example, when one of his ewes presented him
+with a lamb, he would say that it had "queened a pawn"; when he put up
+a new barn against the highway, he called it "castling on the king's
+side"; and when he sent a man with a gun to keep his neighbour's birds
+off his fields, he spoke of it as "attacking his opponent's rooks."
+Everybody in the neighbourhood used to be amused at Farmer Lawrence's
+little jokes, and one boy (the wag of the village) who got his ears
+pulled by the old gentleman for stealing his "chestnuts" went so far
+as to call him "a silly old chess-protector!"</p>
+
+<p>One year he had a large square field divided into forty-nine square
+plots, as shown in the illustration. The white squares were sown with
+wheat and the black squares with barley. When the harvest time came
+round he gave orders that his men were first to cut the corn in the
+patch marked 1, and that each successive cutting should be exactly a
+knight's move from the last one, the thirteenth cutting being in the
+patch marked 13, the twenty-fifth in the patch marked 25, the
+thirty-seventh in the one marked 37, and the last, or forty-ninth
+cutting, in the patch marked 49. This was too much for poor Hodge, and
+each day Farmer Lawrence had to go down to the field and show which
+piece had to be operated upon. But the problem will perhaps present no
+difficulty to my readers.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q335.png" width="400" height="448" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_336_THE_GREYHOUND_PUZZLE" id="X_336_THE_GREYHOUND_PUZZLE"></a><a href="#X_336_THE_GREYHOUND_PUZZLEa"><b>336.&mdash;THE GREYHOUND PUZZLE.</b></a></p>
+
+<p>In this puzzle the twenty kennels do not communicate with one another
+by doors, but are divided off by a low wall. The solitary occupant is
+the greyhound which lives in the kennel in the top left-hand corner.
+When he is allowed his liberty he has to obtain it by visiting every
+kennel once and only once in a series of knight's <span class='pagenum'>Pg 102<a name="Page_102" id="Page_102"></a></span>moves, ending at
+the bottom right-hand corner, which is open to the world. The lines in
+the above diagram show one solution. The puzzle is to discover in how
+many different ways the greyhound may thus make his exit from his
+corner kennel.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q336.png" width="400" height="325" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_337_THE_FOUR_KANGAROOS" id="X_337_THE_FOUR_KANGAROOS"></a><a href="#X_337_THE_FOUR_KANGAROOSa"><b>337.&mdash;THE FOUR KANGAROOS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q337.png" width="400" height="392" alt="" title="" />
+</div>
+
+<p>In introducing a little Commonwealth problem, I must first explain
+that the diagram represents the sixty-four fields, all properly fenced
+off from one another, of an Australian settlement, though I need
+hardly say that our kith and kin "down under" always <i>do</i> set out
+their land in this methodical and exact manner. It will be seen that
+in every one of the four corners is a kangaroo. Why kangaroos have a
+marked preference for corner plots has never been satisfactorily
+explained, and it would be out of place to discuss the point here. I
+should also add that kangaroos, as is well known, always leap in what
+we call "knight's moves." In fact, chess players would probably have
+adopted the better term "kangaroo's move" had not chess been invented
+before kangaroos.</p>
+
+<p>The puzzle is simply this. One morning each kangaroo went for his
+morning hop, and in sixteen consecutive knight's leaps visited just
+fifteen different fields and jumped back to his corner. No field was
+visited by more than one of the kangaroos. The diagram shows how they
+arranged matters. What you are asked to do is to show how they might
+have performed the feat without any kangaroo ever crossing the
+horizontal line in the middle of the square that divides the board
+into two equal parts.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_338_THE_BOARD_IN_COMPARTMENTS" id="X_338_THE_BOARD_IN_COMPARTMENTS"></a><a href="#X_338_THE_BOARD_IN_COMPARTMENTSa"><b>338.&mdash;THE BOARD IN COMPARTMENTS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q338.png" width="400" height="393" alt="" title="" />
+</div>
+
+<p>We cannot divide the ordinary chessboard into four equal square
+compartments, and describe a complete tour, or even path, in each
+compartment. But we may divide it into four compartments, as in the
+illustration, two containing each twenty squares, and the other two
+each twelve squares, and so obtain an interesting puzzle. You are
+asked to describe a complete re-entrant tour on this board, starting
+where you like, but visiting every square in each successive
+compartment before passing into another one, and making the final leap
+back to the square from which the knight set out. It is not difficult,
+but will be found very entertaining and not uninstructive.</p>
+
+<p>Whether a re-entrant "tour" or a complete knight's "path" is possible
+or not on a rectangular board of given dimensions depends not only on
+its dimensions, but also on its shape. A tour is obviously not
+possible on a board containing an odd number of cells, such as 5 by 5
+or 7 by 7, for this reason: Every successive leap of the knight must
+be from a white square to a black and a black to a white alternately.
+But if there be an odd number of cells or squares there must be one
+more square of one colour than of the other, therefore the path must
+begin from a square of the colour that is in excess, and end on a
+similar colour, and as a knight's <span class='pagenum'>Pg 103<a name="Page_103" id="Page_103"></a></span>move from one colour to a similar
+colour is impossible the path cannot be re-entrant. But a perfect tour
+may be made on a rectangular board of any dimensions provided the
+number of squares be even, and that the number of squares on one side
+be not less than 6 and on the other not less than 5. In other words,
+the smallest rectangular board on which a re-entrant tour is possible
+is one that is 6 by 5.</p>
+
+<p>A complete knight's path (not re-entrant) over all the squares of a
+board is never possible if there be only two squares on one side; nor
+is it possible on a square board of smaller dimensions than 5 by 5. So
+that on a board 4 by 4 we can neither describe a knight's tour nor a
+complete knight's path; we must leave one square unvisited. Yet on a
+board 4 by 3 (containing four squares fewer) a complete path may be
+described in sixteen different ways. It may interest the reader to
+discover all these. Every path that starts from and ends at different
+squares is here counted as a different solution, and even reverse
+routes are called different.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_339_THE_FOUR_KNIGHTS_TOURS" id="X_339_THE_FOUR_KNIGHTS_TOURS"></a><a href="#X_339_THE_FOUR_KNIGHTS_TOURSa"><b>339.&mdash;THE FOUR KNIGHTS' TOURS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q339.png" width="400" height="397" alt="" title="" />
+</div>
+
+<p>I will repeat that if a chessboard be cut into four equal parts, as
+indicated by the dark lines in the illustration, it is not possible to
+perform a knight's tour, either re-entrant or not, on one of the
+parts. The best re-entrant attempt is shown, in which each knight has
+to trespass twice on other parts. The puzzle is to cut the board
+differently into four parts, each of the same size and shape, so that
+a re-entrant knight's tour may be made on each part. Cuts along the
+dotted lines will not do, as the four central squares of the board
+would be either detached or hanging on by a mere thread.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_340_THE_CUBIC_KNIGHTS_TOUR" id="X_340_THE_CUBIC_KNIGHTS_TOUR"></a><a href="#X_340_THE_CUBIC_KNIGHTS_TOURa"><b>340.&mdash;THE CUBIC KNIGHT'S TOUR.</b></a></p>
+
+<p>Some few years ago I happened to read somewhere that Abnit
+Vandermonde, a clever mathematician, who was born in 1736 and died in
+1793, had devoted a good deal of study to the question of knight's
+tours. Beyond what may be gathered from a few fragmentary references,
+I am not aware of the exact nature or results of his investigations,
+but one thing attracted my attention, and that was the statement that
+he had proposed the question of a tour of the knight over the six
+surfaces of a cube, each surface being a chessboard. Whether he
+obtained a solution or not I do not know, but I have never seen one
+published. So I at once set to work to master this interesting
+problem. Perhaps the reader may like to attempt it.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_341_THE_FOUR_FROGS" id="X_341_THE_FOUR_FROGS"></a><a href="#X_341_THE_FOUR_FROGSa"><b>341.&mdash;THE FOUR FROGS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q341.png" width="400" height="397" alt="" title="" />
+</div>
+
+<p>In the illustration we have eight toadstools, with white frogs on 1
+and 3 and black frogs on 6 and 8. The puzzle is to move one frog at a
+time, in any order, along one of the straight lines from toadstool to
+toadstool, until they have exchanged places, the white frogs being
+left on 6 and 8 and the black ones on 1 and 3. If you use four
+counters on a simple diagram, you will find this quite easy, but it is
+a little more puzzling to do it in only seven plays, any number of
+successive moves by one frog counting as one play. Of course, more
+than one frog cannot be on a toadstool at the same time.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_342_THE_MANDARINS_PUZZLE" id="X_342_THE_MANDARINS_PUZZLE"></a><a href="#X_342_THE_MANDARINS_PUZZLEa"><b>342.&mdash;THE MANDARIN'S PUZZLE.</b></a></p>
+
+<p>The following puzzle has an added interest from the circumstance that
+a correct solution of it secured for a certain young Chinaman the hand
+of his charming bride. The wealthiest mandarin within a radius of a
+hundred miles of Peking was Hi-Chum-Chop, and his beautiful daughter,
+Peeky-Bo, had innumerable admirers. One of her most ardent lovers was
+Winky-Hi, and when he asked the old mandarin for his consent to their
+marriage, Hi-Chum-Chop presented him with the following puzzle and
+promised his consent if the youth brought him the correct answer
+within a week. Winky-Hi, following a habit which obtains among certain
+<span class='pagenum'>Pg 104<a name="Page_104" id="Page_104"></a></span>solvers to this day, gave it to all his friends, and when he had
+compared their solutions he handed in the best one as his own. Luckily
+it was quite right. The mandarin thereupon fulfilled his promise. The
+fatted pup was killed for the wedding feast, and when Hi-Chum-Chop
+passed Winky-Hi the liver wing all present knew that it was a token of
+eternal goodwill, in accordance with Chinese custom from time
+immemorial.</p>
+
+<p>The mandarin had a table divided into twenty-five squares, as shown in
+the diagram. On each of twenty-four of these squares was placed a
+numbered counter, just as I have indicated. The puzzle is to get the
+counters in numerical order by moving them one at a time in what we
+call "knight's moves." Counter 1 should be where 16 is, 2 where 11 is,
+4 where 13 now is, and so on. It will be seen that all the counters on
+shaded squares are in their proper positions. Of course, two counters
+may never be on a square at the same time. Can you perform the feat in
+the fewest possible moves?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q342.png" width="400" height="411" alt="" title="" />
+</div>
+
+<p>In order to make the manner of moving perfectly clear I will point out
+that the first knight's move can only be made by 1 or by 2 or by 10.
+Supposing 1 moves, then the next move must be by 23, 4, 8, or 21. As
+there is never more than one square vacant, the order in which the
+counters move may be written out as follows: 1&mdash;21&mdash;14&mdash;18&mdash;22, etc. A
+rough diagram should be made on a larger scale for practice, and
+numbered counters or pieces of cardboard used.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_343_EXERCISE_FOR_PRISONERS" id="X_343_EXERCISE_FOR_PRISONERS"></a><a href="#X_343_EXERCISE_FOR_PRISONERSa"><b>343.&mdash;EXERCISE FOR PRISONERS.</b></a></p>
+
+<p>The following is the plan of the north wing of a certain gaol, showing
+the sixteen cells all communicating by open doorways. Fifteen
+prisoners were numbered and arranged in the cells as shown. They were
+allowed to change their cells as much as they liked, but if two
+prisoners were ever in the same cell together there was a severe
+punishment promised them.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q343.png" width="400" height="412" alt="" title="" />
+</div>
+
+<p>Now, in order to reduce their growing obesity, and to combine physical
+exercise with mental recreation, the prisoners decided, on the
+suggestion of one of their number who was interested in knight's
+tours, to try to form themselves into a perfect knight's path without
+breaking the prison regulations, and leaving the bottom right-hand
+corner cell vacant, as originally. The joke of the matter is that the
+arrangement at which they arrived was as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>8</td><td align='right'>3</td><td align='right'>12</td><td align='right'>1</td></tr>
+<tr><td align='right'>11</td><td align='right'>14</td><td align='right'>9</td><td align='right'>6</td></tr>
+<tr><td align='right'>4</td><td align='right'>7</td><td align='right'>2</td><td align='right'>13</td></tr>
+<tr><td align='right'>15</td><td align='right'>10</td><td align='right'>5</td></tr>
+</table></div>
+
+<p>The warders failed to detect the important fact that the men could not
+possibly get into this position without two of them having been at
+some time in the same cell together. Make the attempt with counters on
+a ruled diagram, and you will find that this is so. Otherwise the
+solution is correct enough, each member being, as required, a knight's
+move from the preceding number, and the original corner cell vacant.</p>
+
+<p>The puzzle is to start with the men placed as in the illustration and
+show how it might have been done in the fewest moves, while giving a
+complete rest to as many prisoners as possible.</p>
+
+<p>As there is never more than one vacant cell for a man to enter, it is
+only necessary to write down the numbers of the men in the order in
+which they move. It is clear that very few men can be left throughout
+in their cells undisturbed, but I will leave the solver to discover
+just how many, as this is a very essential part of the puzzle.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 105<a name="Page_105" id="Page_105"></a></span><a name="X_344_THE_KENNEL_PUZZLE" id="X_344_THE_KENNEL_PUZZLE"></a><a href="#X_344_THE_KENNEL_PUZZLEa"><b>344.&mdash;THE KENNEL PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q344.png" width="400" height="397" alt="" title="" />
+</div>
+
+<p>A man has twenty-five dog kennels all communicating with each other by
+doorways, as shown in the illustration. He wishes to arrange his
+twenty dogs so that they shall form a knight's string from dog No. 1
+to dog No. 20, the bottom row of five kennels to be left empty, as at
+present. This is to be done by moving one dog at a time into a vacant
+kennel. The dogs are well trained to obedience, and may be trusted to
+remain in the kennels in which they are placed, except that if two are
+placed in the same kennel together they will fight it out to the
+death. How is the puzzle to be solved in the fewest possible moves
+without two dogs ever being together?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_345_THE_TWO_PAWNS" id="X_345_THE_TWO_PAWNS"></a><a href="#X_345_THE_TWO_PAWNSa"><b>345.&mdash;THE TWO PAWNS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q345.png" width="400" height="428" alt="" title="" />
+</div>
+
+<p>Here is a neat little puzzle in counting. In how many different ways
+may the two pawns advance to the eighth square? You may move them in
+any order you like to form a different sequence. For example, you may
+move the Q R P (one or two squares) first, or the K R P first, or one
+pawn as far as you like before touching the other. Any sequence is
+permissible, only in this puzzle as soon as a pawn reaches the eighth
+square it is dead, and remains there unconverted. Can you count the
+number of different sequences? At first it will strike you as being
+very difficult, but I will show that it is really quite simple when
+properly attacked.</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="VARIOUS_CHESS_PUZZLES" id="VARIOUS_CHESS_PUZZLES"></a><a href="#CONTENTS">VARIOUS CHESS PUZZLES.</a></h2>
+
+<p class='center'>
+"Chesse-play is a good and wittie exercise of
+the minde for some kinde of men."<br />
+<span style="margin-left: 8em;">Burton's <i>Anatomy of Melancholy</i>.<br /></span>
+</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_346_SETTING_THE_BOARD" id="X_346_SETTING_THE_BOARD"></a><a href="#X_346_SETTING_THE_BOARDa"><b>346.&mdash;SETTING THE BOARD.</b></a></p>
+
+<p>I have a single chessboard and a single set of chessmen. In how many
+different ways may the men be correctly set up for the beginning of a
+game? I find that most people slip at a particular point in making the
+calculation.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_347_COUNTING_THE_RECTANGLES" id="X_347_COUNTING_THE_RECTANGLES"></a><a href="#X_347_COUNTING_THE_RECTANGLESa"><b>347.&mdash;COUNTING THE RECTANGLES.</b></a></p>
+
+<p>Can you say correctly just how many squares and other rectangles the
+chessboard contains? In other words, in how great a number of
+different ways is it possible to indicate a square or other rectangle
+enclosed by lines that separate the squares of the board?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_348_THE_ROOKERY" id="X_348_THE_ROOKERY"></a><a href="#X_348_THE_ROOKERYa"><b>348.&mdash;THE ROOKERY.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q348.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p>The White rooks cannot move outside the little square in which they
+are enclosed except on the final move, in giving checkmate. The puzzle
+<span class='pagenum'>Pg 106<a name="Page_106" id="Page_106"></a></span>is how to checkmate Black in the fewest possible moves with No. 8
+rook, the other rooks being left in numerical order round the sides of
+their square with the break between 1 and 7.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_349_STALEMATE" id="X_349_STALEMATE"></a><a href="#X_349_STALEMATEa"><b>349.&mdash;STALEMATE.</b></a></p>
+
+<p>Some years ago the puzzle was proposed to construct an imaginary game
+of chess, in which White shall be stalemated in the fewest possible
+moves with all the thirty-two pieces on the board. Can you build up
+such a position in fewer than twenty moves?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_350_THE_FORSAKEN_KING" id="X_350_THE_FORSAKEN_KING"></a><a href="#X_350_THE_FORSAKEN_KINGa"><b>350.&mdash;THE FORSAKEN KING.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q350.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p>Set up the position shown in the diagram. Then the condition of the
+puzzle is&mdash;White to play and checkmate in six moves. Notwithstanding
+the complexities, I will show how the manner of play may be condensed
+into quite a few lines, merely stating here that the first two moves
+of White cannot be varied.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_351_THE_CRUSADER" id="X_351_THE_CRUSADER"></a><a href="#X_351_THE_CRUSADERa"><b>351.&mdash;THE CRUSADER.</b></a></p>
+
+<p>The following is a prize puzzle propounded by me some years ago.
+Produce a game of chess which, after sixteen moves, shall leave White
+with all his sixteen men on their original squares and Black in
+possession of his king alone (not necessarily on his own square).
+White is then to <i>force</i> mate in three moves.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_352_IMMOVABLE_PAWNS" id="X_352_IMMOVABLE_PAWNS"></a><a href="#X_352_IMMOVABLE_PAWNSa"><b>352.&mdash;IMMOVABLE PAWNS.</b></a></p>
+
+<p>Starting from the ordinary arrangement of the pieces as for a game,
+what is the smallest possible number of moves necessary in order to
+arrive at the following position? The moves for both sides must, of
+course, be played strictly in accordance with the rules of the game,
+though the result will necessarily be a very weird kind of chess.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q352.png" width="400" height="430" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_353_THIRTY-SIX_MATES" id="X_353_THIRTY-SIX_MATES"></a><a href="#X_353_THIRTY-SIX_MATESa"><b>353.&mdash;THIRTY-SIX MATES.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q353.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p>Place the remaining eight White pieces in such a position that White
+shall have the choice of thirty-six different mates on the move. Every
+move that checkmates and leaves a different position is a different
+mate. The pieces already placed must not be moved.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_354_AN_AMAZING_DILEMMA" id="X_354_AN_AMAZING_DILEMMA"></a><a href="#X_354_AN_AMAZING_DILEMMAa"><b>354.&mdash;AN AMAZING DILEMMA.</b></a></p>
+
+<p>In a game of chess between Mr. Black and Mr. White, Black was in
+difficulties, and as usual was obliged to catch a train. So he
+proposed that White should complete the game in his absence on
+condition that no moves whatever <span class='pagenum'>Pg 107<a name="Page_107" id="Page_107"></a></span>should be made for Black, but only
+with the White pieces. Mr. White accepted, but to his dismay found it
+utterly impossible to win the game under such conditions. Try as he
+would, he could not checkmate his opponent. On which square did Mr.
+Black leave his king? The other pieces are in their proper positions
+in the diagram. White may leave Black in check as often as he likes,
+for it makes no difference, as he can never arrive at a checkmate
+position.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q354.png" width="400" height="434" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_355_CHECKMATE" id="X_355_CHECKMATE"></a><a href="#X_355_CHECKMATEa"><b>355.&mdash;CHECKMATE!</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q355.png" width="400" height="430" alt="" title="" />
+</div>
+
+<p>Strolling into one of the rooms of a London club, I noticed a position
+left by two players who had gone. This position is shown in the
+diagram. It is evident that White has checkmated Black. But how did he
+do it? That is the puzzle.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_356_QUEER_CHESS" id="X_356_QUEER_CHESS"></a><a href="#X_356_QUEER_CHESSa"><b>356.&mdash;QUEER CHESS.</b></a></p>
+
+<p>Can you place two White rooks and a White knight on the board so that
+the Black king (who must be on one of the four squares in the middle
+of the board) shall be in check with no possible move open to him? "In
+other words," the reader will say, "the king is to be shown
+checkmated." Well, you can use the term if you wish, though I
+intentionally do not employ it myself. The mere fact that there is no
+White king on the board would be a sufficient reason for my not doing
+so.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_357_ANCIENT_CHINESE_PUZZLE" id="X_357_ANCIENT_CHINESE_PUZZLE"></a><a href="#X_357_ANCIENT_CHINESE_PUZZLEa"><b>357.&mdash;ANCIENT CHINESE PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q357.png" width="400" height="449" alt="" title="" />
+</div>
+
+<p>My next puzzle is supposed to be Chinese, many hundreds of years old,
+and never fails to interest. White to play and mate, moving each of
+the three pieces once, and once only.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_358_THE_SIX_PAWNS" id="X_358_THE_SIX_PAWNS"></a><a href="#X_358_THE_SIX_PAWNSa"><b>358.&mdash;THE SIX PAWNS.</b></a></p>
+
+<p>In how many different ways may I place six pawns on the chessboard so
+that there shall be an even number of unoccupied squares in every row
+and every column? We are not here considering the diagonals at all,
+and every different six squares occupied makes a different solution,
+so we have not to exclude reversals or reflections.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_359_COUNTER_SOLITAIRE" id="X_359_COUNTER_SOLITAIRE"></a><a href="#X_359_COUNTER_SOLITAIREa"><b>359.&mdash;COUNTER SOLITAIRE.</b></a></p>
+
+<p>Here is a little game of solitaire that is quite easy, but not so easy
+as to be uninteresting. You can either rule out the squares on a sheet
+of cardboard or paper, or you can use a portion <span class='pagenum'>Pg 108<a name="Page_108" id="Page_108"></a></span>of your chessboard. I
+have shown numbered counters in the illustration so as to make the
+solution easy and intelligible to all, but chess pawns or draughts
+will serve just as well in practice.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q359.png" width="400" height="156" alt="" title="" />
+</div>
+
+<p>The puzzle is to remove all the counters except one, and this one that
+is left must be No. 1. You remove a counter by jumping over another
+counter to the next space beyond, if that square is vacant, but you
+cannot make a leap in a diagonal direction. The following moves will
+make the play quite clear: 1-9, 2-10, 1-2, and so on. Here 1 jumps
+over 9, and you remove 9 from the board; then 2 jumps over 10, and you
+remove 10; then 1 jumps over 2, and you remove 2. Every move is thus a
+capture, until the last capture of all is made by No. 1.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_360_CHESSBOARD_SOLITAIRE" id="X_360_CHESSBOARD_SOLITAIRE"></a><a href="#X_360_CHESSBOARD_SOLITAIREa"><b>360.&mdash;CHESSBOARD SOLITAIRE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q360.png" width="400" height="397" alt="" title="" />
+</div>
+
+<p>Here is an extension of the last game of solitaire. All you need is a
+chessboard and the thirty-two pieces, or the same number of draughts
+or counters. In the illustration numbered counters are used. The
+puzzle is to remove all the counters except two, and these two must
+have originally been on the same side of the board; that is, the two
+left must either belong to the group 1 to 16 or to the other group, 17
+to 32. You remove a counter by jumping over it with another counter to
+the next square beyond, if that square is vacant, but you cannot make
+a leap in a diagonal direction. The following moves will make the play
+quite clear: 3-11, 4-12, 3-4, 13-3. Here 3 jumps over 11, and you
+remove 11; 4 jumps over 12, and you remove 12; and so on. It will be
+found a fascinating little game of patience, and the solution requires
+the exercise of some ingenuity.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_361_THE_MONSTROSITY" id="X_361_THE_MONSTROSITY"></a><a href="#X_361_THE_MONSTROSITYa"><b>361.&mdash;THE MONSTROSITY.</b></a></p>
+
+<p>One Christmas Eve I was travelling by rail to a little place in one of
+the southern counties. The compartment was very full, and the
+passengers were wedged in very tightly. My neighbour in one of the
+corner seats was closely studying a position set up on one of those
+little folding chessboards that can be carried conveniently in the
+pocket, and I could scarcely avoid looking at it myself. Here is the
+position:&mdash;</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q361.png" width="400" height="445" alt="" title="" />
+</div>
+
+<p>My fellow-passenger suddenly turned his head and caught the look of
+bewilderment on my face.</p>
+
+<p>"Do you play chess?" he asked.</p>
+
+<p>"Yes, a little. What is that? A problem?"</p>
+
+<p>"Problem? No; a game."</p>
+
+<p>"Impossible!" I exclaimed rather rudely. "The position is a perfect
+monstrosity!"</p>
+
+<p>He took from his pocket a postcard and handed it to me. It bore an
+address at one side and on the other the words "43. K to Kt 8."</p>
+
+<p>"It is a correspondence game." he exclaimed. "That is my friend's last
+move, and I am considering my reply."</p>
+
+<p>"But you really must excuse me; the position seems utterly impossible.
+How on earth, for example&mdash;"</p>
+
+<p>"Ah!" he broke in smilingly. "I see; you are a beginner; you play to
+win."</p>
+
+<p><span class='pagenum'>Pg 109<a name="Page_109" id="Page_109"></a></span>"Of course you wouldn't play to lose or draw!"</p>
+
+<p>He laughed aloud.</p>
+
+<p>"You have much to learn. My friend and myself do not play for results
+of that antiquated kind. We seek in chess the wonderful, the
+whimsical, the weird. Did you ever see a position like that?"</p>
+
+<p>I inwardly congratulated myself that I never had.</p>
+
+<p>"That position, sir, materializes the sinuous evolvements and
+syncretic, synthetic, and synchronous concatenations of two cerebral
+individualities. It is the product of an amphoteric and intercalatory
+interchange of&mdash;"</p>
+
+<p>"Have you seen the evening paper, sir?" interrupted the man opposite,
+holding out a newspaper. I noticed on the margin beside his thumb some
+pencilled writing. Thanking him, I took the paper and read&mdash;"Insane,
+but quite harmless. He is in my charge."</p>
+
+<p>After that I let the poor fellow run on in his wild way until both got
+out at the next station.</p>
+
+<p>But that queer position became fixed indelibly in my mind, with
+Black's last move 43. K to Kt 8; and a short time afterwards I found
+it actually possible to arrive at such a position in forty-three
+moves. Can the reader construct such a sequence? How did White get his
+rooks and king's bishop into their present positions, considering
+Black can never have moved his king's bishop? No odds were given, and
+every move was perfectly legitimate.</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="MEASURING_WEIGHING_AND_PACKING_PUZZLES" id="MEASURING_WEIGHING_AND_PACKING_PUZZLES"></a><a href="#CONTENTS">MEASURING, WEIGHING, AND PACKING PUZZLES.</a></h2>
+
+<p class='center'>
+"Measure still for measure."<br />
+<span style="margin-left: 8em;"><i>Measure for Measure</i>, v. 1.</span><br />
+</p>
+
+
+<p>Apparently the first printed puzzle involving the measuring of a given
+quantity of liquid by pouring from one vessel to others of known
+capacity was that propounded by Niccola Fontana, better known as
+"Tartaglia" (the stammerer), 1500-1559. It consists in dividing 24 oz.
+of valuable balsam into three equal parts, the only measures available
+being vessels holding 5, 11, and 13 ounces respectively. There are
+many different solutions to this puzzle in six manipulations, or
+pourings from one vessel to another. Bachet de M&eacute;ziriac reprinted this
+and other of Tartaglia's puzzles in his <i>Probl&egrave;mes plaisans et
+d&eacute;lectables</i> (1612). It is the general opinion that puzzles of this
+class can only be solved by trial, but I think formul&aelig; can be
+constructed for the solution generally of certain related cases. It is
+a practically unexplored field for investigation.</p>
+
+<p>The classic weighing problem is, of course, that proposed by Bachet.
+It entails the determination of the least number of weights that would
+serve to weigh any integral number of pounds from 1 lb. to 40 lbs.
+inclusive, when we are allowed to put a weight in either of the two
+pans. The answer is 1, 3, 9, and 27 lbs. Tartaglia had previously
+propounded the same puzzle with the condition that the weights may
+only be placed in one pan. The answer in that case is 1, 2, 4, 8, 16,
+32 lbs. Major MacMahon has solved the problem quite generally. A full
+account will be found in Ball's <i>Mathematical Recreations</i> (5th
+edition).</p>
+
+<p>Packing puzzles, in which we are required to pack a maximum number of
+articles of given dimensions into a box of known dimensions, are, I
+believe, of quite recent introduction. At least I cannot recall any
+example in the books of the old writers. One would rather expect to
+find in the toy shops the idea presented as a mechanical puzzle, but I
+do not think I have ever seen such a thing. The nearest approach to it
+would appear to be the puzzles of the jig-saw character, where there
+is only one depth of the pieces to be adjusted.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_362_THE_WASSAIL_BOWL" id="X_362_THE_WASSAIL_BOWL"></a><a href="#X_362_THE_WASSAIL_BOWLa"><b>362.&mdash;THE WASSAIL BOWL.</b></a></p>
+
+<p>One Christmas Eve three Weary Willies came into possession of what was
+to them a veritable wassail bowl, in the form of a small barrel,
+containing exactly six quarts of fine ale. One of the men possessed a
+five-pint jug and another a three-pint jug, and the problem for them
+was to divide the liquor equally amongst them without waste. Of
+course, they are not to use any other vessels or measures. If you can
+show how it was to be done at all, then try to find the way that
+requires the fewest possible manipulations, every separate pouring
+from one vessel to another, or down a man's throat, counting as a
+manipulation.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_363_THE_DOCTORS_QUERY" id="X_363_THE_DOCTORS_QUERY"></a><a href="#X_363_THE_DOCTORS_QUERYa"><b>363.&mdash;THE DOCTOR'S QUERY.</b></a></p>
+
+<p>"A curious little point occurred to me in my dispensary this morning,"
+said a doctor. "I had a bottle containing ten ounces of spirits of
+wine, and another bottle containing ten ounces of water. I poured a
+quarter of an ounce of spirits into the water and shook them up
+together. The mixture was then clearly forty to one. Then I poured
+back a quarter-ounce of the mixture, so that the two bottles should
+again each contain the same quantity of fluid. What proportion of
+spirits to water did the spirits of wine bottle then contain?"</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_364_THE_BARREL_PUZZLE" id="X_364_THE_BARREL_PUZZLE"></a><a href="#X_364_THE_BARREL_PUZZLEa"><b>364.&mdash;THE BARREL PUZZLE.</b></a></p>
+
+<p>The men in the illustration are disputing over the liquid contents of
+a barrel. What the particular liquid is it is impossible to say, for
+we are unable to look into the barrel; so we will call it water. One
+man says that the barrel is more than half full, while the other
+insists that it is not half full. What is their easiest way of
+settling the point? It is not necessary to use stick, string, or
+implement of any kind for measuring. I <span class='pagenum'>Pg 110<a name="Page_110" id="Page_110"></a></span>give this merely as one of the
+simplest possible examples of the value of ordinary sagacity in the
+solving of puzzles. What are apparently very difficult problems may
+frequently be solved in a similarly easy manner if we only use a
+little common sense.</p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/q364.png" width="500" height="451" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_365_NEW_MEASURING_PUZZLE" id="X_365_NEW_MEASURING_PUZZLE"></a><a href="#X_365_NEW_MEASURING_PUZZLEa"><b>365.&mdash;NEW MEASURING PUZZLE.</b></a></p>
+
+<p>Here is a new poser in measuring liquids that will be found
+interesting. A man has two ten-quart vessels full of wine, and a
+five-quart and a four-quart measure. He wants to put exactly three
+quarts into each of the two measures. How is he to do it? And how many
+manipulations (pourings from one vessel to another) do you require? Of
+course, waste of wine, tilting, and other tricks are not allowed.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_366_THE_HONEST_DAIRYMAN" id="X_366_THE_HONEST_DAIRYMAN"></a><a href="#X_366_THE_HONEST_DAIRYMANa"><b>366.&mdash;THE HONEST DAIRYMAN.</b></a></p>
+
+<p>An honest dairyman in preparing his milk for public consumption
+employed a can marked B, containing milk, and a can marked A,
+containing water. From can A he poured enough to double the contents
+of can B. Then he poured from can B into can A enough to double its
+contents. Then he finally poured from can A into can B until their
+contents were exactly equal. After these operations he would send the
+can A to London, and the puzzle is to discover what are the relative
+proportions of milk and water that he provides for the Londoners'
+breakfast-tables. Do they get equal proportions of milk and water&mdash;or
+two parts of milk and one of water&mdash;or what? It is an interesting
+question, though, curiously enough, we are not told how much milk or
+water he puts into the cans at the start of his operations.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_367_WINE_AND_WATER" id="X_367_WINE_AND_WATER"></a><a href="#X_367_WINE_AND_WATERa"><b>367.&mdash;WINE AND WATER.</b></a></p>
+
+<p>Mr. Goodfellow has adopted a capital idea of late. When he gives a
+little dinner party and the time arrives to smoke, after the departure
+of the ladies, he sometimes finds that the conversation is apt to
+become too political, too personal, too slow, or too scandalous. Then
+he always manages to introduce to the company some new poser that he
+has secreted up his sleeve for the occasion. This invariably results
+in no end of interesting discussion and debate, and puts everybody in
+a good humour.</p>
+
+<p>Here is a little puzzle that he propounded the other night, and it is
+extraordinary how the company differed in their answers. He filled a
+wine-glass half full of wine, and another glass twice the size
+one-third full of wine. Then he filled up each glass with water and
+emptied the contents of both into a tumbler. "Now," he said, "what
+part of the mixture is wine and what part water?" Can you give the
+correct answer?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_368_THE_KEG_OF_WINE" id="X_368_THE_KEG_OF_WINE"></a><a href="#X_368_THE_KEG_OF_WINEa"><b>368.&mdash;THE KEG OF WINE.</b></a></p>
+
+<p>Here is a curious little problem. A man had a ten-gallon keg full of
+wine and a jug. One day he drew off a jugful of wine and filled up the
+keg with water. Later on, when the wine and water had got thoroughly
+mixed, he drew off <span class='pagenum'>Pg 111<a name="Page_111" id="Page_111"></a></span>another jugful and again filled up the keg with
+water. It was then found that the keg contained equal proportions of
+wine and water. Can you find from these facts the capacity of the jug?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_369_MIXING_THE_TEA" id="X_369_MIXING_THE_TEA"></a><a href="#X_369_MIXING_THE_TEAa"><b>369.&mdash;MIXING THE TEA.</b></a></p>
+
+<p>"Mrs. Spooner called this morning," said the honest grocer to his
+assistant. "She wants twenty pounds of tea at 2<i>s</i>. 4�<i>d</i>. per lb. Of
+course we have a good 2<i>s</i>. 6<i>d</i>. tea, a slightly inferior at 2<i>s</i>. 3<i>d</i>., and
+a cheap Indian at 1<i>s</i>. 9<i>d</i>., but she is very particular always about her
+prices."</p>
+
+<p>"What do you propose to do?" asked the innocent assistant.</p>
+
+<p>"Do?" exclaimed the grocer. "Why, just mix up the three teas in
+different proportions so that the twenty pounds will work out fairly
+at the lady's price. Only don't put in more of the best tea than you
+can help, as we make less profit on that, and of course you will use
+only our complete pound packets. Don't do any weighing."</p>
+
+<p>How was the poor fellow to mix the three teas? Could you have shown
+him how to do it?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_370_A_PACKING_PUZZLE" id="X_370_A_PACKING_PUZZLE"></a><a href="#X_370_A_PACKING_PUZZLEa"><b>370.&mdash;A PACKING PUZZLE.</b></a></p>
+
+<p>As we all know by experience, considerable ingenuity is often
+required in packing articles into a box if space is not to be unduly
+wasted. A man once told me that he had a large number of iron balls, all
+exactly two inches in diameter, and he wished to pack as many of these
+as possible into a rectangular box 24<sup>9</sup>/<sub>10</sub> inches long, 22<sup>4</sup>/<sub>5</sub> inches
+wide, and 14 inches deep. Now, what is the greatest number of the balls
+that he could pack into that box?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_371_GOLD_PACKING_IN_RUSSIA" id="X_371_GOLD_PACKING_IN_RUSSIA"></a><a href="#X_371_GOLD_PACKING_IN_RUSSIAa"><b>371.&mdash;GOLD PACKING IN RUSSIA.</b></a></p>
+
+<p>The editor of the <i>Times</i> newspaper was invited by a high Russian
+official to inspect the gold stored in reserve at St. Petersburg, in
+order that he might satisfy himself that it was not another "Humbert
+safe." He replied that it would be of no use whatever, for although
+the gold might appear to be there, he would be quite unable from a
+mere inspection to declare that what he saw was really gold. A
+correspondent of the <i>Daily Mail</i> thereupon took up the challenge,
+but, although he was greatly impressed by what he saw, he was
+compelled to confess his incompetence (without emptying and counting
+the contents of every box and sack, and assaying every piece of gold)
+to give any assurance on the subject. In presenting the following
+little puzzle, I wish it to be also understood that I do not guarantee
+the real existence of the gold, and the point is not at all material
+to our purpose. Moreover, if the reader says that gold is not usually
+"put up" in slabs of the dimensions that I give, I can only claim
+problematic licence.</p>
+
+<p>Russian officials were engaged in packing 800 gold slabs, each
+measuring 12� inches long, 11 inches wide, and 1 inch deep. What
+are the interior dimensions of a box of equal length and width, and
+necessary depth, that will exactly contain them without any space
+being left over? Not more than twelve slabs may be laid on edge,
+according to the rules of the government. It is an interesting little
+problem in packing, and not at all difficult.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_372_THE_BARRELS_OF_HONEY" id="X_372_THE_BARRELS_OF_HONEY"></a><a href="#X_372_THE_BARRELS_OF_HONEYa"><b>372.&mdash;THE BARRELS OF HONEY.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q372.png" width="400" height="332" alt="" title="" />
+</div>
+
+<p>Once upon a time there was an aged merchant of Bagdad who was much
+respected by all who knew him. He had three sons, and it was a rule of
+his life to treat them all exactly alike. Whenever one received a
+present, the other two were each given one of equal value. One day
+this worthy man fell sick and died, bequeathing all his possessions to
+his three sons in equal shares.</p>
+
+<p>The only difficulty that arose was over the stock of honey. There were
+exactly twenty-one barrels. The old man had left instructions that not
+only should every son receive an equal quantity of honey, but should
+receive exactly the same number of barrels, and that no honey should
+be transferred from barrel to barrel on account of the waste involved.
+Now, as seven of these barrels were full of honey, seven were
+half-full, and seven were empty, this was found to be quite a puzzle,
+especially as each brother objected to taking more than four barrels
+of, the same description&mdash;full, half-full, or empty. Can you show how
+they succeeded in making a correct division of the property?</p>
+
+
+
+<hr style="width: 65%;" /><p><span class='pagenum'>Pg 112<a name="Page_112" id="Page_112"></a></span></p>
+<h2><a name="CROSSING_RIVER_PROBLEMS" id="CROSSING_RIVER_PROBLEMS"></a><a href="#CONTENTS">CROSSING RIVER PROBLEMS</a></h2>
+
+<p class='center'>
+"My boat is on the shore."<br />
+<span style="margin-left: 8em;">BYRON.</span><br />
+</p>
+
+
+<p>This is another medi&aelig;val class of puzzles. Probably the earliest
+example was by Abbot Alcuin, who was born in Yorkshire in 735 and died
+at Tours in 804. And everybody knows the story of the man with the
+wolf, goat, and basket of cabbages whose boat would only take one of
+the three at a time with the man himself. His difficulties arose from
+his being unable to leave the wolf alone with the goat, or the goat
+alone with the cabbages. These puzzles were considered by Tartaglia
+and Bachet, and have been later investigated by Lucas, De Fonteney,
+Delannoy, Tarry, and others. In the puzzles I give there will be found
+one or two new conditions which add to the complexity somewhat. I also
+include a pulley problem that practically involves the same
+principles.</p>
+
+
+<hr style="width: 30%;" />
+<p><a name="X_373_CROSSING_THE_STREAM" id="X_373_CROSSING_THE_STREAM"></a><a href="#X_373_CROSSING_THE_STREAMa"><b>373.&mdash;CROSSING THE STREAM.</b></a></p>
+
+<p>During a country ramble Mr. and Mrs. Softleigh found themselves in a
+pretty little dilemma. They had to cross a stream in a small boat
+which was capable of carrying only 150 lbs. weight. But Mr. Softleigh
+and his wife each weighed exactly 150 lbs., and each of their sons
+weighed 75 lbs. And then there was the dog, who could not be induced
+on any terms to swim. On the principle of "ladies first," they at once
+sent Mrs. Softleigh over; but this was a stupid oversight, because she
+had to come back again with the boat, so nothing was gained by that
+operation. How did they all succeed in getting across? The reader will
+find it much easier than the Softleigh family did, for their greatest
+enemy could not have truthfully called them a brilliant
+quartette&mdash;while the dog was a perfect fool.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_374_CROSSING_THE_RIVER_AXE" id="X_374_CROSSING_THE_RIVER_AXE"></a><a href="#X_374_CROSSING_THE_RIVER_AXEa"><b>374.&mdash;CROSSING THE RIVER AXE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q374.png" width="400" height="456" alt="" title="" />
+</div>
+
+<p>Many years ago, in the days of the smuggler known as "Rob Roy of the
+West," a piratical band buried on the coast of South Devon a quantity
+of treasure which was, of course, abandoned by them in the usual
+inexplicable way. Some time afterwards its whereabouts was discovered
+by three countrymen, who visited the spot one night and divided the
+spoil between them, Giles taking treasure to the value of &pound;800, Jasper
+&pound;500 worth, and Timothy &pound;300 worth. In returning they had to cross the
+river Axe at a point where they had left a small boat in readiness.
+Here, however, was a difficulty they had not anticipated. The boat
+would only carry two men, or one man and a sack, and they had so
+little confidence in one another that no person could be left alone on
+the land or in the boat with more than his share of the spoil, though
+two persons (being a check on each other) might be left with more than
+their shares. The puzzle is to show how they got over the river in the
+fewest possible crossings, taking their treasure with them. No
+<span class='pagenum'>Pg 113<a name="Page_113" id="Page_113"></a></span>tricks, such as ropes, "flying bridges," currents, swimming, or
+similar dodges, may be employed.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_375_FIVE_JEALOUS_HUSBANDS" id="X_375_FIVE_JEALOUS_HUSBANDS"></a><a href="#X_375_FIVE_JEALOUS_HUSBANDSa"><b>375.&mdash;FIVE JEALOUS HUSBANDS.</b></a></p>
+
+<p>During certain local floods five married couples found themselves
+surrounded by water, and had to escape from their unpleasant position
+in a boat that would only hold three persons at a time. Every husband
+was so jealous that he would not allow his wife to be in the boat or
+on either bank with another man (or with other men) unless he was
+himself present. Show the quickest way of getting these five men and
+their wives across into safety.</p>
+
+<p>Call the men A, B, C, D, E, and their respective wives a, b, c, d, e.
+To go over and return counts as two crossings. No tricks such as
+ropes, swimming, currents, etc., are permitted.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_376_THE_FOUR_ELOPEMENTS" id="X_376_THE_FOUR_ELOPEMENTS"></a><a href="#X_376_THE_FOUR_ELOPEMENTSa"><b>376.&mdash;THE FOUR ELOPEMENTS.</b></a></p>
+
+<p>Colonel B&mdash;&mdash; was a widower of a very taciturn disposition. His
+treatment of his four daughters was unusually severe, almost cruel,
+and they not unnaturally felt disposed to resent it. Being charming
+girls with every virtue and many accomplishments, it is not surprising
+that each had a fond admirer. But the father forbade the young men to
+call at his house, intercepted all letters, and placed his daughters
+under stricter supervision than ever. But love, which scorns locks and
+keys and garden walls, was equal to the occasion, and the four youths
+conspired together and planned a general elopement.</p>
+
+<p>At the foot of the tennis lawn at the bottom of the garden ran the
+silver Thames, and one night, after the four girls had been safely
+conducted from a dormitory window to <i>terra firma</i>, they all crept
+softly down to the bank of the river, where a small boat belonging to
+the Colonel was moored. With this they proposed to cross to the
+opposite side and make their way to a lane where conveyances were
+waiting to carry them in their flight. Alas! here at the water's brink
+their difficulties already began.</p>
+
+<p>The young men were so extremely jealous that not one of them would
+allow his prospective bride to remain at any time in the company of
+another man, or men, unless he himself were present also. Now, the
+boat would only hold two persons, though it could, of course, be rowed
+by one, and it seemed impossible that the four couples would ever get
+across. But midway in the stream was a small island, and this seemed
+to present a way out of the difficulty, because a person or persons
+could be left there while the boat was rowed back or to the opposite
+shore. If they had been prepared for their difficulty they could have
+easily worked out a solution to the little poser at any other time.
+But they were now so hurried and excited in their flight that the
+confusion they soon got into was exceedingly amusing&mdash;or would have
+been to any one except themselves.</p>
+
+<p>As a consequence they took twice as long and crossed the river twice
+as often as was really necessary. Meanwhile, the Colonel, who was a
+very light sleeper, thought he heard a splash of oars. He quickly
+raised the alarm among his household, and the young ladies were found
+to be missing. Somebody was sent to the police-station, and a number
+of officers soon aided in the pursuit of the fugitives, who, in
+consequence of that delay in crossing the river, were quickly
+overtaken. The four girls returned sadly to their homes, and
+afterwards broke off their engagements in disgust.</p>
+
+<p>For a considerable time it was a mystery how the party of eight
+managed to cross the river in that little boat without any girl being
+ever left with a man, unless her betrothed was also present. The
+favourite method is to take eight counters or pieces of cardboard and
+mark them A, B, C, D, a, b, c, d, to represent the four men and their
+prospective brides, and carry them from one side of a table to the
+other in a matchbox (to represent the boat), a penny being placed in
+the middle of the table as the island.</p>
+
+<p>Readers are now asked to find the quickest method of getting the party
+across the river. How many passages are necessary from land to land?
+By "land" is understood either shore or island. Though the boat would
+not necessarily call at the island every time of crossing, the
+possibility of its doing so must be provided for. For example, it
+would not do for a man to be alone in the boat (though it were
+understood that he intended merely to cross from one bank to the
+opposite one) if there happened to be a girl alone on the island other
+than the one to whom he was engaged.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_377_STEALING_THE_CASTLE_TREASURE" id="X_377_STEALING_THE_CASTLE_TREASURE"></a><a href="#X_377_STEALING_THE_CASTLE_TREASUREa"><b>377.&mdash;STEALING THE CASTLE TREASURE.</b></a></p>
+
+<p>The ingenious manner in which a box of treasure, consisting
+principally of jewels and precious stones, was stolen from Gloomhurst
+Castle has been handed down as a tradition in the De Gourney family.
+The thieves consisted of a man, a youth, and a small boy, whose only
+mode of escape with the box of treasure was by means of a high window.
+Outside the window was fixed a pulley, over which ran a rope with a
+basket at each end. When one basket was on the ground the other was at
+the window. The rope was so disposed that the persons in the basket
+could neither help themselves by means of it nor receive help from
+others. In short, the only way the baskets could be used was by
+placing a heavier weight in one than in the other.</p>
+
+<p>Now, the man weighed 195 lbs., the youth 105 lbs., the boy 90 lbs.,
+and the box of treasure 75 lbs. The weight in the descending basket
+could not exceed that in the other by more than 15 lbs. without
+causing a descent so rapid as to be most dangerous to a human being,
+though it would not injure the stolen property. Only two persons, or
+one person and the treasure, could be placed in the same basket at one
+time. How did they all manage to escape and take the box of treasure
+with them?</p>
+
+<p><span class='pagenum'>Pg 114<a name="Page_114" id="Page_114"></a></span>The puzzle is to find the shortest way of performing the feat, which
+in itself is not difficult. Remember, a person cannot help himself by
+hanging on to the rope, the only way being to go down "with a bump,"
+with the weight in the other basket as a counterpoise.</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="PROBLEMS_CONCERNING_GAMES" id="PROBLEMS_CONCERNING_GAMES"></a><a href="#CONTENTS">PROBLEMS CONCERNING GAMES.</a></h2>
+
+<p class='center'>
+"The little pleasure of the game."<br />
+<span style="margin-left: 8em;">MATTHEW PRIOR.<br /></span>
+</p>
+
+<p>Every game lends itself to the propounding of a variety of puzzles.
+They can be made, as we have seen, out of the chessboard and the
+peculiar moves of the chess pieces. I will now give just a few
+examples of puzzles with playing cards and dominoes, and also go out
+of doors and consider one or two little posers in the cricket field,
+at the football match, and the horse race and motor-car race.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_378_DOMINOES_IN_PROGRESSION" id="X_378_DOMINOES_IN_PROGRESSION"></a><a href="#X_378_DOMINOES_IN_PROGRESSIONa"><b>378.&mdash;DOMINOES IN PROGRESSION.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q378.png" width="400" height="246" alt="" title="" />
+</div>
+
+<p>It will be seen that I have played six dominoes, in the illustration,
+in accordance with the ordinary rules of the game, 4 against 4, 1
+against 1, and so on, and yet the sum of the spots on the successive
+dominoes, 4, 5, 6, 7, 8, 9, are in arithmetical progression; that is,
+the numbers taken in order have a common difference of 1. In how many
+different ways may we play six dominoes, from an ordinary box of
+twenty-eight, so that the numbers on them may lie in arithmetical
+progression? We must always play from left to right, and numbers in
+decreasing arithmetical progression (such as 9, 8, 7, 6, 5, 4) are not
+admissible.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_379_THE_FIVE_DOMINOES" id="X_379_THE_FIVE_DOMINOES"></a><a href="#X_379_THE_FIVE_DOMINOESa"><b>379.&mdash;THE FIVE DOMINOES.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q379.png" width="400" height="128" alt="" title="" />
+</div>
+
+<p>Here is a new little puzzle that is not difficult, but will probably
+be found entertaining by my readers. It will be seen that the five
+dominoes are so arranged in proper sequence (that is, with 1 against
+1, 2 against 2, and so on), that the total number of pips on the two
+end dominoes is five, and the sum of the pips on the three dominoes in
+the middle is also five. There are just three other arrangements
+giving five for the additions. They are: &mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>(1&mdash;0)</td><td align='center'>(0&mdash;0)</td><td align='center'>(0&mdash;2)</td><td align='center'>(2&mdash;1)</td><td align='left'>(1&mdash;3)</td></tr>
+<tr><td align='center'>(4&mdash;0)</td><td align='center'>(0&mdash;0)</td><td align='center'>(0&mdash;2)</td><td align='center'>(2&mdash;1)</td><td align='left'>(1&mdash;0)</td></tr>
+<tr><td align='center'>(2&mdash;0)</td><td align='center'>(0&mdash;0)</td><td align='center'>(0&mdash;1)</td><td align='center'>(1&mdash;3)</td><td align='left'>(3&mdash;0)</td></tr>
+</table></div>
+
+<p>Now, how many similar arrangements are there of five dominoes that
+shall give six instead of five in the two additions?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_380_THE_DOMINO_FRAME_PUZZLE" id="X_380_THE_DOMINO_FRAME_PUZZLE"></a><a href="#X_380_THE_DOMINO_FRAME_PUZZLEa"><b>380.&mdash;THE DOMINO FRAME PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q380.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p>It will be seen in the illustration that the full set of twenty-eight
+dominoes is arranged in the form of a square frame, with 6 against 6,
+2 against 2, blank against blank, and so on, as in the game. It will
+be found that the pips in the top row and left-hand column both add up
+44. The pips in the other two sides sum to 59 and 32 respectively. The
+puzzle is to rearrange the dominoes in the same form so that all of
+the four sides shall sum to 44. Remember that the dominoes must be
+correctly placed one against another as in the game.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_381_THE_CARD_FRAME_PUZZLE" id="X_381_THE_CARD_FRAME_PUZZLE"></a><a href="#X_381_THE_CARD_FRAME_PUZZLEa"><b>381.&mdash;THE CARD FRAME PUZZLE.</b></a></p>
+
+<p>In the illustration we have a frame constructed from the ten playing
+cards, ace to ten of diamonds. The children who made it wanted the
+pips on all four sides to add up alike, but they failed in their
+attempt and gave it up as impossible. It will be seen that the pips in
+the <span class='pagenum'>Pg 115<a name="Page_115" id="Page_115"></a></span>top row, the bottom row, and the left-hand side all add up 14,
+but the right-hand side sums to 23. Now, what they were trying to do
+is quite possible. Can you rearrange the ten cards in the same
+formation so that all four sides shall add up alike? Of course they
+need not add up 14, but any number you choose to select.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q381.png" width="400" height="510" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_382_THE_CROSS_OF_CARDS" id="X_382_THE_CROSS_OF_CARDS"></a><a href="#X_382_THE_CROSS_OF_CARDSa"><b>382.&mdash;THE CROSS OF CARDS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q382.png" width="400" height="430" alt="" title="" />
+</div>
+
+<p>In this case we use only nine cards&mdash;the ace to nine of diamonds. The
+puzzle is to arrange them in the form of a cross, exactly in the way
+shown in the illustration, so that the pips in the vertical bar and in
+the horizontal bar add up alike. In the example given it will be found
+that both directions add up 23. What I want to know is, how many
+different ways are there of rearranging the cards in order to bring
+about this result? It will be seen that, without affecting the
+solution, we may exchange the 5 with the 6, the 5 with the 7, the 8
+with the 3, and so on. Also we may make the horizontal and the
+vertical bars change places. But such obvious manipulations as these
+are not to be regarded as different solutions. They are all mere
+variations of one fundamental solution. Now, how many of these
+fundamentally different solutions are there? The pips need not, of
+course, always add up 23.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_383_THE_T_CARD_PUZZLE" id="X_383_THE_T_CARD_PUZZLE"></a><a href="#X_383_THE_T_CARD_PUZZLEa"><b>383.&mdash;THE "T" CARD PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q383.png" width="400" height="362" alt="" title="" />
+</div>
+
+<p>An entertaining little puzzle with cards is to take the nine cards of
+a suit, from ace to nine inclusive, and arrange them in the form of
+the letter "T," as shown in the illustration, so that the pips in the
+horizontal line shall count the same as those in the column. In the
+example given they add up twenty-three both ways. Now, it is quite
+easy to get a single correct arrangement. The puzzle is to discover in
+just how many different ways it may be done. Though the number is
+high, the solution is not really difficult if we attack the puzzle in
+the right manner. The reverse way obtained by reflecting the
+illustration in a mirror we will not count as different, but all other
+changes in the relative positions of the cards will here count. How
+many different ways are there?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_384_CARD_TRIANGLES" id="X_384_CARD_TRIANGLES"></a><a href="#X_384_CARD_TRIANGLESa"><b>384.&mdash;CARD TRIANGLES.</b></a></p>
+
+<p>Here you pick out the nine cards, ace to nine of diamonds, and arrange
+them in the form of a triangle, exactly as shown in the illustration,
+so that the pips add up the same on the three sides. In the example
+given it will be seen that they sum to 20 on each side, but the
+particular number is of no importance so long as it is the same on all
+three sides. The puzzle <span class='pagenum'>Pg 116<a name="Page_116" id="Page_116"></a></span>is to find out in just how many different
+ways this can be done.</p>
+
+<p>If you simply turn the cards round so that one of the other two sides
+is nearest to you this will not count as different, for the order will
+be the same. Also, if you make the 4, 9, 5 change places with the 7,
+3, 8, and at the same time exchange the 1 and the 6, it will not be
+different. But if you only change the 1 and the 6 it will be
+different, because the order round the triangle is not the same. This
+explanation will prevent any doubt arising as to the conditions.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q384.png" width="400" height="389" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_385_STRAND_PATIENCE" id="X_385_STRAND_PATIENCE"></a><a href="#X_385_STRAND_PATIENCEa"><b>385.&mdash;"STRAND" PATIENCE.</b></a></p>
+
+<p>The idea for this came to me when considering the game of Patience
+that I gave in the <i>Strand Magazine</i> for December, 1910, which has
+been reprinted in Ernest Bergholt's <i>Second Book of Patience Games</i>,
+under the new name of "King Albert."</p>
+
+<p>Make two piles of cards as follows: 9 D, 8 S, 7 D, 6 S, 5 D, 4 S, 3
+D, 2 S, 1 D, and 9 H, 8 C, 7 H, 6 C, 5 H, 4 C, 3 H, 2 C, 1 H, with the
+9 of diamonds at the bottom of one pile and the 9 of hearts at the
+bottom of the other. The point is to exchange the spades with the
+clubs, so that the diamonds and clubs are still in numerical order in
+one pile and the hearts and spades in the other. There are four vacant
+spaces in addition to the two spaces occupied by the piles, and any
+card may be laid on a space, but a card can only be laid on another of
+the next higher value&mdash;an ace on a two, a two on a three, and so on.
+Patience is required to discover the shortest way of doing this. When
+there are four vacant spaces you can pile four cards in seven moves,
+with only three spaces you can pile them in nine moves, and with two
+spaces you cannot pile more than two cards. When you have a grasp of
+these and similar facts you will be able to remove a number of cards
+bodily and write down 7, 9, or whatever the number of moves may be.
+The gradual shortening of play is fascinating, and first attempts are
+surprisingly lengthy.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_386_A_TRICK_WITH_DICE" id="X_386_A_TRICK_WITH_DICE"></a><a href="#X_386_A_TRICK_WITH_DICEa"><b>386.&mdash;A TRICK WITH DICE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q386.png" width="400" height="151" alt="" title="" />
+</div>
+
+<p>Here is a neat little trick with three dice. I ask you to throw the
+dice without my seeing them. Then I tell you to multiply the points of
+the first die by 2 and add 5; then multiply the result by 5 and add
+the points of the second die; then multiply the result by 10 and add
+the points of the third die. You then give me the total, and I can at
+once tell you the points thrown with the three dice. How do I do it?
+As an example, if you threw 1, 3, and 6, as in the illustration, the
+result you would give me would be 386, from which I could at once say
+what you had thrown.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_387_THE_VILLAGE_CRICKET_MATCH" id="X_387_THE_VILLAGE_CRICKET_MATCH"></a><a href="#X_387_THE_VILLAGE_CRICKET_MATCHa"><b>387.&mdash;THE VILLAGE CRICKET MATCH.</b></a></p>
+
+<p>In a cricket match, Dingley Dell <i>v</i>. All Muggleton, the latter had
+the first innings. Mr. Dumkins and Mr. Podder were at the wickets,
+when the wary Dumkins made a splendid late cut, and Mr. Podder called
+on him to run. Four runs were apparently completed, but the vigilant
+umpires at each end called, "three short," making six short runs in
+all. What number did Mr. Dumkins score? When Dingley Dell took their
+turn at the wickets their champions were Mr. Luffey and Mr. Struggles.
+The latter made a magnificent off-drive, and invited his colleague to
+"come along," with the result that the observant spectators applauded
+them for what was supposed to have been three sharp runs. But the
+umpires declared that there had been two short runs at each end&mdash;four
+in all. To what extent, if any, did this man&#339;uvre increase Mr.
+Struggles's total?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_388_SLOW_CRICKET" id="X_388_SLOW_CRICKET"></a><a href="#X_388_SLOW_CRICKETa"><b>388.&mdash;SLOW CRICKET.</b></a></p>
+
+<p>In the recent county match between Wessex and Nincomshire the former
+team were at the wickets all day, the last man being put out a few
+minutes before the time for drawing stumps. The play was so slow that
+most of the spectators were fast asleep, and, on being awakened by one
+of the officials clearing the ground, we learnt that two men had been
+put out leg-before-wicket for a combined score of 19 runs; four men
+were caught for a combined score or 17 runs; one man was run out for a
+duck's egg; and the others were all bowled for 3 runs each. There were
+no extras. We were not told which of the men was the captain, but he
+made exactly 15 more than the average of his team. What was the
+captain's score?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_389_THE_FOOTBALL_PLAYERS" id="X_389_THE_FOOTBALL_PLAYERS"></a><a href="#X_389_THE_FOOTBALL_PLAYERSa"><b>389.&mdash;THE FOOTBALL PLAYERS.</b></a></p>
+
+<p>"It is a glorious game!" an enthusiast was heard to exclaim. "At the
+close of last season, <span class='pagenum'>Pg 117<a name="Page_117" id="Page_117"></a></span>of the footballers of my acquaintance four had
+broken their left arm, five had broken their right arm, two had the
+right arm sound, and three had sound left arms." Can you discover from
+that statement what is the smallest number of players that the speaker
+could be acquainted with?</p>
+
+<p>It does not at all follow that there were as many as fourteen men,
+because, for example, two of the men who had broken the left arm might
+also be the two who had sound right arms.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_390_THE_HORSE-RACE_PUZZLE" id="X_390_THE_HORSE-RACE_PUZZLE"></a><a href="#X_390_THE_HORSE-RACE_PUZZLEa"><b>390.&mdash;THE HORSE-RACE PUZZLE.</b></a></p>
+
+<p>There are no morals in puzzles. When we are solving the old puzzle of
+the captain who, having to throw half his crew overboard in a storm,
+arranged to draw lots, but so placed the men that only the Turks were
+sacrificed, and all the Christians left on board, we do not stop to
+discuss the questionable morality of the proceeding. And when we are
+dealing with a measuring problem, in which certain thirsty pilgrims
+are to make an equitable division of a barrel of beer, we do not
+object that, as total abstainers, it is against our conscience to have
+anything to do with intoxicating liquor. Therefore I make no apology
+for introducing a puzzle that deals with betting.</p>
+
+<p>Three horses&mdash;Acorn, Bluebottle, and Capsule&mdash;start in a race. The
+odds are 4 to 1, Acorn; 3 to 1, Bluebottle; 2 to 1, Capsule. Now, how
+much must I invest on each horse in order to win &pound;13, no matter which
+horse comes in first? Supposing, as an example, that I betted &pound;5 on
+each horse. Then, if Acorn won, I should receive &pound;20 (four times &pound;5),
+and have to pay &pound;5 each for the other two horses; thereby winning &pound;10.
+But it will be found that if Bluebottle was first I should only win
+&pound;5, and if Capsule won I should gain nothing and lose nothing. This
+will make the question perfectly clear to the novice, who, like
+myself, is not interested in the calling of the fraternity who profess
+to be engaged in the noble task of "improving the breed of horses."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_391_THE_MOTOR-CAR_RACE" id="X_391_THE_MOTOR-CAR_RACE"></a><a href="#X_391_THE_MOTOR-CAR_RACEa"><b>391.&mdash;THE MOTOR-CAR RACE.</b></a></p>
+
+<p>Sometimes a quite simple statement of fact, if worded in an unfamiliar
+manner, will cause considerable perplexity. Here is an example, and it
+will doubtless puzzle some of my more youthful readers just a little.
+I happened to be at a motor-car race at Brooklands, when one spectator
+said to another, while a number of cars were whirling round and round
+the circular track:&mdash;</p>
+
+<p>"There's Gogglesmith&mdash;that man in the white car!"</p>
+
+<p>"Yes, I see," was the reply; "but how many cars are running in this
+race?"</p>
+
+<p>Then came this curious rejoinder:&mdash;</p>
+
+<p>"One-third of the cars in front of Gogglesmith added to three-quarters
+of those behind him will give you the answer."</p>
+
+<p>Now, can you tell how many cars were running in the race?</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="PUZZLE_GAMES" id="PUZZLE_GAMES"></a><a href="#CONTENTS">PUZZLE GAMES.</a></h2>
+
+<p class='center'>
+"He that is beaten may be said<br />
+To lie in honour's truckle bed."<br />
+<span style="margin-left: 8em;">HUDIBRAS.<br /></span>
+</p>
+
+<p>It may be said generally that a game is a contest of skill for two or
+more persons, into which we enter either for amusement or to win a
+prize. A puzzle is something to be done or solved by the individual.
+For example, if it were possible for us so to master the complexities
+of the game of chess that we could be assured of always winning with
+the first or second move, as the case might be, or of always drawing,
+then it would cease to be a game and would become a puzzle. Of course
+among the young and uninformed, when the correct winning play is not
+understood, a puzzle may well make a very good game. Thus there is no
+doubt children will continue to play "Noughts and Crosses," though I
+have shown (No. 109, "<i>Canterbury Puzzles</i>") that between two players
+who both thoroughly understand the play, every game should be drawn.
+Neither player could ever win except through the blundering of his
+opponent. But I am writing from the point of view of the student of
+these things.</p>
+
+<p>The examples that I give in this class are apparently games, but,
+since I show in every case how one player may win if he only play
+correctly, they are in reality puzzles. Their interest, therefore,
+lies in attempting to discover the leading method of play.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_392_THE_PEBBLE_GAME" id="X_392_THE_PEBBLE_GAME"></a><a href="#X_392_THE_PEBBLE_GAMEa"><b>392.&mdash;THE PEBBLE GAME.</b></a></p>
+
+<p>Here is an interesting little puzzle game that I used to play with an
+acquaintance on the beach at Slocomb-on-Sea. Two players place an odd
+number of pebbles, we will say fifteen, between them. Then each takes
+in turn one, two, or three pebbles (as he chooses), and the winner is
+the one who gets the odd number. Thus, if you get seven and your
+opponent eight, you win. If you get six and he gets nine, he wins.
+Ought the first or second player to win, and how? When you have
+settled the question with fifteen pebbles try again with, say,
+thirteen.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_393_THE_TWO_ROOKS" id="X_393_THE_TWO_ROOKS"></a><a href="#X_393_THE_TWO_ROOKSa"><b>393.&mdash;THE TWO ROOKS.</b></a></p>
+
+<p>This is a puzzle game for two players. Each player has a single rook.
+The first player places his rook on any square of the board that he
+may choose to select, and then the second player does the same. They
+now play in turn, the point of each play being to capture the
+opponent's rook. But in this game you cannot play through a line of
+attack without being captured. That is to say, if in the diagram it is
+Black's turn to <span class='pagenum'>Pg 118<a name="Page_118" id="Page_118"></a></span>play, he cannot move his rook to his king's knight's
+square, or to his king's rook's square, because he would enter the
+"line of fire" when passing his king's bishop's square. For the same
+reason he cannot move to his queen's rook's seventh or eighth squares.
+Now, the game can never end in a draw. Sooner or later one of the
+rooks must fall, unless, of course, both players commit the absurdity
+of not trying to win. The trick of winning is ridiculously simple when
+you know it. Can you solve the puzzle?</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q393.png" width="400" height="437" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_394_PUSS_IN_THE_CORNER" id="X_394_PUSS_IN_THE_CORNER"></a><a href="#X_394_PUSS_IN_THE_CORNERa"><b>394.&mdash;PUSS IN THE CORNER.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q394.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>This variation of the last puzzle is also played by two persons. One
+puts a counter on No. 6, and the other puts one on No. 55, and they
+play alternately by removing the counter to any other number in a
+line. If your opponent moves at any time on to one of the lines you
+occupy, or even crosses one of your lines, you immediately capture him
+and win. We will take an illustrative game.</p>
+
+<p>A moves from 55 to 52; B moves from 6 to 13; A advances to 23; B goes
+to 15; A retreats to 26; B retreats to 13; A advances to 21; B
+retreats to 2; A advances to 7; B goes to 3; A moves to 6; B must now
+go to 4; A establishes himself at 11, and B must be captured next move
+because he is compelled to cross a line on which A stands. Play this
+over and you will understand the game directly. Now, the puzzle part
+of the game is this: Which player should win, and how many moves are
+necessary?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_395_A_WAR_PUZZLE_GAME" id="X_395_A_WAR_PUZZLE_GAME"></a><a href="#X_395_A_WAR_PUZZLE_GAMEa"><b>395.&mdash;A WAR PUZZLE GAME.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q395.png" width="400" height="397" alt="" title="" />
+</div>
+
+<p>Here is another puzzle game. One player, representing the British
+general, places a counter at B, and the other player, representing the
+enemy, places his counter at E. The Britisher makes the first advance
+along one of the roads to the next town, then the enemy moves to one
+of his nearest towns, and so on in turns, until the British general
+gets into the same town as the enemy and captures him. Although each
+must always move along a road to the next town only, and the second
+player may do his utmost to avoid capture, the British general (as we
+should suppose, from the analogy of real life) must infallibly win.
+But how? That is the question.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_396_A_MATCH_MYSTERY" id="X_396_A_MATCH_MYSTERY"></a><a href="#X_396_A_MATCH_MYSTERYa"><b>396.&mdash;A MATCH MYSTERY.</b></a></p>
+
+<p>Here is a little game that is childishly simple in its conditions. But
+it is well worth investigation.</p>
+
+<p>Mr. Stubbs pulled a small table between himself and his friend, Mr.
+Wilson, and took a box of matches, from which he counted out thirty.</p>
+
+<p>"Here are thirty matches," he said. "I <span class='pagenum'>Pg 119<a name="Page_119" id="Page_119"></a></span>divide them into three unequal
+heaps. Let me see. We have 14, 11, and 5, as it happens. Now, the two
+players draw alternately any number from any one heap, and he who
+draws the last match loses the game. That's all! I will play with you,
+Wilson. I have formed the heaps, so you have the first draw."</p>
+
+<p>"As I can draw any number," Mr. Wilson said, "suppose I exhibit my
+usual moderation and take all the 14 heap."</p>
+
+<p>"That is the worst you could do, for it loses right away. I take 6
+from the 11, leaving two equal heaps of 5, and to leave two equal
+heaps is a certain win (with the single exception of 1, 1), because
+whatever you do in one heap I can repeat in the other. If you leave 4
+in one heap, I leave 4 in the other. If you then leave 2 in one heap,
+I leave 2 in the other. If you leave only 1 in one heap, then I take
+all the other heap. If you take all one heap, I take all but one in
+the other. No, you must never leave two heaps, unless they are equal
+heaps and more than 1, 1. Let's begin again."</p>
+
+<p>"Very well, then," said Mr. Wilson. "I will take 6 from the 14, and
+leave you 8, 11, 5."</p>
+
+<p>Mr. Stubbs then left 8, 11, 3; Mr. Wilson, 8, 5, 3; Mr. Stubbs, 6, 5,
+3; Mr. Wilson,4, 5, 3; Mr. Stubbs, 4, 5, 1; Mr. Wilson, 4, 3, 1; Mr.
+Stubbs, 2, 3, 1; Mr. Wilson, 2, 1, 1; which Mr. Stubbs reduced to 1,
+1, 1.</p>
+
+<p>"It is now quite clear that I must win," said Mr. Stubbs, because you
+must take 1, and then I take 1, leaving you the last match. You never
+had a chance. There are just thirteen different ways in which the
+matches may be grouped at the start for a certain win. In fact, the
+groups selected, 14, 11, 5, are a certain win, because for whatever
+your opponent may play there is another winning group you can secure,
+and so on and on down to the last match."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_397_THE_MONTENEGRIN_DICE_GAME" id="X_397_THE_MONTENEGRIN_DICE_GAME"></a><a href="#X_397_THE_MONTENEGRIN_DICE_GAMEa"><b>397.&mdash;THE MONTENEGRIN DICE GAME.</b></a></p>
+
+<p>It is said that the inhabitants of Montenegro have a little dice game
+that is both ingenious and well worth investigation. The two players
+first select two different pairs of odd numbers (always higher than 3)
+and then alternately toss three dice. Whichever first throws the dice
+so that they add up to one of his selected numbers wins. If they are
+both successful in two successive throws it is a draw and they try
+again. For example, one player may select 7 and 15 and the other 5 and
+13. Then if the first player throws so that the three dice add up 7 or
+15 he wins, unless the second man gets either 5 or 13 on his throw.</p>
+
+<p>The puzzle is to discover which two pairs of numbers should be
+selected in order to give both players an exactly even chance.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_398_THE_CIGAR_PUZZLE" id="X_398_THE_CIGAR_PUZZLE"></a><a href="#X_398_THE_CIGAR_PUZZLEa"><b>398.&mdash;THE CIGAR PUZZLE.</b></a></p>
+
+<p>I once propounded the following puzzle in a London club, and for a
+considerable period it absorbed the attention of the members. They
+could make nothing of it, and considered it quite impossible of
+solution. And yet, as I shall show, the answer is remarkably simple.</p>
+
+<p>Two men are seated at a square-topped table. One places an ordinary
+cigar (flat at one end, pointed at the other) on the table, then the
+other does the same, and so on alternately, a condition being that no
+cigar shall touch another. Which player should succeed in placing the
+last cigar, assuming that they each will play in the best possible
+manner? The size of the table top and the size of the cigar are not
+given, but in order to exclude the ridiculous answer that the table
+might be so diminutive as only to take one cigar, we will say that the
+table must not be less than 2 feet square and the cigar not more than
+4&frac12; inches long. With those restrictions you may take any dimensions
+you like. Of course we assume that all the cigars are exactly alike in
+every respect. Should the first player, or the second player, win?</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="MAGIC_SQUARE_PROBLEMS" id="MAGIC_SQUARE_PROBLEMS"></a><a href="#CONTENTS">MAGIC SQUARE PROBLEMS.</a></h2>
+
+<p class='center'>
+"By magic numbers."<br />
+<span style="margin-left: 8em;">CONGREVE, <i>The Mourning Bride.</i></span><br />
+</p>
+
+<p>This is a very ancient branch of mathematical puzzledom, and it has an
+immense, though scattered, literature of its own. In their simple form
+of consecutive whole numbers arranged in a square so that every
+column, every row, and each of the two long diagonals shall add up
+alike, these magic squares offer three main lines of investigation:
+Construction, Enumeration, and Classification. Of recent years many
+ingenious methods have been devised for the construction of magics,
+and the law of their formation is so well understood that all the
+ancient mystery has evaporated and there is no longer any difficulty
+in making squares of any dimensions. Almost the last word has been
+said on this subject. The question of the enumeration of all the
+possible squares of a given order stands just where it did over two
+hundred years ago. Everybody knows that there is only one solution for
+the third order, three cells by three; and Fr&eacute;nicle published in 1693
+diagrams of all the arrangements of the fourth order&mdash;880 in
+number&mdash;and his results have been verified over and over again. I may
+here refer to the general solution for this order, for numbers not
+necessarily consecutive, by E. Bergholt in <i>Nature</i>, May 26, 1910, as
+it is of the greatest importance to students of this subject. The
+enumeration of the examples of any higher order is a completely
+unsolved problem.</p>
+
+<p>As to classification, it is largely a matter of individual
+taste&mdash;perhaps an &aelig;sthetic question, for there is beauty in the law
+and order of numbers. A man once said that he divided the human race
+into two great classes: those who take snuff and those who do not. I
+am not <span class='pagenum'>Pg 120<a name="Page_120" id="Page_120"></a></span>sure that some of our classifications of magic squares are not
+almost as valueless. However, lovers of these things seem somewhat
+agreed that Nasik magic squares (so named by Mr. Frost, a student of
+them, after the town in India where he lived, and also called
+Diabolique and Pandiagonal) and Associated magic squares are of
+special interest, so I will just explain what these are for the
+benefit of the novice.</p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/msfig1.png" width="500" height="131" alt="" title="" />
+</div>
+
+<p>I published in <i>The Queen</i> for January 15, 1910, an article that would
+enable the reader to write out, if he so desired, all the 880 magics
+of the fourth order, and the following is the complete classification
+that I gave. The first example is that of a Simple square that fulfils
+the simple conditions and no more. The second example is a Semi-Nasik,
+which has the additional property that the opposite short diagonals of
+two cells each together sum to 34. Thus, 14&nbsp;+&nbsp;4&nbsp;+&nbsp;11&nbsp;+&nbsp;5&nbsp;=&nbsp;34 and 12 +
+6&nbsp;+&nbsp;13&nbsp;+&nbsp;3&nbsp;=&nbsp;34. The third example is not only Semi-Nasik but <span class='pagenum'>Pg 121<a name="Page_121" id="Page_121"></a></span>also
+Associated, because in it every number, if added to the number that is
+equidistant, in a straight line, from the centre gives 17. Thus, 1 +
+16, 2&nbsp;+&nbsp;15, 3&nbsp;+&nbsp;14, etc. The fourth example, considered the most
+"perfect" of all, is a Nasik. Here all the broken diagonals sum to 34.
+Thus, for example, 15&nbsp;+&nbsp;14&nbsp;+&nbsp;2&nbsp;+&nbsp;3, and 10&nbsp;+&nbsp;4&nbsp;+&nbsp;7&nbsp;+&nbsp;13, and 15&nbsp;+&nbsp;5 +
+2&nbsp;+&nbsp;12. As a consequence, its properties are such that if you repeat
+the square in all directions you may mark off a square, 4&nbsp;&times;&nbsp;4,
+wherever you please, and it will be magic.</p>
+
+<p>The following table not only gives a complete enumeration under the
+four forms described, but also a classification under the twelve
+graphic types indicated in the diagrams. The dots at the end of each
+line represent the relative positions of those complementary pairs, 1
++ 16, 2&nbsp;+&nbsp;15, etc., which sum to 17. For example, it will be seen that
+the first and second magic squares given are of Type VI., that the
+third square is of Type III., and that the fourth is of Type I.
+Edouard Lucas indicated these types, but he dropped exactly half of
+them and did not attempt the classification.</p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/msfig2.png" width="500" height="427" alt="" title="" />
+</div>
+
+<div class='center'>
+<table border="0" cellpadding="2" cellspacing="2" summary="">
+<tr><td align='center'>NASIK</td><td align='left'>(Type I.)</td><td>&nbsp;</td><td>&nbsp;</td><td align='right'>48</td></tr>
+<tr><td align='center'>SEMI-NASIK</td><td align='left'>(Type II., Transpositions of Nasik)</td><td>&nbsp;</td><td align='right'>48</td><td>&nbsp;</td></tr>
+<tr><td align='center'>"</td><td align='left'>(Type III., Associated)</td><td>&nbsp;</td><td align='right'>48</td><td>&nbsp;</td></tr>
+<tr><td align='center'>"</td><td align='left'>(Type IV.)</td><td align='right'>96</td><td>&nbsp;</td><td>&nbsp;</td></tr>
+<tr><td align='center'>"</td><td align='left'>(Type V.)</td><td align='right' class='bb'>96</td><td align='right'>192</td><td>&nbsp;</td></tr>
+<tr><td align='center'>"</td><td align='left'>(Type VI.)</td><td>&nbsp;</td><td align='right' class='bb'>96</td><td align='right'>384</td></tr>
+<tr><td align='center'>SIMPLE.</td><td align='left'>(Type VI.)</td><td>&nbsp;</td><td align='right'>208</td><td>&nbsp;</td></tr>
+<tr><td align='center'>"</td><td align='left'>(Type VII.)</td><td align='right'>56</td><td>&nbsp;</td><td>&nbsp;</td></tr>
+<tr><td align='center'>"</td><td align='left'>(Type VIII.)</td><td align='right'>56</td><td>&nbsp;</td><td>&nbsp;</td></tr>
+<tr><td align='center'>"</td><td align='left'>(Type IX.)</td><td align='right'>56</td><td>&nbsp;</td><td>&nbsp;</td></tr>
+<tr><td align='center'>"</td><td align='left'>(Type X.)</td><td align='right' class='bb'>56</td><td align='right'>224</td><td>&nbsp;</td></tr>
+<tr><td align='center'>"</td><td align='left'>(Type XI.)</td><td align='right'>8</td><td>&nbsp;</td><td>&nbsp;</td></tr>
+<tr><td align='center'>"</td><td align='left'>(Type XII.)</td><td align='right' class='bb'>8</td><td align='right' class='bb'>16</td><td align='right' class='bb'>448</td></tr>
+<tr><td align='center'></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td align='left' class='bb'>880</td></tr>
+</table></div>
+
+<p>It is hardly necessary to say that every one of these squares will
+produce seven others by mere reversals and reflections, which we do
+not count as different. So that there are 7,040 squares of this order,
+880 of which are fundamentally different.</p>
+
+<p>An infinite variety of puzzles may be made introducing new conditions
+into the magic square. In <i>The Canterbury Puzzles</i> I have given
+examples of such squares with coins, with postage stamps, with
+cutting-out conditions, and other tricks. I will now give a few
+variants involving further novel conditions.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_399_THE_TROUBLESOME_EIGHT" id="X_399_THE_TROUBLESOME_EIGHT"></a><a href="#X_399_THE_TROUBLESOME_EIGHTa"><b>399.&mdash;THE TROUBLESOME EIGHT.</b></a></p>
+
+<p>Nearly everybody knows that a "magic square" is an arrangement of
+numbers in the form of a square so that every row, every column, and
+each of the two long diagonals adds up alike. For example, you would
+find little difficulty in merely placing a different number in each of
+the nine cells in the illustration so that the rows, columns, and
+diagonals shall all add up 15. And at your first attempt you will
+probably find that you have an 8 in one of the corners. The puzzle is
+to construct the magic square, under the same conditions, with the 8
+in the position shown.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q399.png" width="400" height="333" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_400_THE_MAGIC_STRIPS" id="X_400_THE_MAGIC_STRIPS"></a><a href="#X_400_THE_MAGIC_STRIPSa"><b>400.&mdash;THE MAGIC STRIPS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q400.png" width="400" height="504" alt="" title="" />
+</div>
+
+<p>I happened to have lying on my table a number of strips of cardboard,
+with numbers printed on them from 1 upwards in numerical order. The
+idea suddenly came to me, as ideas have a way of unexpectedly coming,
+to make a little puzzle of this. I wonder whether many readers will
+arrive at the same solution that I did.</p>
+
+<p>Take seven strips of cardboard and lay them together as above. Then
+write on each of them the numbers 1, 2, 3, 4, 5, 6, 7, as shown, so
+that the numbers shall form seven rows and seven columns.</p>
+
+<p>Now, the puzzle is to cut these strips into the fewest possible pieces
+so that they may be placed together and form a magic square, the seven
+rows, seven columns, and two diagonals adding up the same number. No
+figures may <span class='pagenum'>Pg 122<a name="Page_122" id="Page_122"></a></span>be turned upside down or placed on their sides&mdash;that is,
+all the strips must lie in their original direction.</p>
+
+<p>Of course you could cut each strip into seven separate pieces, each
+piece containing a number, and the puzzle would then be very easy, but
+I need hardly say that forty-nine pieces is a long way from being the
+fewest possible.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_401_EIGHT_JOLLY_GAOL_BIRDS" id="X_401_EIGHT_JOLLY_GAOL_BIRDS"></a><a href="#X_401_EIGHT_JOLLY_GAOL_BIRDSa"><b>401.&mdash;EIGHT JOLLY GAOL BIRDS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q401.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>The illustration shows the plan of a prison of nine cells all
+communicating with one another by doorways. The eight prisoners have
+their numbers on their backs, and any one of them is allowed to
+exercise himself in whichever cell may happen to be vacant, subject to
+the rule that at no time shall two prisoners be in the same cell. The
+merry monarch in whose dominions the prison was situated offered them
+special comforts one Christmas Eve if, without breaking that rule,
+they could so place themselves that their numbers should form a magic
+square.</p>
+
+<p>Now, prisoner No. 7 happened to know a good deal about magic squares,
+so he worked out a scheme and naturally selected the method that was
+most expeditious&mdash;that is, one involving the fewest possible moves
+from cell to cell. But one man was a surly, obstinate fellow (quite
+unfit for the society of his jovial companions), and he refused to
+move out of his cell or take any part in the proceedings. But No. 7
+was quite equal to the emergency, and found that he could still do
+what was required in the fewest possible moves without troubling the
+brute to leave his cell. The puzzle is to show how he did it and,
+incidentally, to discover which prisoner was so stupidly obstinate.
+Can you find the fellow?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_402_NINE_JOLLY_GAOL_BIRDS" id="X_402_NINE_JOLLY_GAOL_BIRDS"></a><a href="#X_402_NINE_JOLLY_GAOL_BIRDSa"><b>402.&mdash;NINE JOLLY GAOL BIRDS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q402.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>Shortly after the episode recorded in the last puzzle occurred, a
+ninth prisoner was placed in the vacant cell, and the merry monarch
+then offered them all complete liberty on the following strange
+conditions. They were required so to rearrange themselves in the cells
+that their numbers formed a magic square without their movements
+causing any two of them ever to be in the same cell together, except
+that at the start one man was allowed to be placed on the shoulders of
+another man, and thus add their numbers together, and move as one man.
+For example, No. 8 might be placed on the shoulders of No. 2, and then
+they would move about together as 10. The reader should seek first to
+solve the puzzle in the fewest possible moves, and then see that the
+man who is burdened has the least possible amount of work to do.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_403_THE_SPANISH_DUNGEON" id="X_403_THE_SPANISH_DUNGEON"></a><a href="#X_403_THE_SPANISH_DUNGEONa"><b>403.&mdash;THE SPANISH DUNGEON.</b></a></p>
+
+<p>Not fifty miles from Cadiz stood in the middle ages a castle, all
+traces of which have for centuries disappeared. Among other
+interesting features, this castle contained a particularly unpleasant
+dungeon divided into sixteen cells, all communicating with one
+another, as shown in the illustration.</p>
+
+<p>Now, the governor was a merry wight, and very fond of puzzles withal.
+One day he went to the dungeon and said to the prisoners, "By my
+halidame!" (or its equivalent in Spanish) "you shall all be set free
+if you can solve this puzzle. You must so arrange yourselves in the
+sixteen cells that the numbers on your backs shall form a magic square
+in which every column, every row, and each of the two diagonals shall
+add up the same. Only remember this: that in no case may two of you
+ever be together in the same cell."</p>
+
+<p>One of the prisoners, after working at the problem for two or three
+days, with a piece of chalk, undertook to obtain the liberty of
+himself and his fellow-prisoners if they would follow his directions
+and move through the doorway <span class='pagenum'>Pg 123<a name="Page_123" id="Page_123"></a></span>from cell to cell in the order in which
+he should call out their numbers.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q403.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>He succeeded in his attempt, and, what is more remarkable, it would
+seem from the account of his method recorded in the ancient manuscript
+lying before me, that he did so in the fewest possible moves. The
+reader is asked to show what these moves were.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_404_THE_SIBERIAN_DUNGEONS" id="X_404_THE_SIBERIAN_DUNGEONS"></a><a href="#X_404_THE_SIBERIAN_DUNGEONSa"><b>404.&mdash;THE SIBERIAN DUNGEONS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q404.png" width="400" height="200" alt="" title="" />
+</div>
+
+<p>The above is a trustworthy plan of a certain Russian prison in
+Siberia. All the cells are numbered, and the prisoners are numbered
+the same as the cells they occupy. The prison diet is so fattening
+that these political prisoners are in perpetual fear lest, should
+their pardon arrive, they might not be able to squeeze themselves
+through the narrow doorways and get out. And of course it would be an
+unreasonable thing to ask any government to pull down the walls of a
+prison just to liberate the prisoners, however innocent they might be.
+Therefore these men take all the healthy exercise they can in order to
+retard their increasing obesity, and one of their recreations will
+serve to furnish us with the following puzzle.</p>
+
+<p>Show, in the fewest possible moves, how the sixteen men may form
+themselves into a magic square, so that the numbers on their backs
+shall add up the same in each of the four columns, four rows, and two
+diagonals without two prisoners having been at any time in the same
+cell together. I had better say, for the information of those who have
+not yet been made acquainted with these places, that it is a
+peculiarity of prisons that you are not allowed to go outside their
+walls. Any prisoner may go any distance that is possible in a single
+move.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_405_CARD_MAGIC_SQUARES" id="X_405_CARD_MAGIC_SQUARES"></a><a href="#X_405_CARD_MAGIC_SQUARESa"><b>405.&mdash;CARD MAGIC SQUARES.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q405.png" width="400" height="539" alt="" title="" />
+</div>
+
+<p>Take an ordinary pack of cards and throw out the twelve court cards.
+Now, with nine of the remainder (different suits are of no
+consequence) form the above magic square. It will be seen that the
+pips add up fifteen in every row in every column, and in each of the
+two long diagonals. The puzzle is with the remaining cards (without
+disturbing this arrangement) to form three more such magic squares, so
+that each of the four shall add up to a different sum. There will, of
+course, be four cards in the reduced pack that will not be used. These
+four may be any that you choose. It is not a difficult puzzle, but
+requires just a little thought.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_406_THE_EIGHTEEN_DOMINOES" id="X_406_THE_EIGHTEEN_DOMINOES"></a><a href="#X_406_THE_EIGHTEEN_DOMINOESa"><b>406.&mdash;THE EIGHTEEN DOMINOES.</b></a></p>
+
+<p>The illustration shows eighteen dominoes arranged in the form of a
+square so that the pips in every one of the six columns, six rows, and
+two long diagonals add up 13. This is the smallest summation possible
+with any selection of dominoes from an ordinary box of twenty-eight.
+The greatest possible summation is 23, and a solution for this number
+may be easily obtained by substituting for every number its complement
+to 6. Thus for every blank substitute a 6, for every 1 a 5, for every
+2 a 4, for <span class='pagenum'>Pg 124<a name="Page_124" id="Page_124"></a></span>3 a 3, for 4 a 2, for 5 a 1, and for 6 a blank. But the
+puzzle is to make a selection of eighteen dominoes and arrange them
+(in exactly the form shown) so that the summations shall be 18 in all
+the fourteen directions mentioned.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q406.png" width="400" height="393" alt="" title="" />
+</div>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="SUBTRACTING_MULTIPLYING_AND_DIVIDING_MAGICS" id="SUBTRACTING_MULTIPLYING_AND_DIVIDING_MAGICS"></a><a href="#CONTENTS">SUBTRACTING, MULTIPLYING, AND DIVIDING MAGICS.</a></h2>
+
+<p>Although the adding magic square is of such great antiquity, curiously
+enough the multiplying magic does not appear to have been mentioned
+until the end of the eighteenth century, when it was referred to
+slightly by one writer and then forgotten until I revived it in
+<i>Tit-Bits</i> in 1897. The dividing magic was apparently first discussed
+by me in <i>The Weekly Dispatch</i> in June 1898. The subtracting magic is
+here introduced for the first time. It will now be convenient to deal
+with all four kinds of magic squares together.</p>
+
+<div class="figcenter" style="width: 550px;">
+<img src="images/magicfig1.png" width="550" height="145" alt="" title="" />
+</div>
+
+<p>In these four diagrams we have examples in the third order of adding,
+subtracting, multiplying, and dividing squares. In the first the
+constant, 15, is obtained by the addition of the rows, columns, and
+two diagonals. In the second case you get the constant, 5, by
+subtracting the first number in a line from the second, and the result
+from the third. You can, of course, perform the operation in either
+direction; but, in order to avoid negative numbers, it is more
+convenient simply to deduct the middle number from the sum of the two
+extreme numbers. This is, in effect, the same thing. It will be seen
+that the constant of the adding square is <i>n</i> times that of the
+subtracting square derived from it, where <i>n</i> is the number of cells
+in the side of square. And the manner of derivation here is simply to
+reverse the two diagonals. Both squares are "associated"&mdash;a term I
+have explained in the introductory article to this department.</p>
+
+<p>The third square is a multiplying magic. The constant, 216, is
+obtained by multiplying together the three numbers in any line. It is
+"associated" by multiplication, instead of by addition. It is here
+necessary to remark that in an adding square it is not essential that
+the nine numbers should be consecutive. Write down any nine numbers in
+this way&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>1</td><td align='right'>3</td><td align='right'>5</td></tr>
+<tr><td align='right'>4</td><td align='right'>6</td><td align='right'>8</td></tr>
+<tr><td align='right'>7</td><td align='right'>9</td><td align='right'>11</td></tr>
+</table></div>
+
+<p>so that the horizontal differences are all alike and the vertical
+differences also alike (here 2 and 3), and these numbers will form an
+adding magic square. By making the differences 1 and 3 we, of course,
+get consecutive numbers&mdash;a particular case, and nothing more. Now, in
+the case of the multiplying square we must take these numbers in
+geometrical instead of arithmetical progression, thus&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>1</td><td align='right'>3</td><td align='right'>9</td></tr>
+<tr><td align='right'>2</td><td align='right'>6</td><td align='right'>18</td></tr>
+<tr><td align='right'>4</td><td align='right'>12</td><td align='right'>36</td></tr>
+</table></div>
+
+<p>Here each successive number in the rows is multiplied by 3, and in the
+columns by 2. Had we multiplied by 2 and 8 we should get the regular
+geometrical progression, 1, 2, 4, 8, 16, 32, 64, 128, and 256, but I
+wish to avoid high numbers. The numbers are arranged in the square in
+the same order as in the adding square.</p>
+
+<p>The fourth diagram is a dividing magic square. The constant 6 is here
+obtained by dividing the second number in a line by the first (in
+either direction) and the third number by the quotient. But, again,
+the process is simplified by dividing the product of the two extreme
+numbers by the middle number. This <span class='pagenum'>Pg 125<a name="Page_125" id="Page_125"></a></span>square is also "associated" by
+multiplication. It is derived from the multiplying square by merely
+reversing the diagonals, and the constant of the multiplying square is
+the cube of that of the dividing square derived from it.</p>
+
+<p>The next set of diagrams shows the solutions for the fifth order of
+square. They are all "associated" in the same way as before. The
+subtracting square is derived from the adding square by reversing the
+diagonals and exchanging opposite numbers in the centres of the
+borders, and the constant of one is again <i>n</i> times that of the other.
+The dividing square is derived from the multiplying square in the same
+way, and the constant of the latter is the 5th power (that is the
+<i>n</i>th) of that of the former.</p>
+
+<div class="figcenter" style="width: 550px;">
+<img src="images/magicfig2.png" width="550" height="629" alt="" title="" />
+</div>
+
+<p>These squares are thus quite easy for odd orders. But the reader will
+probably find some difficulty over the even orders, concerning which I
+will leave him to make his own researches, merely propounding two
+little problems.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_407_TWO_NEW_MAGIC_SQUARES" id="X_407_TWO_NEW_MAGIC_SQUARES"></a><a href="#X_407_TWO_NEW_MAGIC_SQUARESa"><b>407.&mdash;TWO NEW MAGIC SQUARES.</b></a></p>
+
+<p>Construct a subtracting magic square with the first sixteen whole
+numbers that shall be "associated" by <i>subtraction</i>. The constant is,
+of course, obtained by subtracting the first number from the second in
+line, the result from the third, and the result again from the fourth.
+Also construct a dividing magic square of the same order that shall be
+"associated" by <i>division</i>. The constant is obtained by dividing the
+second number in a line by the first, the third by the quotient, and
+the fourth by the next quotient.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_408_MAGIC_SQUARES_OF_TWO_DEGREES" id="X_408_MAGIC_SQUARES_OF_TWO_DEGREES"></a><a href="#X_408_MAGIC_SQUARES_OF_TWO_DEGREESa"><b>408.&mdash;MAGIC SQUARES OF TWO DEGREES.</b></a></p>
+
+<p>While reading a French mathematical work I happened to come across,
+the following statement: "A very remarkable magic square of 8, in two
+degrees, has been constructed by M. Pfeffermann. In other words, he
+has managed to dispose the sixty-four first numbers on the squares of
+a chessboard in such a way that the sum of the numbers in every line,
+every column, and in each of the two diagonals, shall be the same; and
+more, that if one substitutes for all the numbers their squares, the
+square still remains magic." I at once set to work to solve this
+problem, and, although it proved a very hard nut, one was rewarded by
+the discovery of some curious and beautiful laws that govern it. The
+reader may like to try his hand at the puzzle.</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="MAGIC_SQUARES_OF_PRIMES" id="MAGIC_SQUARES_OF_PRIMES"></a><a href="#CONTENTS">MAGIC SQUARES OF PRIMES.</a></h2>
+
+<p>The problem of constructing magic squares with prime numbers only was
+first discussed by myself in <i>The Weekly Dispatch</i> for 22nd July and
+5th August 1900; but during the last three or four years it has
+received great attention from American mathematicians. First, they
+have sought to form these squares with the lowest possible constants.
+Thus, the first nine prime numbers, 1 to 23 inclusive, sum to 99,
+which (being divisible by 3) is theoretically a suitable series; yet
+it has been demonstrated that the lowest possible constant is 111, and
+the required series as follows: 1, 7, 13, 31, 37, 43, 61, 67, and 73.
+Similarly, in the case of the fourth order, the lowest series of
+primes that are "theoretically suitable" will not serve. But in every
+other order, up to the 12th inclusive, magic squares have been
+constructed with the lowest series of primes theoretically possible.
+And the 12th is the lowest order in which a straight series of prime
+numbers, unbroken, from 1 upwards has been made to work. In other
+words, the first 144 odd prime numbers have actually been arranged in
+magic form. The following summary is taken from <i>The Monist</i> (Chicago)
+for October 1913:&mdash;</p>
+
+
+
+<div class='center'>
+<table border="1" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>Order <br />of Square.</td><td align='right'>Totals <br />of Series.</td><td align='right'>Lowest <br />Constants.</td><td align='center'>Squares <br />made by&mdash;</td></tr>
+<tr><td align='right'>3rd</td><td align='right'>333</td><td align='right'>111</td><td align='center'>Henry E. <br />Dudeney (1900).</td></tr>
+<tr><td align='right'>4th</td><td align='right'>408</td><td align='right'>102</td><td align='center'>Ernest Bergholt<br />and C. D. Shuldham.</td></tr>
+<tr><td align='right'>5th</td><td align='right'>1065</td><td align='right'>213</td><td align='center'>H. A. Sayles.</td></tr>
+<tr><td align='right'>6th</td><td align='right'>2448</td><td align='right'>408</td><td align='center'>C. D. Shuldham <br />and J. N. Muncey.</td></tr>
+<tr><td align='right'>7th</td><td align='right'>4893</td><td align='right'>699</td><td align='center'>do.</td></tr>
+<tr><td align='right'>8th</td><td align='right'>8912</td><td align='right'>1114</td><td align='center'>do.</td></tr>
+<tr><td align='right'>9th</td><td align='right'>15129</td><td align='right'>1681</td><td align='center'>do.</td></tr>
+<tr><td align='right'>10th</td><td align='right'>24160</td><td align='right'>2416</td><td align='center'>J. N. Muncey.</td></tr>
+<tr><td align='right'>11th</td><td align='right'>36095</td><td align='right'>3355</td><td align='center'>do.</td></tr>
+<tr><td align='right'>12th</td><td align='right'>54168</td><td align='right'>4514</td><td align='center'>do.</td></tr>
+</table></div>
+
+
+<p>For further details the reader should consult the article itself, by
+W. S. Andrews and H. A. Sayles.</p>
+
+<p><span class='pagenum'>Pg 126<a name="Page_126" id="Page_126"></a></span>These same investigators have also performed notable feats in
+constructing associated and bordered prime magics, and Mr. Shuldham has
+sent me a remarkable paper in which he gives examples of Nasik squares
+constructed with primes for all orders from the 4th to the 10th, with
+the exception of the 3rd (which is clearly impossible) and the 9th,
+which, up to the time of writing, has baffled all attempts.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_409_THE_BASKETS_OF_PLUMS" id="X_409_THE_BASKETS_OF_PLUMS"></a><a href="#X_409_THE_BASKETS_OF_PLUMSa"><b>409.&mdash;THE BASKETS OF PLUMS.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q409.png" width="400" height="475" alt="" title="" />
+</div>
+
+<p>This is the form in which I first introduced the question of magic
+squares with prime numbers. I will here warn the reader that there is
+a little trap.</p>
+
+<p>A fruit merchant had nine baskets. Every basket contained plums (all
+sound and ripe), and the number in every basket was different. When
+placed as shown in the illustration they formed a magic square, so
+that if he took any three baskets in a line in the eight possible
+directions there would always be the same number of plums. This part
+of the puzzle is easy enough to understand. But what follows seems at
+first sight a little queer.</p>
+
+<p>The merchant told one of his men to distribute the contents of any
+basket he chose among some children, giving plums to every child so
+that each should receive an equal number. But the man found it quite
+impossible, no matter which basket he selected and no matter how many
+children he included in the treat. Show, by giving contents of the
+nine baskets, how this could come about.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_410_THE_MANDARINS_T_PUZZLE" id="X_410_THE_MANDARINS_T_PUZZLE"></a><a href="#X_410_THE_MANDARINS_T_PUZZLEa"><b>410.&mdash;THE MANDARIN'S "T" PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q410.png" width="600" height="438" alt="" title="" />
+</div>
+
+<p>Before Mr. Beauchamp Cholmondely Marjoribanks set out on his tour in
+the Far East, he prided himself on his knowledge of magic squares, a
+subject that he had made his special hobby; but he soon discovered
+that he had never really touched more than the fringe of the subject,
+and that the wily Chinee could <span class='pagenum'>Pg 127<a name="Page_127" id="Page_127"></a></span>beat him easily. I present a little
+problem that one learned mandarin propounded to our traveller, as
+depicted on the last page.</p>
+
+<p>The Chinaman, after remarking that the construction of the ordinary
+magic square of twenty-five cells is "too velly muchee easy," asked
+our countryman so to place the numbers 1 to 25 in the square that
+every column, every row, and each of the two diagonals should add up
+65, with only prime numbers on the shaded "T." Of course the prime
+numbers available are 1, 2, 3, 5, 7, 11, 13, 17, 19, and 23, so you
+are at liberty to select any nine of these that will serve your
+purpose. Can you construct this curious little magic square?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_411_A_MAGIC_SQUARE_OF_COMPOSITES" id="X_411_A_MAGIC_SQUARE_OF_COMPOSITES"></a><a href="#X_411_A_MAGIC_SQUARE_OF_COMPOSITESa"><b>411.&mdash;A MAGIC SQUARE OF COMPOSITES.</b></a></p>
+
+<p>As we have just discussed the construction of magic squares with prime
+numbers, the following forms an interesting companion problem. Make a
+magic square with nine consecutive composite numbers&mdash;the smallest
+possible.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_412_THE_MAGIC_KNIGHTS_TOUR" id="X_412_THE_MAGIC_KNIGHTS_TOUR"></a><a href="#X_412_THE_MAGIC_KNIGHTS_TOURa"><b>412.&mdash;THE MAGIC KNIGHT'S TOUR.</b></a></p>
+
+<p>Here is a problem that has never yet been solved, nor has its
+impossibility been demonstrated. Play the knight once to every square
+of the chessboard in a complete tour, numbering the squares in the
+order visited, so that when completed the square shall be "magic,"
+adding up to 260 in every column, every row, and each of the two long
+diagonals. I shall give the best answer that I have been able to
+obtain, in which there is a slight error in the diagonals alone. Can a
+perfect solution be found? I am convinced that it cannot, but it is
+only a "pious opinion."</p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="MAZES_AND_HOW_TO_THREAD_THEM" id="MAZES_AND_HOW_TO_THREAD_THEM"></a><a href="#CONTENTS">MAZES AND HOW TO THREAD THEM.</a></h2>
+
+<p class='center'>
+"In wandering mazes lost."<br />
+<span style="margin-left: 8em;"><i>Paradise Lost.</i></span><br />
+</p>
+
+<p>The Old English word "maze," signifying a labyrinth, probably comes
+from the Scandinavian, but its origin is somewhat uncertain. The late
+Professor Skeat thought that the substantive was derived from the
+verb, and as in old times to be mazed or amazed was to be "lost in
+thought," the transition to a maze in whose tortuous windings we are
+lost is natural and easy.</p>
+
+<p>The word "labyrinth" is derived from a Greek word signifying the
+passages of a mine. The ancient mines of Greece and elsewhere inspired
+fear and awe on account of their darkness and the danger of getting
+lost in their intricate passages. Legend was afterwards built round
+these mazes. The most familiar instance is the labyrinth made by
+D&aelig;dalus in Crete for King Minos. In the centre was placed the
+Minotaur, and no one who entered could find his way out again, but
+became the prey of the monster. Seven youths and seven maidens were
+sent regularly by the Athenians, and were duly devoured, until Theseus
+slew the monster and escaped from the maze by aid of the clue of
+thread provided by Ariadne; which accounts for our using to-day the
+expression "threading a maze."</p>
+
+<p>The various forms of construction of mazes include complicated ranges
+of caverns, architectural labyrinths, or sepulchral buildings,
+tortuous devices indicated by coloured marbles and tiled pavements,
+winding paths cut in the turf, and topiary mazes formed by clipped
+hedges. As a matter of fact, they may be said to have descended to us
+in precisely this order of variety.</p>
+
+<p>Mazes were used as ornaments on the state robes of Christian emperors
+before the ninth century, and were soon adopted in the decoration of
+cathedrals and other churches. The original idea was doubtless to
+employ them as symbols of the complicated folds of sin by which man is
+surrounded. They began to abound in the early part of the twelfth
+century, and I give an illustration of one of this period in the
+parish church at St. Quentin (Fig. 1). It formed a pavement of the
+nave, and its diameter is 34&frac12; feet. The path here is the line itself.
+If you place your pencil at the point A and ignore the enclosing line,
+the line leads you to the centre by a long route over the entire area;
+but you never have any option as to direction during your course. As
+we shall find in similar cases, these early ecclesiastical mazes were
+generally not of a puzzle nature, but simply long, winding paths that
+took you over practically all the ground enclosed.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig01.png" width="400" height="413" alt="FIG. 1.&mdash;Maze at St. Quentin." title="" />
+<span class="caption">FIG. 1.&mdash;Maze at St. Quentin.</span>
+</div>
+
+<p>In the abbey church of St. Berlin, at St. Omer, is another of these
+curious floors, representing the Temple of Jerusalem, with stations
+for pilgrims. These mazes were actually visited and traversed by them
+as a compromise for not going to the Holy Land in fulfilment of a vow.
+They were also used as a means of penance, the penitent frequently
+being directed to go the whole course of the maze on hands and knees.</p>
+
+<p><span class='pagenum'>Pg 128<a name="Page_128" id="Page_128"></a></span></p><div class="figcenter" style="width: 400px;">
+<img src="images/mzfig02.png" width="400" height="400" alt="FIG. 2.&mdash;Maze in Chartres Cathedral." title="" />
+<span class="caption">FIG. 2.&mdash;Maze in Chartres Cathedral.</span>
+</div>
+
+<p>The maze in Chartres Cathedral, of which I give an illustration (Fig.
+2), is 40 feet across, and was used by penitents following the
+procession of Calvary. A labyrinth in Amiens Cathedral was octagonal,
+similar to that at St. Quentin, measuring 42 feet across. It bore the
+date 1288, but was destroyed in 1708. In the chapter-house at Bayeux
+is a labyrinth formed of tiles, red, black, and encaustic, with a
+pattern of brown and yellow. Dr. Ducarel, in his "<i>Tour through Part of
+Normandy</i>" (printed in 1767), mentions the floor of the great
+guard-chamber in the abbey of St. Stephen, at Caen, "the middle
+whereof represents a maze or labyrinth about 10 feet diameter, and so
+artfully contrived that, were we to suppose a man following all the
+intricate meanders of its volutes, he could not travel less than a
+mile before he got from one end to the other."</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig03.png" width="400" height="398" alt="FIG. 3.&mdash;Maze in Lucca Cathedral." title="" />
+<span class="caption">FIG. 3.&mdash;Maze in Lucca Cathedral.</span>
+</div>
+
+<p>Then these mazes were sometimes reduced in size and represented on a
+single tile (Fig. 3). I give an example from Lucca Cathedral. It is on
+one of the porch piers, and is 19� inches in diameter. A writer in
+1858 says that, "from the continual attrition it has received from
+thousands of tracing fingers, a central group of Theseus and the
+Minotaur has now been very nearly effaced." Other examples were, and
+perhaps still are, to be found in the Abbey of Toussarts, at
+Ch&acirc;lons-sur-Marne, in the very ancient church of St. Michele at Pavia,
+at Aix in Provence, in the cathedrals of Poitiers, Rheims, and Arras,
+in the church of Santa Maria in Aquiro in Rome, in San Vitale at
+Ravenna, in the Roman mosaic pavement found at Salzburg, and
+elsewhere. These mazes were sometimes called "Chemins de Jerusalem,"
+as being emblematical of the difficulties attending a journey to the
+earthly Jerusalem and of those encountered by the Christian before he
+can reach the heavenly Jerusalem&mdash;where the centre was frequently
+called "Ciel."</p>
+
+<p>Common as these mazes were upon the Continent, it is probable that no
+example is to be found in any English church; at least I am not aware
+of the existence of any. But almost every county has, or has had, its
+specimens of <span class='pagenum'>Pg 129<a name="Page_129" id="Page_129"></a></span>mazes cut in the turf. Though these are frequently known
+as "miz-mazes" or "mize-mazes," it is not uncommon to find them
+locally called "Troy-towns," "shepherds' races," or "Julian's
+Bowers"&mdash;names that are misleading, as suggesting a false origin. From
+the facts alone that many of these English turf mazes are clearly
+copied from those in the Continental churches, and practically all are
+found close to some ecclesiastical building or near the site of an
+ancient one, we may regard it as certain that they were of church
+origin and not invented by the shepherds or other rustics. And
+curiously enough, these turf mazes are apparently unknown on the
+Continent. They are distinctly mentioned by Shakespeare:&mdash;</p>
+
+<div class="poem"><div class="stanza">
+<span class="i0">"The nine men's morris is filled up with mud,<br /></span>
+<span class="i0">And the quaint mazes in the wanton green<br /></span>
+<span class="i0">For lack of tread are undistinguishable."<br /></span>
+</div><div class="stanza">
+<span class="i0"><i>A Midsummer Night's Dream</i>, ii. 1.<br /></span><br /><br />
+</div></div>
+
+<div class="poem"><div class="stanza">
+<span class="i0">"My old bones ache: here's a maze trod indeed,<br /></span>
+<span class="i0">Through forth-rights and meanders!"<br /></span>
+</div><div class="stanza">
+<span class="i0"><i>The Tempest</i>, iii. 3.<br /></span>
+</div></div>
+
+<p>There was such a maze at Comberton, in Cambridgeshire, and another,
+locally called the "miz-maze," at Leigh, in Dorset. The latter was on
+the highest part of a field on the top of a hill, a quarter of a mile
+from the village, and was slightly hollow in the middle and enclosed
+by a bank about 3 feet high. It was circular, and was thirty paces in
+diameter. In 1868 the turf had grown over the little trenches, and it
+was then impossible to trace the paths of the maze. The Comberton one
+was at the same date believed to be perfect, but whether either or
+both have now disappeared I cannot say. Nor have I been able to verify
+the existence or non-existence of the other examples of which I am
+able to give illustrations. I shall therefore write of them all in the
+past tense, retaining the hope that some are still preserved.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig04.png" width="400" height="400" alt="FIG. 4.&mdash;Maze at Saffron Walden, Essex." title="" />
+<span class="caption">FIG. 4.&mdash;Maze at Saffron Walden, Essex.</span>
+</div>
+
+<p>In the next two mazes given&mdash;that at Saffron Walden, Essex (110 feet
+in diameter, Fig. 4), and the one near St. Anne's Well, at Sneinton,
+Nottinghamshire (Fig. 5), which was ploughed up on February 27th, 1797
+(51 feet in diameter, with a path 535 yards long)&mdash;the paths must in
+each case be understood to be on the lines, black or white, as the
+case may be.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig05.png" width="400" height="402" alt="FIG. 5.&mdash;Maze at Sneinton, Nottinghamshire." title="" />
+<span class="caption">FIG. 5.&mdash;Maze at Sneinton, Nottinghamshire.</span>
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig06.png" width="400" height="398" alt="FIG. 6.&mdash;Maze at Alkborough, Lincolnshire." title="" />
+<span class="caption">FIG. 6.&mdash;Maze at Alkborough, Lincolnshire.</span>
+</div>
+
+<p>I give in Fig. 6 a maze that was at Alkborough, Lincolnshire,
+overlooking the Humber. This was 44 feet in diameter, and the
+resem<span class='pagenum'>Pg 130<a name="Page_130" id="Page_130"></a></span>blance between it and the mazes at Chartres and Lucca (Figs. 2
+and 3) will be at once perceived. A maze at Boughton Green, in
+Nottinghamshire, a place celebrated at one time for its fair (Fig. 7),
+was 37 feet in diameter. I also include the plan (Fig. 8) of one that
+used to be on the outskirts of the village of Wing, near Uppingham,
+Rutlandshire. This maze was 40 feet in diameter.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig07.png" width="400" height="396" alt="FIG. 7.&mdash;Maze at Boughton Green, Nottinghamshire." title="" />
+<span class="caption">FIG. 7.&mdash;Maze at Boughton Green, Nottinghamshire.</span>
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig08.png" width="400" height="405" alt="FIG. 8.&mdash;Maze at Wing, Rutlandshire." title="" />
+<span class="caption">FIG. 8.&mdash;Maze at Wing, Rutlandshire.</span>
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig09.png" width="400" height="398" alt="FIG. 9.&mdash;Maze on St. Catherine&#39;s Hill, Winchester." title="" />
+<span class="caption">FIG. 9.&mdash;Maze on St. Catherine&#39;s Hill, Winchester.</span>
+</div>
+
+<p>The maze that was on St. Catherine's Hill, Winchester, in the parish
+of Chilcombe, was a poor specimen (Fig. 9), since, as will be seen,
+there was one short direct route to the centre, unless, as in Fig. 10
+again, the path is the line itself from end to end. This maze was 86
+feet square, cut in the turf, and was locally known as the
+"Mize-maze." It became very indistinct about 1858, and was then recut
+by the Warden of Winchester, with the aid of a plan possessed by a
+lady living in the neighbourhood.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig10.png" width="400" height="397" alt="FIG. 10.&mdash;Maze on Ripon Common." title="" />
+<span class="caption">FIG. 10.&mdash;Maze on Ripon Common.</span>
+</div>
+
+<p>A maze formerly existed on Ripon Common, in Yorkshire (Fig. 10). It
+was ploughed up in 1827, but its plan was fortunately preserved. This
+example was 20 yards in diameter, and its path is said to have been
+407 yards long.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig11.png" width="400" height="400" alt="FIG. 11.&mdash;Maze at Theobalds, Hertfordshire." title="" />
+<span class="caption">FIG. 11.&mdash;Maze at Theobalds, Hertfordshire.</span>
+</div>
+
+<p>In the case of the maze at Theobalds, Hertfordshire, after you have
+found the entrance within the four enclosing hedges, the path is
+<span class='pagenum'>Pg 131<a name="Page_131" id="Page_131"></a></span>forced (Fig. 11). As further illustrations of this class of maze, I
+give one taken from an Italian work on architecture by Serlio,
+published in 1537 (Fig. 12), and one by London and Wise, the designers
+of the Hampton Court maze, from their book, <i>The Retired Gard'ner</i>,
+published in 1706 (Fig. 13). Also, I add a Dutch maze (Fig. 14).</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig12.png" width="400" height="395" alt="FIG. 12.&mdash;Italian Maze of Sixteenth Century." title="" />
+<span class="caption">FIG. 12.&mdash;Italian Maze of Sixteenth Century.</span>
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig13.png" width="400" height="272" alt="FIG. 13.&mdash;By the Designers of Hampton Court Maze." title="" />
+<span class="caption">FIG. 13.&mdash;By the Designers of Hampton Court Maze.</span>
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig14.png" width="400" height="362" alt="FIG. 14.&mdash;A Dutch Maze." title="" />
+<span class="caption">FIG. 14.&mdash;A Dutch Maze.</span>
+</div>
+
+<p>So far our mazes have been of historical interest, but they have
+presented no difficulty in threading. After the Reformation period we
+find mazes converted into mediums for recreation, and they generally
+consisted of labyrinthine paths enclosed by thick and carefully
+trimmed hedges. These topiary hedges were known to the Romans, with
+whom the <i>topiarius</i> was the ornamental gardener. This type of maze
+has of late years degenerated into the seaside "Puzzle Gardens. Teas,
+sixpence, including admission to the Maze." The Hampton Court Maze,
+sometimes called the "Wilderness," at the royal palace, was designed,
+as I have said, by London and Wise for William III., who had a liking
+for such things (Fig. 15). I have before me some three or four
+versions of it, all slightly different from one another; but the plan
+I select is taken from an old guide-book to the palace, and therefore
+ought to be trustworthy. The meaning of the dotted lines, etc., will
+be explained later on.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/mzfig15.png" width="600" height="248" alt="FIG. 15.&mdash;Maze at Hampton Court Palace." title="" />
+<span class="caption">FIG. 15.&mdash;Maze at Hampton Court Palace.</span>
+</div>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/mzfig16.png" width="600" height="369" alt="FIG. 16.&mdash;Maze at Hatfield House, Herts." title="" />
+<span class="caption">FIG. 16.&mdash;Maze at Hatfield House, Herts.</span>
+</div>
+
+<p>The maze at Hatfield House (Fig. 16), the seat of the Marquis of
+Salisbury, like so many labyrinths, is not difficult on paper; but
+both <span class='pagenum'>Pg 132<a name="Page_132" id="Page_132"></a></span>this and the Hampton Court Maze may prove very puzzling to
+actually thread without knowing the plan. One reason is that one is so
+apt to go down the same blind alleys over and over again, if one
+proceeds without method. The maze planned by the desire of the Prince
+Consort for the Royal Horticultural Society's Gardens at South
+Kensington was allowed to go to ruin, and was then destroyed&mdash;no great
+loss, for it was a feeble thing. It will be seen that there were three
+entrances from the outside (Fig. 17), but the way to the centre is
+very easy to discover. I include a German maze that is curious, but
+not difficult to thread on paper (Fig. 18). The example of a labyrinth
+formerly existing at Pimperne, in Dorset, is in a class by itself
+(Fig. 19). It was formed of small ridges about a foot high, and
+covered nearly an <span class='pagenum'>Pg 133<a name="Page_133" id="Page_133"></a></span>acre of ground; but it was, unfortunately, ploughed
+up in 1730.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig17.png" width="400" height="590" alt="FIG. 17.&mdash;Maze formerly at South Kensington." title="" />
+<span class="caption">FIG. 17.&mdash;Maze formerly at South Kensington.</span>
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig18.png" width="400" height="278" alt="FIG. 18.&mdash;A German Maze." title="" />
+<span class="caption">FIG. 18.&mdash;A German Maze.</span>
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig19.png" width="400" height="404" alt="FIG. 19.&mdash;Maze at Pimperne, Dorset." title="" />
+<span class="caption">FIG. 19.&mdash;Maze at Pimperne, Dorset.</span>
+</div>
+
+<p>We will now pass to the interesting subject of how to thread any maze.
+While being necessarily brief, I will try to make the matter clear to
+readers who have no knowledge of mathematics. And first of all we will
+assume that we are trying to enter a maze (that is, get to the
+"centre") of which we have no plan and about which we know nothing.
+The first rule is this: If a maze has no parts of its hedges detached
+from the rest, then if we always keep in touch with the hedge with the
+right hand (or always touch it with the left), going down to the stop
+in every blind alley and coming back on the other side, we shall pass
+through every part of the maze and make our exit where we went in.
+Therefore we must at one time or another enter the centre, and every
+alley will be traversed twice.</p>
+
+<p>Now look at the Hampton Court plan. Follow, say to the right, the path
+indicated by the <span class='pagenum'>Pg 134<a name="Page_134" id="Page_134"></a></span>dotted line, and what I have said is clearly correct
+if we obliterate the two detached parts, or "islands," situated on
+each side of the star. But as these islands are there, you cannot by
+this method traverse every part of the maze; and if it had been so
+planned that the "centre" was, like the star, between the two islands,
+you would never pass through the "centre" at all. A glance at the
+Hatfield maze will show that there are three of these detached hedges
+or islands at the centre, so this method will never take you to the
+"centre" of that one. But the rule will at least always bring you
+safely out again unless you blunder in the following way. Suppose,
+when you were going in the direction of the arrow in the Hampton Court
+Maze, that you could not distinctly see the turning at the bottom,
+that you imagined you were in a blind alley and, to save time, crossed
+at once to the opposite hedge, then you would go round and round that
+U-shaped island with your right hand still always on the hedge&mdash;for
+ever after!</p>
+
+<p>This blunder happened to me a few years ago in a little maze on the
+isle of Caldy, South Wales. I knew the maze was a small one, but after
+a very long walk I was amazed to find that I did not either reach the
+"centre" or get out again. So I threw a piece of paper on the ground,
+and soon came round to it; from which I knew that I had blundered over
+a supposed blind alley and was going round and round an island.
+Crossing to the opposite hedge and using more care, I was quickly at
+the centre and out again. Now, if I had made a similar mistake at
+Hampton Court, and discovered the error when at the star, I should
+merely have passed from one island to another! And if I had again
+discovered that I was on a detached part, I might with ill luck have
+recrossed to the first island again! We thus see that this "touching
+the hedge" method should always bring us safely out of a maze that we
+have entered; it may happen to take us through the "centre," and if we
+miss the centre we shall know there must be islands. But it has to be
+<span class='pagenum'>Pg 135<a name="Page_135" id="Page_135"></a></span>done with a little care, and in no case can we be sure that we have
+traversed every alley or that there are no detached parts.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/mzfig20.png" width="400" height="374" alt="FIG. 20.&mdash;M. Tremaux&#39;s Method of Solution." title="" />
+<span class="caption">FIG. 20.&mdash;M. Tremaux&#39;s Method of Solution.</span>
+</div>
+
+<p>If the maze has many islands, the traversing of the whole of it may be
+a matter of considerable difficulty. Here is a method for solving any
+maze, due to M. Tr&eacute;maux, but it necessitates carefully marking in some
+way your entrances and exits where the galleries fork. I give a
+diagram of an imaginary maze of a very simple character that will
+serve our purpose just as well as something more complex (Fig. 20).
+The circles at the regions where we have a choice of turnings we may
+call nodes. A "new" path or node is one that has not been entered
+before on the route; an "old" path or node is one that has already
+been entered, 1. No path may be traversed more than twice. 2. When you
+come to a new node, take any path you like. 3. When by a new path you
+come to an old node or to the stop of a blind alley, return by the
+path you came. 4. When by an old path you come to an old node, take a
+new path if there is one; if not, an old path. The route indicated by
+the dotted line in the diagram is taken in accordance with these
+simple rules, and it will be seen that it leads us to the centre,
+although the maze consists of four islands.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/mzfig21.png" width="600" height="367" alt="FIG. 21.&mdash;How to thread the Hatfield Maze." title="" />
+<span class="caption">FIG. 21.&mdash;How to thread the Hatfield Maze.</span>
+</div>
+
+<p><span class='pagenum'>Pg 136<a name="Page_136" id="Page_136"></a></span>Neither of the methods I have given will disclose to us the shortest
+way to the centre, nor the number of the different routes. But we can
+easily settle these points with a plan. Let us take the Hatfield maze
+(Fig. 21). It will be seen that I have suppressed all the blind alleys
+by the shading. I begin at the stop and work backwards until the path
+forks. These shaded parts, therefore, can never be entered without our
+having to retrace our steps. Then it is very clearly seen that if we
+enter at A we must come out at B; if we enter at C we must come out at
+D. Then we have merely to determine whether A, B, E, or C, D, E, is
+the shorter route. As a matter of fact, it will be found by rough
+measurement or calculation that the shortest route to the centre is by
+way of C, D, E, F.</p>
+
+<p>I will now give three mazes that are simply puzzles on paper, for, so
+far as I know, they have never been constructed in any other way. The
+first I will call the Philadelphia maze (Fig. 22). Fourteen years ago
+a travelling salesman, <span class='pagenum'>Pg 137<a name="Page_137" id="Page_137"></a></span>living in Philadelphia, U.S.A., developed a
+curiously unrestrained passion for puzzles. He neglected his business,
+and soon his position was taken from him. His days and nights were now
+passed with the subject that fascinated him, and this little maze
+seems to have driven him into insanity. He had been puzzling over it
+for some time, and finally it sent him mad and caused him to fire a
+bullet through his brain. Goodness knows what his difficulties could
+have been! But there can be little doubt that he had a disordered
+mind, and that if this little puzzle had not caused him to lose his
+mental balance some other more or less trivial thing would in time
+have done so. There is no moral in the story, unless it be that of the
+Irish maxim, which applies to every occupation of life as much as to
+the solving of puzzles: "Take things aisy; if you can't take them
+aisy, take them as aisy as you can." And it is a bad and empirical way
+of solving any puzzle&mdash;by blowing your brains out.</p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/mzfig22.png" width="500" height="516" alt="FIG. 22. The Philadelphia Maze, and its Solution." title="" />
+<span class="caption">FIG. 22. The Philadelphia Maze, and its Solution.</span>
+</div>
+
+<p>Now, how many different routes are there from A to B in this maze if
+we must never in any route go along the same passage twice? The four
+open spaces where four passages end are not reckoned as "passages." In
+the diagram (Fig. 22) it will be seen that I have again suppressed the
+blind alleys. It will be found that, in any case, we must go from A to
+C, and also from F to B. But when we have arrived at C there are three
+ways, marked 1, 2, 3, of getting to D. Similarly, when we get to E
+there are three ways, marked 4, 5, 6, of getting to F. We have also
+the dotted route from C to E, the other dotted route from D to F, and
+the passage from D to E, indicated by stars. We can, therefore,
+express the position of affairs by the little diagram annexed (Fig.
+23). Here every condition of route exactly corresponds to that in the
+circular maze, only it is much less confusing to the eye. Now, the
+number of routes, under the conditions, from A to B on this simplified
+diagram is 640, and that is the required answer to the maze puzzle.</p>
+
+<div class="figcenter" style="width: 396px;">
+<img src="images/mzfig23.png" width="396" height="93" alt="FIG. 23.&mdash;Simplified Diagram of Fig. 22." title="" />
+<span class="caption">FIG. 23.&mdash;Simplified Diagram of Fig. 22.</span>
+</div>
+<p><br /></p>
+<div class="figcenter" style="width: 500px;">
+<img src="images/mzfig24.png" width="500" height="510" alt="FIG. 24.&mdash;Can you find the Shortest Way to Centre?" title="" />
+<span class="caption">FIG. 24.&mdash;Can you find the Shortest Way to Centre?</span>
+</div>
+
+<p>Finally, I will leave two easy maze puzzles (Figs. 24, 25) for my
+readers to solve for themselves. The puzzle in each case is to find
+the shortest possible route to the centre. Everybody knows the story
+of Fair Rosamund and the Woodstock maze. What the maze was like or
+whether it ever existed except in imagination is not known, many
+writers believing that it was simply a badly-constructed house with a
+large number of confusing rooms and passages. At any rate, my sketch
+lacks the authority of the other mazes in this article. My "Rosamund's
+Bower" is simply designed to show that where you have the plan before
+you it often happens that the easiest way to find a route into a maze
+is by working backwards and first finding a way out.</p>
+
+<div class="figcenter" style="width: 550px;">
+<img src="images/mzfig25.png" width="550" height="669" alt="FIG. 25.&mdash;Rosamund&#39;s Bower." title="" />
+<span class="caption">FIG. 25.&mdash;Rosamund&#39;s Bower.</span>
+</div>
+
+<hr style="width: 65%;" />
+<h2><a name="THE_PARADOX_PARTY" id="THE_PARADOX_PARTY"></a><a href="#CONTENTS">THE PARADOX PARTY.</a></h2>
+
+<p class='center'>
+"Is not life itself a paradox?"<br />
+<span style="margin-left: 8em;">C.L. DODGSON, <i>Pillow Problems</i>.</span><br />
+</p>
+
+
+<p>"It is a wonderful age!" said Mr. Allgood, and everybody at the table
+turned towards him and assumed an attitude of expectancy.</p>
+
+<p>This was an ordinary Christmas dinner of the Allgood family, with a
+sprinkling of local friends. Nobody would have supposed that the above
+remark would lead, as it did, to a succession of curious puzzles and
+paradoxes, to which every member of the party contributed something of
+interest. The little symposium was quite unpremeditated, so we must
+not be too critical respecting a few of the posers that were
+forthcoming. The varied character of the contributions is just what we
+would expect on such an occasion, for it was a gathering not of expert
+mathematicians and logicians, but of quite ordinary folk.</p>
+
+<p>"It is a wonderful age!" repeated Mr. Allgood. "A man has just
+designed a square house in such a cunning manner that all the windows
+on the four sides have a south aspect."</p>
+
+<p>"That would appeal to me," said Mrs. Allgood, "for I cannot endure a
+room with a north aspect."</p>
+
+<p>"I cannot conceive how it is done," Uncle John confessed. "I suppose
+he puts bay windows on the east and west sides; but how on earth can
+be contrive to look south from the north side? Does he use mirrors, or
+something of that kind?"</p>
+
+<p>"No," replied Mr. Allgood, "nothing of the sort. All the windows are
+flush with the walls, and yet you get a southerly prospect from every
+one of them. You see, there is no real difficulty in designing the
+house if you select the proper spot for its erection. Now, this house
+is designed for a gentleman who proposes to build it exactly at the
+North Pole. If you think a moment you will realize that when you stand
+at the North Pole it is impossible, no matter which way you may turn,
+to look elsewhere than due south! There are no such directions as
+north, east, or west when you are exactly at the North Pole.
+Everything is due south!"</p>
+
+<p>"I am afraid, mother," said her son George, after the laughter had
+subsided, "that, however much you might like the aspect, the situation
+would be a little too bracing for you."</p>
+
+<p>"Ah, well!" she replied. "Your Uncle John fell also into the trap. I
+am no good at catches and puzzles. I suppose I haven't the right sort
+of brain. Perhaps some one will explain this to me. Only last week I
+remarked to my hairdresser that it had been said that there are more
+persons in the world than any one of them has hairs on his head. He
+replied, 'Then it follows, madam, that two persons, at least, must
+have exactly the same number of <span class='pagenum'>Pg 138<a name="Page_138" id="Page_138"></a></span>hairs on their heads.' If this is a
+fact, I confess I cannot see it."</p>
+
+<p>"How do the bald-headed affect the question?" asked Uncle John.</p>
+
+<p>"If there are such persons in existence," replied Mrs. Allgood, "who
+haven't a solitary hair on their heads discoverable under a
+magnifying-glass, we will leave them out of the question. Still, I
+don't see how you are to prove that at least two persons have exactly
+the same number to a hair."</p>
+
+<p>"I think I can make it clear," said Mr. Filkins, who had dropped in
+for the evening. "Assume the population of the world to be only one
+million. Any number will do as well as another. Then your statement
+was to the effect that no person has more than nine hundred and
+ninety-nine thousand nine hundred and ninety-nine hairs on his head.
+Is that so?"</p>
+
+<p>"Let me think," said Mrs. Allgood. "Yes&mdash;yes&mdash;that is correct."</p>
+
+<p>"Very well, then. As there are only nine hundred and ninety-nine
+thousand nine hundred and ninety-nine <i>different</i> ways of bearing
+hair, it is clear that the millionth person must repeat one of those
+ways. Do you see?"</p>
+
+<p>"Yes; I see that&mdash;at least I think I see it."</p>
+
+<p>"Therefore two persons at least must have the same number of hairs on
+their heads; and as the number of people on the earth so greatly
+exceeds the number of hairs on any one person's head, there must, of
+course, be an immense number of these repetitions."</p>
+
+<p>"But, Mr. Filkins," said little Willie Allgood, "why could not the
+millionth man have, say, ten thousand hairs and a half?"</p>
+
+<p>"That is mere hair-splitting, Willie, and does not come into the
+question."</p>
+
+<p>"Here is a curious paradox," said George. "If a thousand soldiers are
+drawn up in battle array on a plane"&mdash;they understood him to mean
+"plain"&mdash;"only one man will stand upright."</p>
+
+<p>Nobody could see why. But George explained that, according to Euclid,
+a plane can touch a sphere only at one point, and that person only who
+stands at that point, with respect to the centre of the earth, will
+stand upright.</p>
+
+<p>"In the same way," he remarked, "if a billiard-table were quite
+level&mdash;that is, a perfect plane&mdash;the balls ought to roll to the
+centre."</p>
+
+<p>Though he tried to explain this by placing a visiting-card on an
+orange and expounding the law of gravitation, Mrs. Allgood declined to
+accept the statement. She could not see that the top of a true
+billiard-table must, theoretically, be spherical, just like a portion
+of the orange-peel that George cut out. Of course, the table is so
+small in proportion to the surface of the earth that the curvature is
+not appreciable, but it is nevertheless true in theory. A surface that
+we call level is not the same as our idea of a true geometrical plane.</p>
+
+<p>"Uncle John," broke in Willie Allgood, "there is a certain island
+situated between England and France, and yet that island is farther
+from France than England is. What is the island?"</p>
+
+<p>"That seems absurd, my boy; because if I place this tumbler, to
+represent the island, between these two plates, it seems impossible
+that the tumbler can be farther from either of the plates than they
+are from each other."</p>
+
+<p>"But isn't Guernsey between England and France?" asked Willie.</p>
+
+<p>"Yes, certainly."</p>
+
+<p>"Well, then, I think you will find, uncle, that Guernsey is about
+twenty-six miles from France, and England is only twenty-one miles
+from France, between Calais and Dover."</p>
+
+<p>"My mathematical master," said George, "has been trying to induce me
+to accept the axiom that 'if equals be multiplied by equals the
+products are equal.'"</p>
+
+<p>"It is self-evident," pointed out Mr. Filkins. "For example, if 3 feet
+equal 1 yard, then twice 3 feet will equal 2 yards. Do you see?"</p>
+
+<p>"But, Mr. Filkins," asked George, "is this tumbler half full of water
+equal to a similar glass half empty?"</p>
+
+<p>"Certainly, George."</p>
+
+<p>"Then it follows from the axiom that a glass full must equal a glass
+empty. Is that correct?"</p>
+
+<p>"No, clearly not. I never thought of it in that light."</p>
+
+<p>"Perhaps," suggested Mr. Allgood, "the rule does not apply to
+liquids."</p>
+
+<p>"Just what I was thinking, Allgood. It would seem that we must make an
+exception in the case of liquids."</p>
+
+<p>"But it would be awkward," said George, with a smile, "if we also had
+to except the case of solids. For instance, let us take the solid
+earth. One mile square equals one square mile. Therefore two miles
+square must equal two square miles. Is this so?"</p>
+
+<p>"Well, let me see! No, of course not," Mr. Filkins replied, "because
+two miles square is four square miles."</p>
+
+<p>"Then," said George, "if the axiom is not true in these cases, when is
+it true?"</p>
+
+<p>Mr. Filkins promised to look into the matter, and perhaps the reader
+will also like to give it consideration at leisure.</p>
+
+<p>"Look here, George," said his cousin Reginald Woolley: "by what
+fractional part does four-fourths exceed three-fourths?"</p>
+
+<p>"By one-fourth!" shouted everybody at once.</p>
+
+<p>"Try another one," George suggested.</p>
+
+<p>"With pleasure, when you have answered that one correctly," was
+Reginald's reply.</p>
+
+<p>"Do you mean to say that it isn't one-fourth?"</p>
+
+<p>"Certainly I do."</p>
+
+<p>Several members of the company failed to see that the correct answer
+is "one-third," although Reginald tried to explain that three of
+anything, if increased by one-third, becomes four.</p>
+
+<p>"Uncle John, how do you pronounce 't-o-o'?" asked Willie.</p>
+
+<p>"'Too," my boy."</p>
+
+<p>"And how do you pronounce 't-w-o'?"</p>
+
+<p>"That is also 'too.'"</p>
+
+<p><span class='pagenum'>Pg 139<a name="Page_139" id="Page_139"></a></span>"Then how do you pronounce the second day of the week?"</p>
+
+<p>"Well, that I should pronounce 'Tuesday,' not 'Toosday.'"</p>
+
+<p>"Would you really? I should pronounce it 'Monday.'"</p>
+
+<p>"If you go on like this, Willie," said Uncle John, with mock severity,
+"you will soon be without a friend in the world."</p>
+
+<p>"Can any of you write down quickly in figures 'twelve thousand twelve
+hundred and twelve pounds'?" asked Mr. Allgood.</p>
+
+<p>His eldest daughter, Miss Mildred, was the only person who happened to
+have a pencil at hand.</p>
+
+<p>"It can't be done," she declared, after making an attempt on the white
+table-cloth; but Mr. Allgood showed her that it should be written,
+"&pound;13,212."</p>
+
+<p>"Now it is my turn," said Mildred. "I have been waiting to ask you all
+a question. In the Massacre of the Innocents under Herod, a number of
+poor little children were buried in the sand with only their feet
+sticking out. How might you distinguish the boys from the girls?"</p>
+
+<p>"I suppose," said Mrs. Allgood, "it is a conundrum&mdash;something to do
+with their poor little 'souls.'"</p>
+
+<p>But after everybody had given it up, Mildred reminded the company that
+only boys were put to death.</p>
+
+<p>"Once upon a time," began George, "Achilles had a race with a
+tortoise&mdash;"</p>
+
+<p>"Stop, George!" interposed Mr. Allgood. "We won't have that one. I
+knew two men in my youth who were once the best of friends, but they
+quarrelled over that infernal thing of Zeno's, and they never spoke to
+one another again for the rest of their lives. I draw the line at
+that, and the other stupid thing by Zeno about the flying arrow. I
+don't believe anybody understands them, because I could never do so
+myself."</p>
+
+<p>"Oh, very well, then, father. Here is another. The Post-Office people
+were about to erect a line of telegraph-posts over a high hill from
+Turmitville to Wurzleton; but as it was found that a railway company
+was making a deep level cutting in the same direction, they arranged
+to put up the posts beside the line. Now, the posts were to be a
+hundred yards apart, the length of the road over the hill being five
+miles, and the length of the level cutting only four and a half miles.
+How many posts did they save by erecting them on the level?"</p>
+
+<p>"That is a very simple matter of calculation," said Mr. Filkins.
+"Find how many times one hundred yards will go in five miles, and how
+many times in four and a half miles. Then deduct one from the other,
+and you have the number of posts saved by the shorter route."</p>
+
+<p>"Quite right," confirmed Mr. Allgood. "Nothing could be easier."</p>
+
+<p>"That is just what the Post-Office people said," replied George, "but
+it is quite wrong. If you look at this sketch that I have just made,
+you will see that there is no difference whatever. If the posts are a
+hundred yards apart, just the same number will be required on the
+level as over the surface of the hill."</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/pdoxprty.png" width="400" height="106" alt="" title="" />
+</div>
+
+<p>"Surely you must be wrong, George," said Mrs. Allgood, "for if the
+posts are a hundred yards apart and it is half a mile farther over the
+hill, you have to put up posts on that extra half-mile."</p>
+
+<p>"Look at the diagram, mother. You will see that the distance from post
+to post is not the distance from base to base measured along the
+ground. I am just the same distance from you if I stand on this spot
+on the carpet or stand immediately above it on the chair."</p>
+
+<p>But Mrs. Allgood was not convinced.</p>
+
+<p>Mr. Smoothly, the curate, at the end of the table, said at this point
+that he had a little question to ask.</p>
+
+<p>"Suppose the earth were a perfect sphere with a smooth surface, and a
+girdle of steel were placed round the Equator so that it touched at
+every point."</p>
+
+<p>"'I'll put a girdle round about the earth in forty minutes,'" muttered
+George, quoting the words of Puck in <i>A Midsummer Night's Dream</i>.</p>
+
+<p>"Now, if six yards were added to the length of the girdle, what would
+then be the distance between the girdle and the earth, supposing that
+distance to be equal all round?"</p>
+
+<p>"In such a great length," said Mr. Allgood, "I do not suppose the
+distance would be worth mentioning."</p>
+
+<p>"What do you say, George?" asked Mr. Smoothly.</p>
+
+<p>"Well, without calculating I should imagine it would be a very minute
+fraction of an inch."</p>
+
+<p>Reginald and Mr. Filkins were of the same opinion.</p>
+
+<p>"I think it will surprise you all," said the curate, "to learn that
+those extra six yards would make the distance from the earth all round
+the girdle very nearly a yard!"</p>
+
+<p>"Very nearly a yard!" everybody exclaimed, with astonishment; but Mr.
+Smoothly was quite correct. The increase is independent of the
+original length of the girdle, which may be round the earth or round
+an orange; in any case the additional six yards will give a distance
+of nearly a yard all round. This is apt to surprise the
+non-mathematical mind.</p>
+
+<p>"Did you hear the story of the extraordinary precocity of Mrs.
+Perkins's baby that died last week?" asked Mrs. Allgood. "It was only
+three months old, and lying at the point of death, when the
+grief-stricken mother asked the doctor if nothing could save it.
+'Absolutely nothing!' said the doctor. Then the infant looked up
+pitifully into its mother's face and said&mdash;absolutely nothing!"</p>
+
+<p>"Impossible!" insisted Mildred. "And only three months old!"</p>
+
+<p><span class='pagenum'>Pg 140<a name="Page_140" id="Page_140"></a></span>"There have been extraordinary cases of infantile precocity," said
+Mr. Filkins, "the truth of which has often been carefully attested.
+But are you sure this really happened, Mrs. Allgood?"</p>
+
+<p>"Positive," replied the lady. "But do you really think it astonishing
+that a child of three months should say absolutely nothing? What would
+you expect it to say?"</p>
+
+<p>"Speaking of death," said Mr. Smoothly, solemnly, "I knew two men,
+father and son, who died in the same battle during the South African
+War. They were both named Andrew Johnson and buried side by side, but
+there was some difficulty in distinguishing them on the headstones.
+What would you have done?"</p>
+
+<p>"Quite simple," said Mr. Allgood. "They should have described one as
+'Andrew Johnson, Senior,' and the other as 'Andrew Johnson, Junior.'"</p>
+
+<p>"But I forgot to tell you that the father died first."</p>
+
+<p>"What difference can that make?"</p>
+
+<p>"Well, you see, they wanted to be absolutely exact, and that was the
+difficulty."</p>
+
+<p>"But I don't see any difficulty," said Mr. Allgood, nor could anybody
+else.</p>
+
+<p>"Well," explained Mr. Smoothly, "it is like this. If the father died
+first, the son was then no longer 'Junior.' Is that so?"</p>
+
+<p>"To be strictly exact, yes."</p>
+
+<p>"That is just what they wanted&mdash;to be strictly exact. Now, if he was
+no longer 'Junior,' then he did not die 'Junior." Consequently it must
+be incorrect so to describe him on the headstone. Do you see the
+point?"</p>
+
+<p>"Here is a rather curious thing," said Mr. Filkins, "that I have just
+remembered. A man wrote to me the other day that he had recently
+discovered two old coins while digging in his garden. One was dated
+'51 B.C.,' and the other one marked 'George I.' How do I know that he
+was not writing the truth?"</p>
+
+<p>"Perhaps you know the man to be addicted to lying," said Reginald.</p>
+
+<p>"But that would be no proof that he was not telling the truth in this
+instance."</p>
+
+<p>"Perhaps," suggested Mildred, "you know that there were no coins made
+at those dates.</p>
+
+<p>"On the contrary, they were made at both periods."</p>
+
+<p>"Were they silver or copper coins?" asked Willie.</p>
+
+<p>"My friend did not state, and I really cannot see, Willie, that it
+makes any difference."</p>
+
+<p>"I see it!" shouted Reginald. "The letters 'B.C.' would never be used
+on a coin made before the birth of Christ. They never anticipated the
+event in that way. The letters were only adopted later to denote dates
+previous to those which we call 'A.D.' That is very good; but I cannot
+see why the other statement could not be correct."</p>
+
+<p>"Reginald is quite right," said Mr. Filkins, "about the first coin.
+The second one could not exist, because the first George would never
+be described in his lifetime as 'George I.'"</p>
+
+<p>"Why not?" asked Mrs. Allgood. "He <i>was</i> George I."</p>
+
+<p>"Yes; but they would not know it until there was a George II."</p>
+
+<p>"Then there was no George II. until George III. came to the throne?"</p>
+
+<p>"That does not follow. The second George becomes 'George II.' on
+account of there having been a 'George I.'"</p>
+
+<p>"Then the first George was 'George I.' on account of there having been
+no king of that name before him."</p>
+
+<p>"Don't you see, mother," said George Allgood, "we did not call Queen
+Victoria 'Victoria I.;' but if there is ever a 'Victoria II.,' then
+she will be known that way."</p>
+
+<p>"But there <i>have</i> been several Georges, and therefore he was 'George
+I.' There <i>haven't</i> been several Victorias, so the two cases are not
+similar."</p>
+
+<p>They gave up the attempt to convince Mrs. Allgood, but the reader
+will, of course, see the point clearly.</p>
+
+<p>"Here is a question," said Mildred Allgood, "that I should like some
+of you to settle for me. I am accustomed to buy from our greengrocer
+bundles of asparagus, each 12 inches in circumference. I always put a
+tape measure round them to make sure I am getting the full quantity.
+The other day the man had no large bundles in stock, but handed me
+instead two small ones, each 6 inches in circumference. 'That is the
+same thing,' I said, 'and, of course, the price will be the same;' but
+he insisted that the two bundles together contained more than the
+large one, and charged me a few pence extra. Now, what I want to know
+is, which of us was correct? Would the two small bundles contain the
+same quantity as the large one? Or would they contain more?"</p>
+
+<p>"That is the ancient puzzle," said Reginald, laughing, "of the sack of
+corn that Sempronius borrowed from Caius, which your greengrocer,
+perhaps, had been reading about somewhere. He caught you beautifully."</p>
+
+<p>"Then they were equal?"</p>
+
+<p>"On the contrary, you were both wrong, and you were badly cheated. You
+only got half the quantity that would have been contained in a large
+bundle, and therefore ought to have been charged half the original
+price, instead of more."</p>
+
+<p>Yes, it was a bad swindle, undoubtedly. A circle with a circumference
+half that of another must have its area a quarter that of the other.
+Therefore the two small bundles contained together only half as much
+asparagus as a large one.</p>
+
+<p>"Mr. Filkins, can you answer this?" asked Willie. "There is a man in
+the next village who eats two eggs for breakfast every morning."</p>
+
+<p>"Nothing very extraordinary in that," George broke in. "If you told us
+that the two eggs ate the man it would be interesting."</p>
+
+<p>"Don't interrupt the boy, George," said his mother.</p>
+
+<p>"Well," Willie continued, "this man neither buys, borrows, barters,
+begs, steals, nor finds the eggs. He doesn't keep hens, and the eggs
+<span class='pagenum'>Pg 141<a name="Page_141" id="Page_141"></a></span>are not given to him. How does he get the eggs?"</p>
+
+<p>"Does he take them in exchange for something else?" asked Mildred.</p>
+
+<p>"That would be bartering them," Willie replied.</p>
+
+<p>"Perhaps some friend sends them to him," suggested Mrs. Allgood.</p>
+
+<p>"I said that they were not given to him."</p>
+
+<p>"I know," said George, with confidence. "A strange hen comes into his
+place and lays them."</p>
+
+<p>"But that would be finding them, wouldn't it?"</p>
+
+<p>"Does he hire them?" asked Reginald.</p>
+
+<p>"If so, he could not return them after they were eaten, so that would
+be stealing them."</p>
+
+<p>"Perhaps it is a pun on the word 'lay,'" Mr. Filkins said. "Does he
+lay them on the table?"</p>
+
+<p>"He would have to get them first, wouldn't he? The question was, How
+does he get them?"</p>
+
+<p>"Give it up!" said everybody. Then little Willie crept round to the
+protection of his mother, for George was apt to be rough on such
+occasions.</p>
+
+<p>"The man keeps ducks!" he cried, "and his servant collects the eggs
+every morning."</p>
+
+<p>"But you said he doesn't keep birds!" George protested.</p>
+
+<p>"I didn't, did I, Mr. Filkins? I said he doesn't keep hens."</p>
+
+<p>"But he finds them," said Reginald.</p>
+
+<p>"No; I said his servant finds them."</p>
+
+<p>"Well, then," Mildred interposed, "his servant gives them to him."</p>
+
+<p>"You cannot give a man his own property, can you?"</p>
+
+<p>All agreed that Willie's answer was quite satisfactory. Then Uncle
+John produced a little fallacy that "brought the proceedings to a
+close," as the newspapers say.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_413_A_CHESSBOARD_FALLACY" id="X_413_A_CHESSBOARD_FALLACY"></a><a href="#X_413_A_CHESSBOARD_FALLACYa"><b>413.&mdash;A CHESSBOARD FALLACY.</b></a></p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/q413.png" width="500" height="286" alt="" title="" />
+</div>
+
+<p>"Here is a diagram of a chessboard," he said. "You see there are
+sixty-four squares&mdash;eight by eight. Now I draw a straight line from
+the top left-hand corner, where the first and second squares meet, to
+the bottom right-hand corner. I cut along this line with the scissors,
+slide up the piece that I have marked B, and then clip off the little
+corner C by a cut along the first upright line. This little piece will
+exactly fit into its place at the top, and we now have an oblong with
+seven squares on one side and nine squares on the other. There are,
+therefore, now only sixty-three squares, because seven multiplied by
+nine makes sixty-three. Where on earth does that lost square go to? I
+have tried over and over again to catch the little beggar, but he
+always eludes me. For the life of me I cannot discover where he hides
+himself."</p>
+
+<p>"It seems to be like the other old chessboard fallacy, and perhaps the
+explanation is the same," said Reginald&mdash;"that the pieces do not
+exactly fit."</p>
+
+<p>"But they <i>do</i> fit," said Uncle John. "Try it, and you will see."</p>
+
+<p>Later in the evening Reginald and George, were seen in a corner with
+their heads together, trying to catch that elusive little square, and
+it is only fair to record that before they retired for the night they
+succeeded in securing their prey, though some others of the company
+failed to see it when captured. Can the reader solve the little
+mystery?</p>
+
+
+
+<hr style="width: 65%;" /><p><span class='pagenum'>Pg 142<a name="Page_142" id="Page_142"></a></span></p>
+<h2><a name="UNCLASSIFIED_PROBLEMS" id="UNCLASSIFIED_PROBLEMS"></a><a href="#CONTENTS">UNCLASSIFIED PROBLEMS.</a></h2>
+
+
+<p class='center'>
+"A snapper up of unconsidered trifles."<br />
+<span style="margin-left: 8em;"><i>Winter's Tale</i>, iv. 2.</span><br />
+</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_414_WHO_WAS_FIRST" id="X_414_WHO_WAS_FIRST"></a><a href="#X_414_WHO_WAS_FIRSTa"><b>414.&mdash;WHO WAS FIRST?</b></a></p>
+
+<p>Anderson, Biggs, and Carpenter were staying together at a place by the
+seaside. One day they went out in a boat and were a mile at sea when a
+rifle was fired on shore in their direction. Why or by whom the shot
+was fired fortunately does not concern us, as no information on these
+points is obtainable, but from the facts I picked up we can get
+material for a curious little puzzle for the novice.</p>
+
+<p>It seems that Anderson only heard the report of the gun, Biggs only
+saw the smoke, and Carpenter merely saw the bullet strike the water
+near them. Now, the question arises: Which of them first knew of the
+discharge of the rifle?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_415_A_WONDERFUL_VILLAGE" id="X_415_A_WONDERFUL_VILLAGE"></a><a href="#X_415_A_WONDERFUL_VILLAGEa"><b>415.&mdash;A WONDERFUL VILLAGE.</b></a></p>
+
+<p>There is a certain village in Japan, situated in a very low valley,
+and yet the sun is nearer to the inhabitants every noon, by 3,000
+miles and upwards, than when he either rises or sets to these people.
+In what part of the country is the village situated?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_416_A_CALENDAR_PUZZLE" id="X_416_A_CALENDAR_PUZZLE"></a><a href="#X_416_A_CALENDAR_PUZZLEa"><b>416.&mdash;A CALENDAR PUZZLE.</b></a></p>
+
+<p>If the end of the world should come on the first day of a new century,
+can you say what are the chances that it will happen on a Sunday?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_417_THE_TIRING_IRONS" id="X_417_THE_TIRING_IRONS"></a><a href="#X_417_THE_TIRING_IRONSa"><b>417.&mdash;THE TIRING IRONS.</b></a></p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q417a.png" width="600" height="174" alt="" title="" />
+</div>
+
+<p>The illustration represents one of the most ancient of all mechanical
+puzzles. Its origin is unknown. Cardan, the mathematician, wrote about
+it in 1550, and Wallis in 1693; while it is said still to be found in
+obscure English villages (sometimes deposited in strange places, such
+as a church belfry), made of iron, and appropriately called
+"tiring-irons," and to be used by the Norwegians to-day as a lock for
+boxes and bags. In the toyshops it is sometimes called the "Chinese
+rings," though there seems to be no authority for the description, and
+it more frequently goes by the unsatisfactory name of "the puzzling
+rings." The French call it "Baguenaudier."</p>
+
+<p>The puzzle will be seen to consist of a simple <i>loop</i> of wire fixed in
+a handle to be held in the left hand, and a certain number of <i>rings</i>
+secured by <i>wires</i> which pass through holes in the <i>bar</i> and are kept
+there by their blunted ends. The wires work freely in the bar, but
+cannot come apart from it, nor can the wires be removed from the
+rings. The general puzzle is to detach the loop completely from all
+the rings, and then to put them all on again.</p>
+
+<p>Now, it will be seen at a glance that the first ring (to the right)
+can be taken off at any time by sliding it over the end and dropping
+it through the loop; or it may be put on by reversing the operation.
+With this exception, the only ring that can ever be removed is the one
+that happens to be a contiguous second on the loop at the right-hand
+end. Thus, with all the rings on, the second can be dropped at once;
+with the first ring down, you cannot drop the second, but may remove
+the third; with the first three rings down, you cannot drop the
+fourth, but may remove the fifth; and so on. It will be found that the
+first and second rings can be dropped together or put on together; but
+to prevent confusion we will throughout disallow this exceptional
+double move, and say that only one ring may be put on or removed at a
+time.</p>
+
+<p>We can thus take off one ring in 1 move; two rings in 2 moves; three
+rings in 5 moves; four rings in 10 moves; five rings in 21 moves; and
+if we keep on doubling (and adding one where the number of rings is
+odd) we may easily ascertain the number of moves for completely
+removing any number of rings. To get off all the seven rings requires
+85 moves. Let us look at the five moves made in removing the first
+three rings, the circles above the line standing for rings on the loop
+and those under for rings off the loop.</p>
+
+<p>Drop the first ring; drop the third; put up the first; drop the
+second; and drop the first&mdash;5 moves, as shown clearly in the diagrams.
+The dark circles show at each stage, from the starting position to the
+finish, which rings it is possible to drop. After move 2 it will be
+noticed that no ring can be dropped until one has been put on, because
+the first and second rings from the right now on the loop are not
+together. After the fifth move, if we wish to remove all seven <span class='pagenum'>Pg 143<a name="Page_143" id="Page_143"></a></span>rings
+we must now drop the fifth. But before we can then remove the fourth
+it is necessary to put on the first three and remove the first two. We
+shall then have 7, 6, 4, 3 on the loop, and may therefore drop the
+fourth. When we have put on 2 and 1 and removed 3, 2, 1, we may drop
+the seventh ring. The next operation then will be to get 6, 5, 4, 3,
+2, 1 on the loop and remove 4, 3, 2, 1, when 6 will come off; then get
+5, 4, 3, 2, 1 on the loop, and remove 3, 2, 1, when 5 will come off;
+then get 4, 3, 2, 1 on the loop and remove 2, 1, when 4 will come off;
+then get 3, 2, 1 on the loop and remove 1, when 3 will come off; then
+get 2, 1 on the loop, when 2 will come off; and 1 will fall through on
+the 85th move, leaving the loop quite free. The reader should now be
+able to understand the puzzle, whether or not he has it in his hand in
+a practical form.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q417b.png" width="400" height="755" alt="" title="" />
+</div>
+
+<p>The particular problem I propose is simply this. Suppose there are
+altogether fourteen rings on the tiring-irons, and we proceed to take
+them all off in the correct way so as not to waste any moves. What
+will be the position of the rings after the 9,999th move has been
+made?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_418_SUCH_A_GETTING_UPSTAIRS" id="X_418_SUCH_A_GETTING_UPSTAIRS"></a><a href="#X_418_SUCH_A_GETTING_UPSTAIRSa"><b>418.&mdash;SUCH A GETTING UPSTAIRS.</b></a></p>
+
+<p>In a suburban villa there is a small staircase with eight steps, not
+counting the landing. The little puzzle with which Tommy Smart
+perplexed his family is this. You are required to start from the
+bottom and land twice on the floor above (stopping there at the
+finish), having returned once to the ground floor. But you must be
+careful to use every tread the same number of times. In how few steps
+can you make the ascent? It seems a very simple matter, but it is more
+than likely that at your first attempt you will make a great many more
+steps than are necessary. Of course you must not go more than one
+riser at a time.</p>
+
+<p>Tommy knows the trick, and has shown it to his father, who professes
+to have a contempt for such things; but when the children are in bed
+the pater will often take friends out into the hall and enjoy a good
+laugh at their bewilderment. And yet it is all so very simple when you
+know how it is done.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_419_THE_FIVE_PENNIES" id="X_419_THE_FIVE_PENNIES"></a><a href="#X_419_THE_FIVE_PENNIESa"><b>419.&mdash;THE FIVE PENNIES.</b></a></p>
+
+<p>Here is a really hard puzzle, and yet its conditions are so absurdly
+simple. Every reader knows how to place four pennies so that they are
+equidistant from each other. All you have to do is to arrange three of
+them flat on the table so that they touch one another in the form of a
+triangle, and lay the fourth penny on top in the centre. Then, as
+every penny touches every other penny, they are all at equal distances
+from one another. Now try to do the same thing with five
+pennies&mdash;place them so that every penny shall touch every other
+penny&mdash;and you will find it a different matter altogether.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_420_THE_INDUSTRIOUS_BOOKWORM" id="X_420_THE_INDUSTRIOUS_BOOKWORM"></a><a href="#X_420_THE_INDUSTRIOUS_BOOKWORMa"><b>420.&mdash;THE INDUSTRIOUS BOOKWORM.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q420.png" width="400" height="441" alt="" title="" />
+</div>
+
+<p>Our friend Professor Rackbrane is seen in the illustration to be
+propounding another of his <span class='pagenum'>Pg 144<a name="Page_144" id="Page_144"></a></span>little posers. He is explaining that since
+he last had occasion to take down those three volumes of a learned
+book from their place on his shelves a bookworm has actually bored a
+hole straight through from the first page to the last. He says that
+the leaves are together three inches thick in each volume, and that
+every cover is exactly one-eighth of an inch thick, and he asks how
+long a tunnel had the industrious worm to bore in preparing his new
+tube railway. Can you tell him?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_421_A_CHAIN_PUZZLE" id="X_421_A_CHAIN_PUZZLE"></a><a href="#X_421_A_CHAIN_PUZZLEa"><b>421.&mdash;A CHAIN PUZZLE.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q421.png" width="400" height="335" alt="" title="" />
+</div>
+
+<p>This is a puzzle based on a pretty little idea first dealt with by the
+late Mr. Sam Loyd. A man had nine pieces of chain, as shown in the
+illustration. He wanted to join these fifty links into one endless
+chain. It will cost a penny to open any link and twopence to weld a
+link together again, but he could buy a new endless chain of the same
+character and quality for 2<i>s</i>. 2<i>d</i>. What was the cheapest course for him
+to adopt? Unless the reader is cunning he may find himself a good way
+out in his answer.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_422_THE_SABBATH_PUZZLE" id="X_422_THE_SABBATH_PUZZLE"></a><a href="#X_422_THE_SABBATH_PUZZLEa"><b>422.&mdash;THE SABBATH PUZZLE.</b></a></p>
+
+<p>I have come across the following little poser in an old book. I wonder
+how many readers will see the author's intended solution to the
+riddle.</p>
+
+<p>
+<span style="margin-left: 2em;">Christians the week's <i>first</i> day for Sabbath hold;</span><br />
+<span style="margin-left: 2em;">The Jews the <i>seventh</i>, as they did of old;</span><br />
+<span style="margin-left: 2em;">The Turks the <i>sixth</i>, as we have oft been told.</span><br />
+<span style="margin-left: 2em;">How can these three, in the same place and day,</span><br />
+<span style="margin-left: 2em;">Have each his own true Sabbath? tell, I pray.</span><br />
+</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_423_THE_RUBY_BROOCH" id="X_423_THE_RUBY_BROOCH"></a><a href="#X_423_THE_RUBY_BROOCHa"><b>423.&mdash;THE RUBY BROOCH.</b></a></p>
+
+<p>The annals of Scotland Yard contain some remarkable cases of jewel
+robberies, but one of the most perplexing was the theft of Lady
+Littlewood's rubies. There have, of course, been many greater
+robberies in point of value, but few so artfully conceived. Lady
+Littlewood, of Romley Manor, had a beautiful but rather eccentric
+heirloom in the form of a ruby brooch. While staying at her town house
+early in the eighties she took the jewel to a shop in Brompton for
+some slight repairs.</p>
+
+<p>"A fine collection of rubies, madam," said the shopkeeper, to whom her
+ladyship was a stranger.</p>
+
+<p>"Yes," she replied; "but curiously enough I have never actually
+counted them. My mother once pointed out to me that if you start from
+the centre and count up one line, along the outside and down the next
+line, there are always eight rubies. So I should always know if a
+stone were missing."</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q423.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p>Six months later a brother of Lady Little<span class='pagenum'>Pg 145<a name="Page_145" id="Page_145"></a></span>wood's, who had returned
+from his regiment in India, noticed that his sister was wearing the
+ruby brooch one night at a county ball, and on their return home asked
+to look at it more closely. He immediately detected the fact that four
+of the stones were gone.</p>
+
+<p>"How can that possibly be?" said Lady Littlewood. "If you count up one
+line from the centre, along the edge, and down the next line, in any
+direction, there are always eight stones. This was always so and is so
+now. How, therefore, would it be possible to remove a stone without my
+detecting it?"</p>
+
+<p>"Nothing could be simpler," replied the brother. "I know the brooch
+well. It originally contained forty-five stones, and there are now
+only forty-one. Somebody has stolen four rubies, and then reset as
+small a number of the others as possible in such a way that there
+shall always be eight in any of the directions you have mentioned."</p>
+
+<p>There was not the slightest doubt that the Brompton jeweller was the
+thief, and the matter was placed in the hands of the police. But the
+man was wanted for other robberies, and had left the neighbourhood
+some time before. To this day he has never been found.</p>
+
+<p>The interesting little point that at first baffled the police, and
+which forms the subject of our puzzle, is this: How were the
+forty-five rubies originally arranged on the brooch? The illustration
+shows exactly how the forty-one were arranged after it came back from
+the jeweller; but although they count eight correctly in any of the
+directions mentioned, there are four stones missing.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_424_THE_DOVETAILED_BLOCK" id="X_424_THE_DOVETAILED_BLOCK"></a><a href="#X_424_THE_DOVETAILED_BLOCKa"><b>424.&mdash;THE DOVETAILED BLOCK.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q424.png" width="400" height="320" alt="" title="" />
+</div>
+
+<p>Here is a curious mechanical puzzle that was given to me some years
+ago, but I cannot say who first invented it. It consists of two solid
+blocks of wood securely dovetailed together. On the other two vertical
+sides that are not visible the appearance is precisely the same as on
+those shown. How were the pieces put together? When I published this
+little puzzle in a London newspaper I received (though they were
+unsolicited) quite a stack of models, in oak, in teak, in mahogany,
+rosewood, satinwood, elm, and deal; some half a foot in length, and
+others varying in size right down to a delicate little model about
+half an inch square. It seemed to create considerable interest.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_425_JACK_AND_THE_BEANSTALK" id="X_425_JACK_AND_THE_BEANSTALK"></a><a href="#X_425_JACK_AND_THE_BEANSTALKa"><b>425.&mdash;JACK AND THE BEANSTALK.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q425.png" width="400" height="633" alt="" title="" />
+</div>
+
+<p>The illustration, by a British artist, is a sketch of Jack climbing
+the beanstalk. Now, the artist has made a serious blunder in this
+drawing. Can you find out what it is?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_426_THE_HYMN-BOARD_POSER" id="X_426_THE_HYMN-BOARD_POSER"></a><a href="#X_426_THE_HYMN-BOARD_POSERa"><b>426.&mdash;THE HYMN-BOARD POSER.</b></a></p>
+
+<p>The worthy vicar of Chumpley St. Winifred is in great distress. A
+little church difficulty has arisen that all the combined intelligence
+of the parish seems unable to surmount. What this difficulty is I will
+state hereafter, but it may add to the interest of the problem if I
+first give a short account of the curious position that has been
+brought about. It all has to do with the church hymn-boards, the
+plates of which have become so damaged that they have ceased to fulfil
+the purpose for which they were devised. A generous parishioner has
+promised to pay for a new set of plates at a certain rate of cost; but
+strange as it may seem, no agreement can be come to as to what that
+cost should be. The proposed maker of the plates has named <span class='pagenum'>Pg 146<a name="Page_146" id="Page_146"></a></span>a price
+which the donor declares to be absurd. The good vicar thinks they are
+both wrong, so he asks the schoolmaster to work out the little sum.
+But this individual declares that he can find no rule bearing on the
+subject in any of his arithmetic books. An application having been
+made to the local medical practitioner, as a man of more than average
+intellect at Chumpley, he has assured the vicar that his practice is
+so heavy that he has not had time even to look at it, though his
+assistant whispers that the doctor has been sitting up unusually late
+for several nights past. Widow Wilson has a smart son, who is reputed
+to have once won a prize for puzzle-solving. He asserts that as he
+cannot find any solution to the problem it must have something to do
+with the squaring of the circle, the duplication of the cube, or the
+trisection of an angle; at any rate, he has never before seen a puzzle
+on the principle, and he gives it up.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q426.png" width="400" height="523" alt="" title="" />
+</div>
+
+<p>This was the state of affairs when the assistant curate (who, I should
+say, had frankly confessed from the first that a profound study of
+theology had knocked out of his head all the knowledge of mathematics
+he ever possessed) kindly sent me the puzzle.</p>
+
+<p>A church has three hymn-boards, each to indicate the numbers of five
+different hymns to be sung at a service. All the boards are in use at
+the same service. The hymn-book contains 700 hymns. A new set of
+numbers is required, and a kind parishioner offers to present a set
+painted on metal plates, but stipulates that only the smallest number
+of plates necessary shall be purchased. The cost of each plate is to
+be 6<i>d</i>., and for the painting of each plate the charges are to be: For
+one plate, 1<i>s</i>.; for two plates alike, 11�<i>d</i>. each; for three plates
+alike, 11�<i>d</i>. each, and so on, the charge being one farthing less
+per plate for each similarly painted plate. Now, what should be the
+lowest cost?</p>
+
+<p>Readers will note that they are required to use every legitimate and
+practical method of economy. The illustration will make clear the
+nature of the three hymn-boards and plates. The five hymns are here
+indicated by means of twelve plates. These plates slide in separately
+at the back, and in the illustration there is room, of course, for
+three more plates.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_427_PHEASANT-SHOOTING" id="X_427_PHEASANT-SHOOTING"></a><a href="#X_427_PHEASANT-SHOOTINGa"><b>427.&mdash;PHEASANT-SHOOTING.</b></a></p>
+
+<p>A Cockney friend, who is very apt to draw the long bow, and is
+evidently less of a sportsman than he pretends to be, relates to me
+the following not very credible yarn:&mdash;</p>
+
+<p>"I've just been pheasant-shooting with my friend the duke. We had
+splendid sport, and I made some wonderful shots. What do you think of
+this, for instance? Perhaps you can twist it into a puzzle. The duke
+and I were crossing a field when suddenly twenty-four pheasants rose
+on the wing right in front of us. I fired, and two-thirds of them
+dropped dead at my feet. Then the duke had a shot at what were left,
+and brought down three-twenty-fourths of them, wounded in the wing.
+Now, out of those twenty-four birds, how many still remained?"</p>
+
+<p>It seems a simple enough question, but can the reader give a correct
+answer?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_428_THE_GARDENER_AND_THE_COOK" id="X_428_THE_GARDENER_AND_THE_COOK"></a><a href="#X_428_THE_GARDENER_AND_THE_COOKa"><b>428.&mdash;THE GARDENER AND THE COOK.</b></a></p>
+
+<p>A correspondent, signing himself "Simple Simon," suggested that I
+should give a special catch puzzle in the issue of <i>The Weekly
+Dispatch</i> for All Fools' Day, 1900. So I gave the following, and it
+caused considerable amusement; for out of a very large body of
+competitors, many quite expert, not a single person solved it, though
+it ran for nearly a month.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q428.png" width="400" height="365" alt="" title="" />
+</div>
+
+<p>"The illustration is a fancy sketch of my correspondent, 'Simple
+Simon,' in the act of trying to solve the following innocent little
+<span class='pagenum'>Pg 147<a name="Page_147" id="Page_147"></a></span>arithmetical puzzle. A race between a man and a woman that I happened
+to witness one All Fools' Day has fixed itself indelibly on my memory.
+It happened at a country-house, where the gardener and the cook
+decided to run a race to a point 100 feet straight away and return. I
+found that the gardener ran 3 feet at every bound and the cook only 2
+feet, but then she made three bounds to his two. Now, what was the
+result of the race?"</p>
+
+<p>A fortnight after publication I added the following note: "It has been
+suggested that perhaps there is a catch in the 'return,' but there is
+not. The race is to a point 100 feet away and home again&mdash;that is, a
+distance of 200 feet. One correspondent asks whether they take exactly
+the same time in turning, to which I reply that they do. Another seems
+to suspect that it is really a conundrum, and that the answer is that
+'the result of the race was a (matrimonial) tie.' But I had no such
+intention. The puzzle is an arithmetical one, as it purports to be."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_429_PLACING_HALFPENNIES" id="X_429_PLACING_HALFPENNIES"></a><a href="#X_429_PLACING_HALFPENNIESa"><b>429.&mdash;PLACING HALFPENNIES.</b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/q429.png" width="400" height="261" alt="" title="" />
+</div>
+
+<p>Here is an interesting little puzzle suggested to me by Mr. W. T.
+Whyte. Mark off on a sheet of paper a rectangular space 5 inches by 3
+inches, and then find the greatest number of halfpennies that can be
+placed within the enclosure under the following conditions. A
+halfpenny is exactly an inch in diameter. Place your first halfpenny
+where you like, then place your second coin at exactly the distance of
+an inch from the first, the third an inch distance from the second,
+and so on. No halfpenny may touch another halfpenny or cross the
+boundary. Our illustration will make the matter perfectly clear. No. 2
+coin is an inch from No. 1; No. 3 an inch from No. 2; No. 4 an inch
+from No. 3; but after No. 10 is placed we can go no further in this
+attempt. Yet several more halfpennies might have been got in. How many
+can the reader place?</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_430_FIND_THE_MANS_WIFE" id="X_430_FIND_THE_MANS_WIFE"></a><a href="#X_430_FIND_THE_MANS_WIFEa"><b>430.&mdash;FIND THE MAN'S WIFE.</b></a></p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/q430.png" width="600" height="217" alt="" title="" />
+</div>
+
+<p>One summer day in 1903 I was loitering on the Brighton front, watching
+the people strolling about on the beach, when the friend who was with
+me suddenly drew my attention to an individual who was standing alone,
+and said, "Can you point out that man's wife? They are stopping at the
+same hotel as I am, and the lady is one of those in view." After a few
+minutes' observation, I was successful in indicating the lady
+correctly. My friend was curious to know by what method of reasoning I
+had arrived at the result. This was my answer:&mdash;</p>
+
+<p>"We may at once exclude that Sister of Mercy and the girl in the short
+frock; also the woman selling oranges. It cannot be the lady in
+widows' weeds. It is not the lady in the bath chair, because she is
+not staying at your hotel, for I happened to see her come out of a
+private house this morning assisted by her maid. The two ladies in red
+breakfasted at my hotel this morning, and as they were not wearing
+outdoor dress I conclude they are staying there. It therefore rests
+between the lady in blue and the one with the green parasol. But the
+left hand that holds the parasol is, you see, ungloved and bears no
+wedding-ring. Consequently I am driven to the conclusion that the lady
+in blue is the man's wife&mdash;and you say this is correct."</p>
+
+<p>Now, as my friend was an artist, and as I thought an amusing puzzle
+might be devised on the lines of his question, I asked him to make me
+a drawing according to some directions that I gave him, and I have
+pleasure in presenting his production to my readers. It will be seen
+that the picture shows six men and six ladies: Nos. 1, 3, 5, 7, 9, and
+11 are ladies, and Nos. 2, <span class='pagenum'>Pg 148<a name="Page_148" id="Page_148"></a></span>4, 6, 8, 10, and 12 are men. These twelve
+individuals represent six married couples, all strangers to one
+another, who, in walking aimlessly about, have got mixed up. But we
+are only concerned with the man that is wearing a straw hat&mdash;Number
+10. The puzzle is to find this man's wife. Examine the six ladies
+carefully, and see if you can determine which one of them it is.</p>
+
+<p>I showed the picture at the time to a few friends, and they expressed
+very different opinions on the matter. One said, "I don't believe he
+would marry a girl like Number 7." Another said, "I am sure a nice
+girl like Number 3 would not marry such a fellow!" Another said, "It
+must be Number 1, because she has got as far away as possible from the
+brute!" It was suggested, again, that it must be Number 11, because
+"he seems to be looking towards her;" but a cynic retorted, "For that
+very reason, if he is really looking at her, I should say that she is
+not his wife!"</p>
+
+<p>I now leave the question in the hands of my readers. Which is really
+Number 10's wife?</p>
+
+<p>The illustration is of necessity considerably reduced from the large
+scale on which it originally appeared in <i>The Weekly Dispatch</i> (24th
+May 1903), but it is hoped that the details will be sufficiently clear
+to allow the reader to derive entertainment from its examination. In
+any case the solution given will enable him to follow the points with
+interest.<br /><br /><br /><br /></p>
+
+
+
+<hr style="width: 65%;" />
+<h2><a name="SOLUTIONS" id="SOLUTIONS"></a><a href="#CONTENTS">SOLUTIONS.</a><br /><br /><br /></h2>
+
+<hr style="width: 30%;" />
+<p><a name="X_1_A_POST-OFFICE_PERPLEXITYa" id="X_1_A_POST-OFFICE_PERPLEXITYa"></a><a href="#X_1_A_POST-OFFICE_PERPLEXITY"><b>1.&mdash;A POST-OFFICE PERPLEXITY.&mdash;<i>solution</i></b></a></p>
+
+<p>The young lady supplied 5 twopenny stamps, 30 penny stamps, and 8
+twopence-halfpenny stamps, which delivery exactly fulfils the
+conditions and represents a cost of five shillings.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_2_YOUTHFUL_PRECOCITYa" id="X_2_YOUTHFUL_PRECOCITYa"></a><a href="#X_2_YOUTHFUL_PRECOCITY"><b>2.&mdash;YOUTHFUL PRECOCITY.&mdash;<i>solution</i></b></a></p>
+
+<p>The price of the banana must have been one penny farthing. Thus, 960
+bananas would cost &pound;5, and 480 sixpences would buy 2,304 bananas.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_3_AT_A_CATTLE_MARKETa" id="X_3_AT_A_CATTLE_MARKETa"></a><a href="#X_3_AT_A_CATTLE_MARKET"><b>3.&mdash;AT A CATTLE MARKET.&mdash;<i>solution</i></b></a></p>
+
+<p>Jakes must have taken 7 animals to market, Hodge must have taken 11,
+and Durrant must have taken 21. There were thus 39 animals altogether.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_4_THE_BEANFEAST_PUZZLEa" id="X_4_THE_BEANFEAST_PUZZLEa"></a><a href="#X_4_THE_BEANFEAST_PUZZLE"><b>4.&mdash;THE BEANFEAST PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The cobblers spent 35<i>s</i>., the tailors spent also 35<i>s</i>., the hatters
+spent 42<i>s</i>., and the glovers spent 21<i>s</i>. Thus, they spent altogether
+&pound;6,13<i>s</i>., while it will be found that the five cobblers spent as much
+as four tailors, twelve tailors as much as nine hatters, and six
+hatters as much as eight glovers.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_5_A_QUEER_COINCIDENCEa" id="X_5_A_QUEER_COINCIDENCEa"></a><a href="#X_5_A_QUEER_COINCIDENCE"><b>5.&mdash;A QUEER COINCIDENCE.&mdash;<i>solution</i></b></a></p>
+
+<p>Puzzles of this class are generally solved in the old books by the
+tedious process of "working backwards." But a simple general solution
+is as follows: If there are <i>n</i> players, the amount held by every
+player at the end will be <i>m</i>(2<sup><i>n</i></sup>), the last winner must have held
+<i>m</i>(<i>n</i>+1) at the start, the next <i>m</i>(2<i>n</i>+1), the next <i>m</i>(4<i>n</i>+1),
+the next <i>m</i>(8<i>n</i>+1), and so on to the first player, who must have
+held <i>m</i>(2<sup><i>n</i>-1</sup><i>n</i>+1).</p>
+
+<p>Thus, in this case, <i>n</i> = 7, and the amount held by every player at the
+end was 2<sup>7</sup> farthings. Therefore <i>m</i> = 1, and G started with 8
+farthings, F with 15, E with 29, D with 57, C with 113, B with 225,
+and A with 449 farthings.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_6_A_CHARITABLE_BEQUESTa" id="X_6_A_CHARITABLE_BEQUESTa"></a><a href="#X_6_A_CHARITABLE_BEQUEST"><b>6.&mdash;A CHARITABLE BEQUEST.&mdash;<i>solution</i></b></a></p>
+
+<p>There are seven different ways in which the money may be distributed:
+5 women and 19 men, 10 women and 16 men, 15 women and 13 men, 20 women
+and 10 men, 25 women and 7 men, 30 women and 4 men, and 35 women and 1
+man. But the last case must not be counted, because the condition was
+that there should be "men," and a single man is not men. Therefore the
+answer is six years.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_7_THE_WIDOWS_LEGACYa" id="X_7_THE_WIDOWS_LEGACYa"></a><a href="#X_7_THE_WIDOWS_LEGACY"><b>7.&mdash;THE WIDOW'S LEGACY.&mdash;<i>solution</i></b></a></p>
+
+<p>The widow's share of the legacy must be &pound;205, 2<i>s</i>. 6<i>d</i>. and <sup>10</sup>/<sub>13</sub> of a
+penny.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_8_INDISCRIMINATE_CHARITYa" id="X_8_INDISCRIMINATE_CHARITYa"></a><a href="#X_8_INDISCRIMINATE_CHARITY"><b>8.&mdash;INDISCRIMINATE CHARITY&mdash;<i>solution</i></b></a></p>
+
+<p>The gentleman must have had 3<i>s</i>. 6<i>d</i>. in his pocket when he set out for
+home.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_9_THE_TWO_AEROPLANESa" id="X_9_THE_TWO_AEROPLANESa"></a><a href="#X_9_THE_TWO_AEROPLANES"><b>9.&mdash;THE TWO AEROPLANES.&mdash;<i>solution</i></b></a></p>
+
+<p>The man must have paid &pound;500 and &pound;750 for the two machines, making
+together &pound;1,250; but as he sold them for only &pound;1,200, he lost &pound;50 by
+the transaction.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_10_BUYING_PRESENTSa" id="X_10_BUYING_PRESENTSa"></a><a href="#X_10_BUYING_PRESENTS"><b>10.&mdash;BUYING PRESENTS.&mdash;<i>solution</i></b></a></p>
+
+<p>Jorkins had originally &pound;19, 18<i>s</i>. in his pocket, and spent &pound;9, 19<i>s</i>.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_11_THE_CYCLISTS_FEASTa" id="X_11_THE_CYCLISTS_FEASTa"></a><a href="#X_11_THE_CYCLISTS_FEAST"><b>11.&mdash;THE CYCLISTS' FEAST.&mdash;<i>solution</i></b></a></p>
+
+<p>There were ten cyclists at the feast. They should have paid 8<i>s</i>. each;
+but, owing to the departure of two persons, the remaining eight would
+pay 10<i>s</i>. each.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_12_A_QUEER_THING_IN_MONEYa" id="X_12_A_QUEER_THING_IN_MONEYa"></a><a href="#X_12_A_QUEER_THING_IN_MONEY"><b>12.&mdash;A QUEER THING IN MONEY.&mdash;<i>solution</i></b></a></p>
+
+<p>The answer is as follows: &pound;44,444, 4<i>s</i>. 4<i>d</i>. = 28, and, reduced to pence,
+10,666,612 = 28.</p>
+
+<p>It is a curious little coincidence that in the answer 10,666,612 the
+four central figures indicate the only other answer, &pound;66, 6<i>s</i>. 6<i>d</i>.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_13_A_NEW_MONEY_PUZZLEa" id="X_13_A_NEW_MONEY_PUZZLEa"></a><a href="#X_13_A_NEW_MONEY_PUZZLE"><b>13.&mdash;A NEW MONEY PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The smallest sum of money, in pounds, shillings, pence, and farthings,
+containing all the nine digits once, and once only, is &pound;2,567, 18<i>s</i>.
+9�<i>d</i>.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 149<a name="Page_149" id="Page_149"></a></span><a name="X_14_SQUARE_MONEYa" id="X_14_SQUARE_MONEYa"></a><a href="#X_14_SQUARE_MONEY"><b>14.&mdash;SQUARE MONEY.&mdash;<i>solution</i></b></a></p>
+
+<p>The answer is 1�<i>d</i>. and 3<i>d</i>. Added together they make 4�<i>d</i>., and
+1�<i>d</i>. multiplied by 3 is also 4�<i>d</i>.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_15_POCKET_MONEYa" id="X_15_POCKET_MONEYa"></a><a href="#X_15_POCKET_MONEY"><b>15.&mdash;POCKET MONEY.&mdash;<i>solution</i></b></a></p>
+
+<p>The largest possible sum is 15<i>s</i>. 9<i>d</i>., composed of a crown and a
+half-crown (or three half-crowns), four florins, and a threepenny
+piece.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_16_THE_MILLIONAIRES_PERPLEXITYa" id="X_16_THE_MILLIONAIRES_PERPLEXITYa"></a><a href="#X_16_THE_MILLIONAIRES_PERPLEXITY"><b>16.&mdash;THE MILLIONAIRE'S PERPLEXITY.&mdash;<i>solution</i></b></a></p>
+
+<p>The answer to this quite easy puzzle may, of course, be readily
+obtained by trial, deducting the largest power of 7 that is contained
+in one million dollars, then the next largest power from the
+remainder, and so on. But the little problem is intended to illustrate
+a simple direct method. The answer is given at once by converting
+1,000,000 to the septenary scale, and it is on this subject of scales
+of notation that I propose to write a few words for the benefit of
+those who have never sufficiently considered the matter.</p>
+
+<p>Our manner of figuring is a sort of perfected arithmetical shorthand,
+a system devised to enable us to manipulate numbers as rapidly and
+correctly as possible by means of symbols. If we write the number
+2,341 to represent two thousand three hundred and forty-one dollars,
+we wish to imply 1 dollar, added to four times 10 dollars, added to
+three times 100 dollars, added to two times 1,000 dollars. From the
+number in the units place on the right, every figure to the left is
+understood to represent a multiple of the particular power of 10 that
+its position indicates, while a cipher (0) must be inserted where
+necessary in order to prevent confusion, for if instead of 207 we
+wrote 27 it would be obviously misleading. We thus only require ten
+figures, because directly a number exceeds 9 we put a second figure to
+the left, directly it exceeds 99 we put a third figure to the left,
+and so on. It will be seen that this is a purely arbitrary method. It
+is working in the denary (or ten) scale of notation, a system
+undoubtedly derived from the fact that our forefathers who devised it
+had ten fingers upon which they were accustomed to count, like our
+children of to-day. It is unnecessary for us ordinarily to state that
+we are using the denary scale, because this is always understood in
+the common affairs of life.</p>
+
+<p>But if a man said that he had 6,553 dollars in the septenary (or
+seven) scale of notation, you will find that this is precisely the
+same amount as 2,341 in our ordinary denary scale. Instead of using
+powers of ten, he uses powers of 7, so that he never needs any figure
+higher than 6, and 6,553 really stands for 3, added to five times 7,
+added to five times 49, added to six times 343 (in the ordinary
+notation), or 2,341. To reverse the operation, and convert 2,341 from
+the denary to the septenary scale, we divide it by 7, and get 334 and
+remainder 3; divide 334 by 7, and get 47 and remainder 5; and so keep
+on dividing by 7 as long as there is anything to divide. The
+remainders, read backwards, 6, 5, 5, 3, give us the answer, 6,553.</p>
+
+<p>Now, as I have said, our puzzle may be solved at once by merely
+converting 1,000,000 dollars to the septenary scale. Keep on dividing
+this number by 7 until there is nothing more left to divide, and the
+remainders will be found to be 11333311 which is 1,000,000 expressed
+in the septenary scale. Therefore, 1 gift of 1 dollar, 1 gift of 7
+dollars, 3 gifts of 49 dollars, 3 gifts of 343 dollars, 3 gifts of
+2,401 dollars, 3 gifts of 16,807 dollars, 1 gift of 117,649 dollars,
+and one substantial gift of 823,543 dollars, satisfactorily solves our
+problem. And it is the only possible solution. It is thus seen that no
+"trials" are necessary; by converting to the septenary scale of
+notation we go direct to the answer.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_17_THE_PUZZLING_MONEY_BOXESa" id="X_17_THE_PUZZLING_MONEY_BOXESa"></a><a href="#X_17_THE_PUZZLING_MONEY_BOXES"><b>17.&mdash;THE PUZZLING MONEY BOXES.&mdash;<i>solution</i></b></a></p>
+
+<p>The correct answer to this puzzle is as follows: John put into his
+money-box two double florins (8<i>s</i>.), William a half-sovereign and a
+florin (12<i>s</i>.), Charles a crown (5<i>s</i>.), and Thomas a sovereign (20<i>s</i>.).
+There are six coins in all, of a total value of 45<i>s</i>. If John had 2<i>s</i>.
+more, William 2<i>s</i>. less, Charles twice as much, and Thomas half as much
+as they really possessed, they would each have had exactly 10<i>s</i>.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_18_THE_MARKET_WOMENa" id="X_18_THE_MARKET_WOMENa"></a><a href="#X_18_THE_MARKET_WOMEN"><b>18.&mdash;THE MARKET WOMEN.&mdash;<i>solution</i></b></a></p>
+
+<p>The price received was in every case 105 farthings. Therefore the
+greatest number of women is eight, as the goods could only be sold at
+the following rates: 105 lbs. at 1 farthing, 35 at 3, 21 at 5, 15 at
+7, 7 at 15, 5 at 21, 3 at 35, and 1 lb. at 105 farthings.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_19_THE_NEW_YEARS_EVE_SUPPERSa" id="X_19_THE_NEW_YEARS_EVE_SUPPERSa"></a><a href="#X_19_THE_NEW_YEARS_EVE_SUPPERS"><b>19.&mdash;THE NEW YEAR'S EVE SUPPERS.&mdash;<i>solution</i></b></a></p>
+
+<p>The company present on the occasion must have consisted of seven
+pairs, ten single men, and one single lady. Thus, there were
+twenty-five persons in all, and at the prices stated they would pay
+exactly &pound;5 together.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_20_BEEF_AND_SAUSAGESa" id="X_20_BEEF_AND_SAUSAGESa"></a><a href="#X_20_BEEF_AND_SAUSAGES"><b>20.&mdash;BEEF AND SAUSAGES.&mdash;<i>solution</i></b></a></p>
+
+<p>The lady bought 48 lbs. of beef at 2<i>s</i>., and the same quantity of
+sausages at 1<i>s</i>. 6<i>d</i>., thus spending &pound;8, 8<i>s</i>. Had she bought 42 lbs. of
+beef and 56 lbs. of sausages she would have spent &pound;4, 4<i>s</i>. on each, and
+have obtained 98 lbs. instead of 96 lbs.&mdash;a gain in weight of 2 lbs.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_21_A_DEAL_IN_APPLESa" id="X_21_A_DEAL_IN_APPLESa"></a><a href="#X_21_A_DEAL_IN_APPLES"><b>21.&mdash;A DEAL IN APPLES.&mdash;<i>solution</i></b></a></p>
+
+<p>I was first offered sixteen apples for my shilling, which would be at
+the rate of ninepence a dozen. The two extra apples gave me eighteen
+for a shilling, which is at the rate of eightpence a dozen, or one
+penny a dozen less than the first price asked.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_22_A_DEAL_IN_EGGSa" id="X_22_A_DEAL_IN_EGGSa"></a><a href="#X_22_A_DEAL_IN_EGGS"><b>22.&mdash;A DEAL IN EGGS.&mdash;<i>solution</i></b></a></p>
+
+<p>The man must have bought ten eggs at fivepence, ten eggs at one penny,
+and eighty eggs <span class='pagenum'>Pg 150<a name="Page_150" id="Page_150"></a></span>at a halfpenny. He would then have one hundred eggs
+at a cost of eight shillings and fourpence, and the same number of
+eggs of two of the qualities.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_23_THE_CHRISTMAS-BOXESa" id="X_23_THE_CHRISTMAS-BOXESa"></a><a href="#X_23_THE_CHRISTMAS-BOXES"><b>23.&mdash;THE CHRISTMAS-BOXES.&mdash;<i>solution</i></b></a></p>
+
+<p>The distribution took place "some years ago," when the fourpenny-piece
+was in circulation. Nineteen persons must each have received nineteen
+pence. There are five different ways in which this sum may have been
+paid in silver coins. We need only use two of these ways. Thus if
+fourteen men each received four four-penny-pieces and one
+threepenny-piece, and five men each received five threepenny-pieces
+and one fourpenny-piece, each man would receive nineteen pence, and
+there would be exactly one hundred coins of a total value of &pound;1, 10<i>s</i>.
+1<i>d</i>.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_24_A_SHOPPING_PERPLEXITYa" id="X_24_A_SHOPPING_PERPLEXITYa"></a><a href="#X_24_A_SHOPPING_PERPLEXITY"><b>24.&mdash;A SHOPPING PERPLEXITY.&mdash;<i>solution</i></b></a></p>
+
+<p>The first purchase amounted to 1<i>s</i>. 5�<i>d</i>., the second to 1<i>s</i>.
+11�<i>d</i>., and together they make 3<i>s</i>. 5�<i>d</i>. Not one of these three
+amounts can be paid in fewer than six current coins of the realm.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_25_CHINESE_MONEYa" id="X_25_CHINESE_MONEYa"></a><a href="#X_25_CHINESE_MONEY"><b>25.&mdash;CHINESE MONEY.&mdash;<i>solution</i></b></a></p>
+
+<p>As a ching-chang is worth twopence and four-fifteenths of a
+ching-chang, the remaining eleven-fifteenths of a ching-chang must be
+worth twopence. Therefore eleven ching-changs are worth exactly thirty
+pence, or half a crown. Now, the exchange must be made with seven
+round-holed coins and one square-holed coin. Thus it will be seen that
+7 round-holed coins are worth seven-elevenths of 15 ching-changs, and
+1 square-holed coin is worth one-eleventh of 16 ching-changs&mdash;that is,
+77 rounds equal 105 ching-changs and 11 squares equal 16 ching-changs.
+Therefore 77 rounds added to 11 squares equal 121 ching-changs; or 7
+rounds and 1 square equal 11 ching-changs, or its equivalent, half a
+crown. This is more simple in practice than it looks here.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_26_THE_JUNIOR_CLERKS_PUZZLEa" id="X_26_THE_JUNIOR_CLERKS_PUZZLEa"></a><a href="#X_26_THE_JUNIOR_CLERKS_PUZZLE"><b>26.&mdash;THE JUNIOR CLERKS' PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Although Snoggs's <i>reason</i> for wishing to take his rise at &pound;2, 10<i>s</i>.
+half-yearly did not concern our puzzle, the <i>fact</i> that he was duping
+his employer into paying him more than was intended did concern it.
+Many readers will be surprised to find that, although Moggs only
+received &pound;350 in five years, the artful Snoggs actually obtained &pound;362,
+10<i>s</i>. in the same time. The rest is simplicity itself. It is evident
+that if Moggs saved &pound;87, 10<i>s</i>. and Snoggs &pound;181, 5<i>s</i>., the latter would
+be saving twice as great a proportion of his salary as the former
+(namely, one-half as against one-quarter), and the two sums added
+together make &pound;268, 15<i>s</i>.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_27_GIVING_CHANGEa" id="X_27_GIVING_CHANGEa"></a><a href="#X_27_GIVING_CHANGE"><b>27.&mdash;GIVING CHANGE.&mdash;<i>solution</i></b></a></p>
+
+<p>The way to help the American tradesman out of his dilemma is this.
+Describing the coins by the number of cents that they represent, the
+tradesman puts on the counter 50 and 25; the buyer puts down 100, 3,
+and 2; the stranger adds his 10, 10, 5, 2, and 1. Now, considering
+that the cost of the purchase amounted to 34 cents, it is clear that
+out of this pooled money the tradesman has to receive 109, the buyer
+71, and the stranger his 28 cents. Therefore it is obvious at a glance
+that the 100-piece must go to the tradesman, and it then follows that
+the 50-piece must go to the buyer, and then the 25-piece can only go
+to the stranger. Another glance will now make it clear that the two
+10-cent pieces must go to the buyer, because the tradesman now only
+wants 9 and the stranger 3. Then it becomes obvious that the buyer
+must take the 1 cent, that the stranger must take the 3 cents, and the
+tradesman the 5, 2, and 2. To sum up, the tradesman takes 100, 5, 2,
+and 2; the buyer, 50, 10, 10, and 1; the stranger, 25 and 3. It will
+be seen that not one of the three persons retains any one of his own
+coins.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_28_DEFECTIVE_OBSERVATIONa" id="X_28_DEFECTIVE_OBSERVATIONa"></a><a href="#X_28_DEFECTIVE_OBSERVATION"><b>28.&mdash;DEFECTIVE OBSERVATION.&mdash;<i>solution</i></b></a></p>
+
+<p>Of course the date on a penny is on the same side as Britannia&mdash;the
+"tail" side. Six pennies may be laid around another penny, all flat on
+the table, so that every one of them touches the central one. The
+number of threepenny-pieces that may be laid on the surface of a
+half-crown, so that no piece lies on another or overlaps the edge of
+the half-crown, is one. A second threepenny-piece will overlap the
+edge of the larger coin. Few people guess fewer than three, and many
+persons give an absurdly high number.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_29_THE_BROKEN_COINSa" id="X_29_THE_BROKEN_COINSa"></a><a href="#X_29_THE_BROKEN_COINS"><b>29.&mdash;THE BROKEN COINS.&mdash;<i>solution</i></b></a></p>
+
+<p>If the three broken coins when perfect were worth 253 pence, and are
+now in their broken condition worth 240 pence, it should be obvious
+that <sup>13</sup>/<sub>253</sub> of the original value has been lost. And as the same
+fraction of each coin has been broken away, each coin has lost <sup>13</sup>/<sub>253</sub>
+of its original bulk.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_30_TWO_QUESTIONS_IN_PROBABILITIESa" id="X_30_TWO_QUESTIONS_IN_PROBABILITIESa"></a><a href="#X_30_TWO_QUESTIONS_IN_PROBABILITIES"><b>30.&mdash;TWO QUESTIONS IN PROBABILITIES.&mdash;<i>solution</i></b></a></p>
+
+<p>In tossing with the five pennies all at the same time, it is obvious
+that there are 32 different ways in which the coins may fall, because
+the first coin may fall in either of two ways, then the second coin
+may also fall in either of two ways, and so on. Therefore five 2's
+multiplied together make 32. Now, how are these 32 ways made up? Here
+they are:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>(<i>a</i>)</td><td align='right'>5</td><td align='left'>heads</td><td align='right'>1</td><td align='left'>way</td></tr>
+<tr><td align='center'>(<i>b</i>)</td><td align='right'>5</td><td align='left'>tails</td><td align='right'>1</td><td align='left'>way</td></tr>
+<tr><td align='center'>(<i>c</i>)</td><td align='right'>4</td><td align='left'>heads and 1 tail</td><td align='right'>5</td><td align='left'>ways</td></tr>
+<tr><td align='center'>(<i>d</i>)</td><td align='right'>4</td><td align='left'>tails and 1 head</td><td align='right'>5</td><td align='left'>ways</td></tr>
+<tr><td align='center'>(<i>e</i>)</td><td align='right'>3</td><td align='left'>heads and 2 tails</td><td align='right'>10</td><td align='left'>ways</td></tr>
+<tr><td align='center'>(<i>f</i>)</td><td align='right'>3</td><td align='left'>tails and 2 heads</td><td align='right'>10</td><td align='left'>ways</td></tr>
+</table></div>
+
+<p>Now, it will be seen that the only favourable cases are <i>a</i>, <i>b</i>, <i>c</i>,
+and <i>d</i>&mdash;12 cases. The remaining 20 cases are unfavourable, because
+they do <span class='pagenum'>Pg 151<a name="Page_151" id="Page_151"></a></span>not give at least four heads or four tails. Therefore the
+chances are only 12 to 20 in your favour, or (which is the same thing)
+3 to 5. Put another way, you have only 3 chances out of 8.</p>
+
+<p>The amount that should be paid for a draw from the bag that contains
+three sovereigns and one shilling is 15<i>s</i>. 3<i>d</i>. Many persons will say
+that, as one's chances of drawing a sovereign were 3 out of 4, one
+should pay three-fourths of a pound, or 15<i>s</i>., overlooking the fact
+that one must draw at least a shilling&mdash;there being no blanks.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_31_DOMESTIC_ECONOMYa" id="X_31_DOMESTIC_ECONOMYa"></a><a href="#X_31_DOMESTIC_ECONOMY"><b>31.&mdash;DOMESTIC ECONOMY.&mdash;<i>solution</i></b></a></p>
+
+<p>Without the hint that I gave, my readers would probably have been
+unanimous in deciding that Mr. Perkins's income must have been &pound;1,710.
+But this is quite wrong. Mrs. Perkins says, "We have spent a third of
+his yearly income in rent," etc., etc.&mdash;that is, in two years they
+have spent an amount in rent, etc., equal to one-third of his yearly
+income. Note that she does <i>not</i> say that they have spent <i>each year</i>
+this sum, whatever it is, but that <i>during the two years</i> that amount
+has been spent. The only possible answer, according to the exact
+reading of her words, is, therefore, that his income was &pound;180 per
+annum. Thus the amount spent in two years, during which his income has
+amounted to &pound;360, will be &pound;60 in rent, etc., &pound;90 in domestic expenses,
+&pound;20 in other ways, leaving the balance of &pound;190 in the bank as stated.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_32_THE_EXCURSION_TICKET_PUZZLEa" id="X_32_THE_EXCURSION_TICKET_PUZZLEa"></a><a href="#X_32_THE_EXCURSION_TICKET_PUZZLE"><b>32.&mdash;THE EXCURSION TICKET PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Nineteen shillings and ninepence may be paid in 458,908,622 different
+ways.</p>
+
+<p>I do not propose to give my method of solution. Any such explanation
+would occupy an amount of space out of proportion to its interest or
+value. If I could give within reasonable limits a general solution for
+all money payments, I would strain a point to find room; but such a
+solution would be extremely complex and cumbersome, and I do not
+consider it worth the labour of working out.</p>
+
+<p>Just to give an idea of what such a solution would involve, I will
+merely say that I find that, dealing only with those sums of money
+that are multiples of threepence, if we only use bronze coins any sum
+can be paid in (<i>n</i>+1)<sup>2</sup> ways where <i>n</i> always represents the number
+of pence. If threepenny-pieces are admitted, there are</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' class='bb'>2<i>n</i><sup>3</sup>+15<i>n</i><sup>2</sup>+33<i>n</i></td><td rowspan='2'>&nbsp;+&nbsp;1 </td></tr>
+<tr><td align='center'>18</td></tr>
+</table></div>
+
+<p>ways. If sixpences are also used there are</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' class='bb'><i>n</i><sup>4</sup>+22<i>n</i><sup>3</sup>+159<i>n</i><sup>2</sup>+414<i>n</i>+216</td></tr>
+<tr><td align='center'>216</td></tr>
+</table></div>
+
+<p>ways, when the sum is a multiple of sixpence, and the constant, 216,
+changes to 324 when the money is not such a multiple. And so the
+formulas increase in complexity in an accelerating ratio as we go on
+to the other coins.</p>
+
+<p>I will, however, add an interesting little table of the possible ways
+of changing our current coins which I believe has never been given in
+a book before. Change may be given for a</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='left'>Farthing in</td><td align='left'>0 way.</td></tr>
+<tr><td align='left'>Halfpenny in</td><td align='left'>1 way.</td></tr>
+<tr><td align='left'>Penny in</td><td align='left'>3 ways.</td></tr>
+<tr><td align='left'>Threepenny-piece in</td><td align='left'>16 ways.</td></tr>
+<tr><td align='left'>Sixpence in</td><td align='left'>66 ways.</td></tr>
+<tr><td align='left'>Shilling in</td><td align='left'>402 ways.</td></tr>
+<tr><td align='left'>Florin in</td><td align='left'>3,818 ways.</td></tr>
+<tr><td align='left'>Half-crown in</td><td align='left'>8,709 ways.</td></tr>
+<tr><td align='left'>Double florin in</td><td align='left'>60,239 ways.</td></tr>
+<tr><td align='left'>Crown in</td><td align='left'>166,651 ways.</td></tr>
+<tr><td align='left'>Half-sovereign in</td><td align='left'>6,261,622 ways.</td></tr>
+<tr><td align='left'>Sovereign in</td><td align='left'>500,291,833 ways.</td></tr>
+</table></div>
+
+<p>It is a little surprising to find that a sovereign may be changed in
+over five hundred million different ways. But I have no doubt as to
+the correctness of my figures.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_33_A_PUZZLE_IN_REVERSALSa" id="X_33_A_PUZZLE_IN_REVERSALSa"></a><a href="#X_33_A_PUZZLE_IN_REVERSALS"><b>33.&mdash;A PUZZLE IN REVERSALS.&mdash;<i>solution</i></b></a></p>
+
+<p>(i) &pound;13. (2) &pound;23, 19<i>s</i>. 11<i>d</i>. The words "the number of pounds exceeds
+that of the pence" exclude such sums of money as &pound;2, 16<i>s</i>. 2<i>d</i>. and all
+sums under &pound;1.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_34_THE_GROCER_AND_DRAPERa" id="X_34_THE_GROCER_AND_DRAPERa"></a><a href="#X_34_THE_GROCER_AND_DRAPER"><b>34.&mdash;THE GROCER AND DRAPER.&mdash;<i>solution</i></b></a></p>
+
+<p>The grocer was delayed half a minute and the draper eight minutes and
+a half (seventeen times as long as the grocer), making together nine
+minutes. Now, the grocer took twenty-four minutes to weigh out the
+sugar, and, with the half-minute delay, spent 24 min. 30 sec. over the
+task; but the draper had only to make <i>forty-seven</i> cuts to divide the
+roll of cloth, containing forty-eight yards, into yard pieces! This
+took him 15 min. 40 sec., and when we add the eight minutes and a half
+delay we get 24 min. 10 sec., from which it is clear that the draper
+won the race by twenty seconds. The majority of solvers make
+forty-eight cuts to divide the roll into forty-eight pieces!</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_35_JUDKINSS_CATTLEa" id="X_35_JUDKINSS_CATTLEa"></a><a href="#X_35_JUDKINSS_CATTLE"><b>35.&mdash;JUDKINS'S CATTLE.&mdash;<i>solution</i></b></a></p>
+
+<p>As there were five droves with an equal number of animals in each
+drove, the number must be divisible by 5; and as every one of the
+eight dealers bought the same number of animals, the number must be
+divisible by 8. Therefore the number must be a multiple of 40. The
+highest possible multiple of 40 that will work will be found to be
+120, and this number could be made up in one of two ways&mdash;1 ox, 23
+pigs, and 96 sheep, or 3 oxen, 8 pigs, and 109 sheep. But the first is
+excluded by the statement that the animals consisted of "oxen, pigs,
+and sheep," because a single ox is not oxen. Therefore the second
+grouping is the correct answer.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_36_BUYING_APPLESa" id="X_36_BUYING_APPLESa"></a><a href="#X_36_BUYING_APPLES"><b>36.&mdash;BUYING APPLES.&mdash;<i>solution</i></b></a></p>
+
+<p>As there were the same number of boys as girls, it is clear that the
+number of children must be even, and, apart from a careful and exact
+reading of the question, there would be three different answers. There
+might be two, six, or fourteen children. In the first of these cases
+there <span class='pagenum'>Pg 152<a name="Page_152" id="Page_152"></a></span>are ten different ways in which the apples could be bought. But
+we were told there was an equal number of "boys and girls," and one
+boy and one girl are not boys and girls, so this case has to be
+excluded. In the case of fourteen children, the only possible
+distribution is that each child receives one halfpenny apple. But we
+were told that each child was to receive an equal distribution of
+"apples," and one apple is not apples, so this case has also to be
+excluded. We are therefore driven back on our third case, which
+exactly fits in with all the conditions. Three boys and three girls
+each receive 1 halfpenny apple and 2 third-penny apples. The value of
+these 3 apples is one penny and one-sixth, which multiplied by six
+makes sevenpence. Consequently, the correct answer is that there were
+six children&mdash;three girls and three boys.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_37_BUYING_CHESTNUTSa" id="X_37_BUYING_CHESTNUTSa"></a><a href="#X_37_BUYING_CHESTNUTS"><b>37.&mdash;BUYING CHESTNUTS.&mdash;<i>solution</i></b></a></p>
+
+<p>In solving this little puzzle we are concerned with the exact
+interpretation of the words used by the buyer and seller. I will give
+the question again, this time adding a few words to make the matter
+more clear. The added words are printed in italics.</p>
+
+<p>"A man went into a shop to buy chestnuts. He said he wanted a
+pennyworth, and was given five chestnuts. 'It is not enough; I ought
+to have a sixth <i>of a chestnut more</i>,' he remarked. 'But if I give you
+one chestnut more,' the shopman replied, 'you will have <i>five-sixths</i>
+too many.' Now, strange to say, they were both right. How many
+chestnuts should the buyer receive for half a crown?"</p>
+
+<p>The answer is that the price was 155 chestnuts for half a crown.
+Divide this number by 30, and we find that the buyer was entitled to
+5<sup>1</sup>/<sub>6</sub> chestnuts in exchange for his penny. He was, therefore, right
+when he said, after receiving five only, that he still wanted a sixth.
+And the salesman was also correct in saying that if he gave one
+chestnut more (that is, six chestnuts in all) he would be giving
+five-sixths of a chestnut in excess.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_38_THE_BICYCLE_THIEFa" id="X_38_THE_BICYCLE_THIEFa"></a><a href="#X_38_THE_BICYCLE_THIEF"><b>38.&mdash;THE BICYCLE THIEF.&mdash;<i>solution</i></b></a></p>
+
+<p>People give all sorts of absurd answers to this question, and yet it
+is perfectly simple if one just considers that the salesman cannot
+possibly have lost more than the cyclist actually stole. The latter
+rode away with a bicycle which cost the salesman eleven pounds, and
+the ten pounds "change;" he thus made off with twenty-one pounds, in
+exchange for a worthless bit of paper. This is the exact amount of the
+salesman's loss, and the other operations of changing the cheque and
+borrowing from a friend do not affect the question in the slightest.
+The loss of prospective profit on the sale of the bicycle is, of
+course, not direct loss of money out of pocket.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_39_THE_COSTERMONGERS_PUZZLEa" id="X_39_THE_COSTERMONGERS_PUZZLEa"></a><a href="#X_39_THE_COSTERMONGERS_PUZZLE"><b>39.&mdash;THE COSTERMONGER'S PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Bill must have paid 8<i>s</i>. per hundred for his oranges&mdash;that is, 125 for
+10<i>s</i>. At 8<i>s</i>. 4<i>d</i>. per hundred, he would only have received 120 oranges
+for 10<i>s</i>. This exactly agrees with Bill's statement.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_40_MAMMAS_AGEa" id="X_40_MAMMAS_AGEa"></a><a href="#X_40_MAMMAS_AGE"><b>40.&mdash;MAMMA'S AGE.&mdash;<i>solution</i></b></a></p>
+
+<p>The age of Mamma must have been 29 years 2 months; that of Papa, 35
+years; and that of the child, Tommy, 5 years 10 months. Added
+together, these make seventy years. The father is six times the age of
+the son, and, after 23 years 4 months have elapsed, their united ages
+will amount to 140 years, and Tommy will be just half the age of his
+father.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_41_THEIR_AGESa" id="X_41_THEIR_AGESa"></a><a href="#X_41_THEIR_AGES"><b>41.&mdash;THEIR AGES.&mdash;<i>solution</i></b></a></p>
+
+<p>The gentleman's age must have been 54 years and that of his wife 45
+years.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_42_THE_FAMILY_AGESa" id="X_42_THE_FAMILY_AGESa"></a><a href="#X_42_THE_FAMILY_AGES"><b>42.&mdash;THE FAMILY AGES.&mdash;<i>solution</i></b></a></p>
+
+<p>The ages were as follows: Billie, 3� years; Gertrude, 1� year;
+Henrietta, 5� years; Charlie, 10�; years; and Janet, 21 years.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_43_MRS_TIMPKINSS_AGEa" id="X_43_MRS_TIMPKINSS_AGEa"></a><a href="#X_43_MRS_TIMPKINSS_AGE"><b>43.&mdash;MRS. TIMPKINS'S AGE.&mdash;<i>solution</i></b></a></p>
+
+<p>The age of the younger at marriage is always the same as the number of
+years that expire before the elder becomes twice her age, if he was
+three times as old at marriage. In our case it was eighteen years
+afterwards; therefore Mrs. Timpkins was eighteen years of age on the
+wedding-day, and her husband fifty-four.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_44_A_CENSUS_PUZZLEa" id="X_44_A_CENSUS_PUZZLEa"></a><a href="#X_44_A_CENSUS_PUZZLE"><b>44.&mdash;A CENSUS PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Miss Ada Jorkins must have been twenty-four and her little brother
+Johnnie three years of age, with thirteen brothers and sisters
+between. There was a trap for the solver in the words "seven times
+older than little Johnnie." Of course, "seven times older" is equal to
+eight times as old. It is surprising how many people hastily assume
+that it is the same as "seven times as old." Some of the best writers
+have committed this blunder. Probably many of my readers thought that
+the ages 24� and 3� were correct.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_45_MOTHER_AND_DAUGHTERa" id="X_45_MOTHER_AND_DAUGHTERa"></a><a href="#X_45_MOTHER_AND_DAUGHTER"><b>45.&mdash;MOTHER AND DAUGHTER.&mdash;<i>solution</i></b></a></p>
+
+<p>In four and a half years, when the daughter will be sixteen years and
+a half and the mother forty-nine and a half years of age.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_46_MARY_AND_MARMADUKEa" id="X_46_MARY_AND_MARMADUKEa"></a><a href="#X_46_MARY_AND_MARMADUKE"><b>46.&mdash;MARY AND MARMADUKE.&mdash;<i>solution</i></b></a></p>
+
+<p>Marmaduke's age must have been twenty-nine years and two-fifths, and
+Mary's nineteen years and three-fifths. When Marmaduke was aged
+nineteen and three-fifths, Mary was only nine and four-fifths; so
+Marmaduke was at that time twice her age.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_47_ROVERS_AGEa" id="X_47_ROVERS_AGEa"></a><a href="#X_47_ROVERS_AGE"><b>47.&mdash;ROVER'S AGE.&mdash;<i>solution</i></b></a></p>
+
+<p>Rover's present age is ten years and Mildred's thirty years. Five
+years ago their respective <span class='pagenum'>Pg 153<a name="Page_153" id="Page_153"></a></span>ages were five and twenty-five. Remember
+that we said "four times older than the dog," which is the same as
+"five times as old." (See answer to No. <a href="#X_44_A_CENSUS_PUZZLEa">44</a>.)</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_48_CONCERNING_TOMMYS_AGEa" id="X_48_CONCERNING_TOMMYS_AGEa"></a><a href="#X_48_CONCERNING_TOMMYS_AGE"><b>48.&mdash;CONCERNING TOMMY'S AGE.&mdash;<i>solution</i></b></a></p>
+
+<p>Tommy Smart's age must have been nine years and three-fifths. Ann's
+age was sixteen and four-fifths, the mother's thirty-eight and
+two-fifths, and the father's fifty and two-fifths.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_49_NEXT-DOOR_NEIGHBOURSa" id="X_49_NEXT-DOOR_NEIGHBOURSa"></a><a href="#X_49_NEXT-DOOR_NEIGHBOURS"><b>49.&mdash;NEXT-DOOR NEIGHBOURS.&mdash;<i>solution</i></b></a></p>
+
+<p>Mr. Jupp 39, Mrs. Jupp 34, Julia 14, and Joe 13; Mr. Simkin 42; Mrs.
+Simkin 40; Sophy 10; and Sammy 8.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_50_THE_BAG_OF_NUTSa" id="X_50_THE_BAG_OF_NUTSa"></a><a href="#X_50_THE_BAG_OF_NUTS"><b>50.&mdash;THE BAG OF NUTS.&mdash;<i>solution</i></b></a></p>
+
+<p>It will be found that when Herbert takes twelve, Robert and
+Christopher will take nine and fourteen respectively, and that they
+will have together taken thirty-five nuts. As 35 is contained in 770
+twenty-two times, we have merely to multiply 12, 9, and 14 by 22 to
+discover that Herbert's share was 264, Robert's 198, and Christopher's
+308. Then, as the total of their ages is 17&frac12; years or half the sum of
+12, 9, and 14, their respective ages must be 6, 4&frac12;, and 7 years.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_51_HOW_OLD_WAS_MARYa" id="X_51_HOW_OLD_WAS_MARYa"></a><a href="#X_51_HOW_OLD_WAS_MARY"><b>51.&mdash;HOW OLD WAS MARY?&mdash;<i>solution</i></b></a></p>
+
+<p>The age of Mary to that of Ann must be as 5 to 3. And as the sum of
+their ages was 44, Mary was 27&frac12; and Ann 16&frac12;. One is exactly 11 years
+older than the other. I will now insert in brackets in the original
+statement the various ages specified: "Mary is (27&frac12;) twice as old as
+Ann was (13&frac34;) when Mary was half as old (24&frac34;) as Ann will be (49&frac12;)
+when Ann is three times as old (49&frac12;) as Mary was (16&frac12;) when Mary was
+(16&frac12;) three times as old as Ann (5&frac12;)." Now, check this backwards. When
+Mary was three times as old as Ann, Mary was 16&frac12; and Ann 5&frac12; (11 years
+younger). Then we get 49&frac12; for the age Ann will be when she is three
+times as old as Mary was then. When Mary was half this she was 24&frac34;.
+And at that time Ann must have been 13&frac34; (11 years younger). Therefore
+Mary is now twice as old&mdash;27&frac12;, and Ann 11 years younger&mdash;16&frac12;.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_52_QUEER_RELATIONSHIPSa" id="X_52_QUEER_RELATIONSHIPSa"></a><a href="#X_52_QUEER_RELATIONSHIPS"><b>52.&mdash;QUEER RELATIONSHIPS.&mdash;<i>solution</i></b></a></p>
+
+<p>If a man marries a woman, who dies, and he then marries his deceased
+wife's sister and himself dies, it may be correctly said that he had
+(previously) married the sister of his widow.</p>
+
+<p>The youth was not the nephew of Jane Brown, because he happened to be
+her son. Her surname was the same as that of her brother, because she
+had married a man of the same name as herself.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_53_HEARD_ON_THE_TUBE_RAILWAYa" id="X_53_HEARD_ON_THE_TUBE_RAILWAYa"></a><a href="#X_53_HEARD_ON_THE_TUBE_RAILWAY"><b>53.&mdash;HEARD ON THE TUBE RAILWAY.&mdash;<i>solution</i></b></a></p>
+
+<p>The gentleman was the second lady's uncle.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_54_A_FAMILY_PARTYa" id="X_54_A_FAMILY_PARTYa"></a><a href="#X_54_A_FAMILY_PARTY"><b>54.&mdash;A FAMILY PARTY.&mdash;<i>solution</i></b></a></p>
+
+<p>The party consisted of two little girls and a boy, their father and
+mother, and their father's father and mother.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_55_A_MIXED_PEDIGREEa" id="X_55_A_MIXED_PEDIGREEa"></a><a href="#X_55_A_MIXED_PEDIGREE"><b>55.&mdash;A MIXED PEDIGREE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a055.png" width="400" height="280" alt="" title="" />
+</div>
+
+<p>The letter <i>m</i> stands for "married." It will be seen that John Snoggs
+can say to Joseph Bloggs, "You are my <i>father's brother-in-law</i>,
+because my father married your sister Kate; you are my <i>brother's
+father-in-law</i>, because my brother Alfred married your daughter Mary;
+and you are my <i>father-in-law's brother</i>, because my wife Jane was
+your brother Henry's daughter."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_56_WILSONS_POSERa" id="X_56_WILSONS_POSERa"></a><a href="#X_56_WILSONS_POSER"><b>56.&mdash;WILSON'S POSER.&mdash;<i>solution</i></b></a></p>
+
+<p>If there are two men, each of whom marries the mother of the other,
+and there is a son of each marriage, then each of such sons will be at
+the same time uncle and nephew of the other. There are other ways in
+which the relationship may be brought about, but this is the simplest.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_57_WHAT_WAS_THE_TIMEa" id="X_57_WHAT_WAS_THE_TIMEa"></a><a href="#X_57_WHAT_WAS_THE_TIME"><b>57.&mdash;WHAT WAS THE TIME?&mdash;<i>solution</i></b></a></p>
+
+<p>The time must have been 9.36 p.m. A quarter of the time since noon is
+2 hr. 24 min., and a half of the time till noon next day is 7 hr. 12
+min. These added together make 9 hr. 36 min.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_58_A_TIME_PUZZLEa" id="X_58_A_TIME_PUZZLEa"></a><a href="#X_58_A_TIME_PUZZLE"><b>58.&mdash;A TIME PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Twenty-six minutes.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_59_A_PUZZLING_WATCHa" id="X_59_A_PUZZLING_WATCHa"></a><a href="#X_59_A_PUZZLING_WATCH"><b>59.&mdash;A PUZZLING WATCH.&mdash;<i>solution</i></b></a></p>
+
+<p>If the 65 minutes be counted on the face of the same watch, then the
+problem would be impossible: for the hands must coincide every 65<sup>5</sup>/<sub>11</sub>
+minutes as shown by its face, and it matters not whether it runs fast
+or slow; but if it is measured by true time, it gains <sup>5</sup>/<sub>11</sub> of a minute
+in 65 minutes, or <sup>60</sup>/<sub>143</sub> of a minute per hour.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_60_THE_WAPSHAWS_WHARF_MYSTERYa" id="X_60_THE_WAPSHAWS_WHARF_MYSTERYa"></a><a href="#X_60_THE_WAPSHAWS_WHARF_MYSTERY"><b>60.&mdash;THE WAPSHAW'S WHARF MYSTERY.&mdash;<i>solution</i></b></a></p>
+
+<p>There are eleven different times in twelve hours when the hour and
+minute hands of a clock are exactly one above the other. If we divide
+12 <span class='pagenum'>Pg 154<a name="Page_154" id="Page_154"></a></span>hours by 11 we get 1 hr. 5 min. 27<sup>3</sup>/<sub>11</sub> sec., and this is the time
+after twelve o'clock when they are first together, and also the time
+that elapses between one occasion of the hands being together and the
+next. They are together for the second time at 2 hr. 10 min. 54<sup>6</sup>/<sub>11</sub>
+sec. (twice the above time); next at 3 hr. 16 min. 21<sup>9</sup>/<sub>11</sub> sec.; next
+at 4 hr. 21 min. 49<sup>1</sup>/<sub>11</sub> sec. This last is the only occasion on which
+the two hands are together with the second hand "just past the
+forty-ninth second." This, then, is the time at which the watch must
+have stopped. Guy Boothby, in the opening sentence of his <i>Across the
+World for a Wife</i>, says, "It was a cold, dreary winter's afternoon,
+and by the time the hands of the clock on my mantelpiece joined forces
+and stood at twenty minutes past four, my chambers were well-nigh as
+dark as midnight." It is evident that the author here made a slip,
+for, as we have seen above, he is 1 min. 49<sup>1</sup>/<sub>11</sub> sec. out in his
+reckoning.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_61_CHANGING_PLACESa" id="X_61_CHANGING_PLACESa"></a><a href="#X_61_CHANGING_PLACES"><b>61.&mdash;CHANGING PLACES.&mdash;<i>solution</i></b></a></p>
+
+<p>There are thirty-six pairs of times when the hands exactly change
+places between three p.m. and midnight. The number of pairs of times
+from any hour (<i>n</i>) to midnight is the sum of 12&nbsp;-&nbsp;(<i>n</i>&nbsp;+&nbsp;1) natural
+numbers. In the case of the puzzle <i>n</i>&nbsp;=&nbsp;3; therefore 12&nbsp;-&nbsp;(3&nbsp;+&nbsp;1)&nbsp;=&nbsp;8
+and 1&nbsp;+&nbsp;2&nbsp;+&nbsp;3&nbsp;+&nbsp;4&nbsp;+&nbsp;5&nbsp;+&nbsp;6&nbsp;+&nbsp;7&nbsp;+&nbsp;8&nbsp;=&nbsp;36, the required answer.</p>
+
+<p>The first pair of times is 3 hr. 21<sup>57</sup>/<sub>143</sub> min. and 4 hr. 16<sup>112</sup>/<sub>143</sub>
+min., and the last pair is 10 hr. 59<sup>83</sup>/<sub>143</sub> min. and 11 hr. 54<sup>138</sup>/<sub>143</sub>
+min. I will not give all the remainder of the thirty-six pairs of
+times, but supply a formula by which any of the sixty-six pairs that
+occur from midday to midnight may be at once found:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' rowspan='2'>a hr</td><td align='center' class='bb'>720b&nbsp;+&nbsp;60a</td><td align='center' rowspan='2'>min. and b hr.</td><td align='center' class='bb'>720a&nbsp;+&nbsp;60b min.</td></tr>
+<tr><td align='center'>143</td><td align='center'>143</td></tr>
+</table></div>
+
+<p>For the letter a may be substituted any hour from 0, 1, 2, 3 up to 10
+(where nought stands for 12 o'clock midday); and b may represent any
+hour, later than a, up to 11.</p>
+
+<p>By the aid of this formula there is no difficulty in discovering the
+answer to the second question: a&nbsp;=&nbsp;8 and b&nbsp;=&nbsp;11 will give the pair 8
+hr. 58<sup>106</sup>/<sub>143</sub> min. and 11 hr. 44<sup>128</sup>/<sub>143</sub> min., the latter being the
+time when the minute hand is nearest of all to the point IX&mdash;in fact,
+it is only <sup>15</sup>/<sub>143</sub> of a minute distant.</p>
+
+<p>Readers may find it instructive to make a table of all the sixty-six
+pairs of times when the hands of a clock change places. An easy way is
+as follows: Make a column for the first times and a second column for
+the second times of the pairs. By making a&nbsp;=&nbsp;0 and b&nbsp;=&nbsp;1 in the above
+expressions we find the first case, and enter hr. 5<sup>5</sup>/<sub>143</sub> min. at the
+head of the first column, and 1 hr. 0<sup>60</sup>/<sub>143</sub> min. at the head of the
+second column. Now, by successively adding 5<sup>5</sup>/<sub>143</sub> min. in the first,
+and 1 hr. 0<sup>60</sup>/<sub>143</sub> min. in the second column, we get all the <i>eleven</i>
+pairs in which the first time is a certain number of minutes after
+nought, or mid-day. Then there is a "jump" in the times, but you can
+find the next pair by making a&nbsp;=&nbsp;1 and b&nbsp;=&nbsp;2, and then by successively
+adding these two times as before you will get all the <i>ten</i> pairs
+after 1 o'clock. Then there is another "jump," and you will be able to
+get by addition all the <i>nine</i> pairs after 2 o'clock. And so on to the
+end. I will leave readers to investigate for themselves the nature and
+cause of the "jumps." In this way we get under the successive hours,
+11&nbsp;+&nbsp;10&nbsp;+&nbsp;9&nbsp;+&nbsp;8&nbsp;+&nbsp;7&nbsp;+&nbsp;6&nbsp;+&nbsp;5&nbsp;+&nbsp;4&nbsp;+&nbsp;3&nbsp;+&nbsp;2&nbsp;+&nbsp;1&nbsp;=&nbsp;66 pairs of times, which
+result agrees with the formula in the first paragraph of this article.</p>
+
+<p>Some time ago the principal of a Civil Service Training College, who
+conducts a "Civil Service Column" in one of the periodicals, had the
+query addressed to him, "How soon after XII o'clock will a clock with
+both hands of the same length be ambiguous?" His first answer was,
+"Some time past one o'clock," but he varied the answer from issue to
+issue. At length some of his readers convinced him that the answer is,
+"At 5<sup>5</sup>/<sub>143</sub> min. past XII;" and this he finally gave as correct,
+together with the reason for it that at that time <i>the time indicated
+is the same whichever hand you may assume as hour hand!</i></p>
+
+<hr style="width: 30%;" />
+<p><a name="X_62_THE_CLUB_CLOCKa" id="X_62_THE_CLUB_CLOCKa"></a><a href="#X_62_THE_CLUB_CLOCK"><b>62.&mdash;THE CLUB CLOCK.&mdash;<i>solution</i></b></a></p>
+
+<p>The positions of the hands shown in the illustration could only
+indicate that the clock stopped at 44 min. 51<sup>1143</sup>/<sub>1427</sub> sec. after
+eleven o'clock. The second hand would next be "exactly midway between
+the other two hands" at 45 min. 52<sup>496</sup>/<sub>1427</sub> sec. after eleven o'clock.
+If we had been dealing with the points on the circle to which the
+three hands are directed, the answer would be 45 min. 22<sup>106</sup>/<sub>1427</sub> sec.
+after eleven; but the question applied to the hands, and the second
+hand would not be between the others at that time, but outside them.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_63_THE_STOP-WATCHa" id="X_63_THE_STOP-WATCHa"></a><a href="#X_63_THE_STOP-WATCH"><b>63.&mdash;THE STOP-WATCH.&mdash;<i>solution</i></b></a></p>
+
+<p>The time indicated on the watch was 5<sup>5</sup>/<sub>11</sub> min. past 9, when the
+second hand would be at 27<sup>3</sup>/<sub>11</sub> sec. The next time the hands would be
+similar distances apart would be 54<sup>6</sup>/<sub>11</sub> min. past 2, when the second
+hand would be at 32<sup>8</sup>/<sub>11</sub> sec. But you need only hold the watch (or our
+previous illustration of it) in front of a mirror, when you will see
+the second time reflected in it! Of course, when reflected, you will
+read XI as I, X as II, and so on.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_64_THE_THREE_CLOCKSa" id="X_64_THE_THREE_CLOCKSa"></a><a href="#X_64_THE_THREE_CLOCKS"><b>64.&mdash;THE THREE CLOCKS.&mdash;<i>solution</i></b></a></p>
+
+<p>As a mere arithmetical problem this question presents no difficulty.
+In order that the hands shall all point to twelve o'clock at the same
+time, it is necessary that B shall gain at least twelve hours and that
+C shall lose twelve hours. As B gains a minute in a day of twenty-four
+hours, and C loses a minute in precisely the same time, it is evident
+that one will have gained 720 minutes (just twelve hours) in 720 days,
+and the other will have lost 720 minutes in 720 days. Clock A keeping
+perfect time, all three clocks must indicate twelve o'clock
+simultaneously at noon on the 720th day from April 1, 1898. What day
+of the month will that be?</p>
+
+<p>I published this little puzzle in 1898 to see <span class='pagenum'>Pg 155<a name="Page_155" id="Page_155"></a></span>how many people were
+aware of the fact that 1900 would not be a leap year. It was
+surprising how many were then ignorant on the point. Every year that
+can be divided by four without a remainder is bissextile or leap year,
+with the exception that one leap year is cut off in the century. 1800
+was not a leap year, nor was 1900. On the other hand, however, to make
+the calendar more nearly agree with the sun's course, every fourth
+hundred year is still considered bissextile. Consequently, 2000, 2400,
+2800, 3200, etc., will all be leap years. May my readers live to see
+them. We therefore find that 720 days from noon of April 1, 1898,
+brings us to noon of March 22, 1900.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_65_THE_RAILWAY_STATION_CLOCKa" id="X_65_THE_RAILWAY_STATION_CLOCKa"></a><a href="#X_65_THE_RAILWAY_STATION_CLOCK"><b>65.&mdash;THE RAILWAY STATION CLOCK.&mdash;<i>solution</i></b></a></p>
+
+<p>The time must have been 43<sup>7</sup>/<sub>11</sub> min. past two o'clock.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_66_THE_VILLAGE_SIMPLETONa" id="X_66_THE_VILLAGE_SIMPLETONa"></a><a href="#X_66_THE_VILLAGE_SIMPLETON"><b>66.&mdash;THE VILLAGE SIMPLETON.&mdash;<i>solution</i></b></a></p>
+
+<p>The day of the week on which the conversation took place was Sunday.
+For when the day after to-morrow (Tuesday) is "yesterday," "to-day"
+will be Wednesday; and when the day before yesterday (Friday) was
+"to-morrow," "to-day" was Thursday. There are two days between
+Thursday and Sunday, and between Sunday and Wednesday.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_67_AVERAGE_SPEEDa" id="X_67_AVERAGE_SPEEDa"></a><a href="#X_67_AVERAGE_SPEED"><b>67.&mdash;AVERAGE SPEED.&mdash;<i>solution</i></b></a></p>
+
+<p>The average speed is twelve miles an hour, not twelve and a half, as
+most people will hastily declare. Take any distance you like, say
+sixty miles. This would have taken six hours going and four hours
+returning. The double journey of 120 miles would thus take ten hours,
+and the average speed is clearly twelve miles an hour.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_68_THE_TWO_TRAINSa" id="X_68_THE_TWO_TRAINSa"></a><a href="#X_68_THE_TWO_TRAINS"><b>68.&mdash;THE TWO TRAINS.&mdash;<i>solution</i></b></a></p>
+
+<p>One train was running just twice as fast as the other.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_69_THE_THREE_VILLAGESa" id="X_69_THE_THREE_VILLAGESa"></a><a href="#X_69_THE_THREE_VILLAGES"><b>69.&mdash;THE THREE VILLAGES.&mdash;<i>solution</i></b></a></p>
+
+<p>Calling the three villages by their initial letters, it is clear that
+the three roads form a triangle, A, B, C, with a perpendicular,
+measuring twelve miles, dropped from C to the base A, B. This divides
+our triangle into two right-angled triangles with a twelve-mile side
+in common. It is then found that the distance from A to C is 15 miles,
+from C to B 20 miles, and from A to B 25 (that is 9 and 16) miles.
+These figures are easily proved, for the square of 12 added to the
+square of 9 equals the square of 15, and the square of 12 added to the
+square of 16 equals the square of 20.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_70_DRAWING_HER_PENSIONa" id="X_70_DRAWING_HER_PENSIONa"></a><a href="#X_70_DRAWING_HER_PENSION"><b>70.&mdash;DRAWING HER PENSION.&mdash;<i>solution</i></b></a></p>
+
+<p>The distance must be 6� miles.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_71_SIR_EDWYN_DE_TUDORa" id="X_71_SIR_EDWYN_DE_TUDORa"></a><a href="#X_71_SIR_EDWYN_DE_TUDOR"><b>71.&mdash;SIR EDWYN DE TUDOR.&mdash;<i>solution</i></b></a></p>
+
+<p>The distance must have been sixty miles. If Sir Edwyn left at noon and
+rode 15 miles an hour, he would arrive at four o'clock&mdash;an hour too
+soon. If he rode 10 miles an hour, he would arrive at six o'clock&mdash;an
+hour too late. But if he went at 12 miles an hour, he would reach the
+castle of the wicked baron exactly at five o'clock&mdash;the time
+appointed.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_72_THE_HYDROPLANE_QUESTIONa" id="X_72_THE_HYDROPLANE_QUESTIONa"></a><a href="#X_72_THE_HYDROPLANE_QUESTION"><b>72.&mdash;THE HYDROPLANE QUESTION.&mdash;<i>solution</i></b></a></p>
+
+<p>The machine must have gone at the rate of seven-twenty-fourths of a
+mile per minute and the wind travelled five-twenty-fourths of a mile
+per minute. Thus, going, the wind would help, and the machine would do
+twelve-twenty-fourths, or half a mile a minute, and returning only
+two-twenty-fourths, or one-twelfth of a mile per minute, the wind
+being against it. The machine without any wind could therefore do the
+ten miles in thirty-four and two-sevenths minutes, since it could do
+seven miles in twenty-four minutes.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_73_DONKEY_RIDINGa" id="X_73_DONKEY_RIDINGa"></a><a href="#X_73_DONKEY_RIDING"><b>73.&mdash;DONKEY RIDING.&mdash;<i>solution</i></b></a></p>
+
+<p>The complete mile was run in nine minutes. From the facts stated we
+cannot determine the time taken over the first and second
+quarter-miles separately, but together they, of course, took four and
+a half minutes. The last two quarters were run in two and a quarter
+minutes each.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_74_THE_BASKET_OF_POTATOESa" id="X_74_THE_BASKET_OF_POTATOESa"></a><a href="#X_74_THE_BASKET_OF_POTATOES"><b>74.&mdash;THE BASKET OF POTATOES.&mdash;<i>solution</i></b></a></p>
+
+<p>Multiply together the number of potatoes, the number less one, and
+twice the number less one, then divide by 3. Thus 50, 49, and 99
+multiplied together make 242,550, which, divided by 3, gives us 80,850
+yards as the correct answer. The boy would thus have to travel 45
+miles and fifteen-sixteenths&mdash;a nice little recreation after a day's
+work.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_75_THE_PASSENGERS_FAREa" id="X_75_THE_PASSENGERS_FAREa"></a><a href="#X_75_THE_PASSENGERS_FARE"><b>75.&mdash;THE PASSENGER'S FARE.&mdash;<i>solution</i></b></a></p>
+
+<p>Mr. Tompkins should have paid fifteen shillings as his correct share
+of the motor-car fare. He only shared half the distance travelled for
+&pound;3, and therefore should pay half of thirty shillings, or fifteen
+shillings.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_76_THE_BARREL_OF_BEERa" id="X_76_THE_BARREL_OF_BEERa"></a><a href="#X_76_THE_BARREL_OF_BEER"><b>76.&mdash;THE BARREL OF BEER.&mdash;<i>solution</i></b></a></p>
+
+<p>Here the digital roots of the six numbers are 6, 4, 1, 2, 7, 9, which
+together sum to 29, whose digital root is 2. As the contents of the
+barrels sold must be a number divisible by 3, if one buyer purchased
+twice as much as the other, we must find a barrel with root 2, 5, or 8
+to set on one side. There is only one barrel, that containing 20
+gallons, that fulfils these conditions. So the man must have kept
+these 20 gallons of beer for his own use and sold one man 33 gallons
+(the 18-gallon and 15-gallon barrels) and sold the other man 66
+gallons (the 16, 19, and 31 gallon barrels).</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_77_DIGITS_AND_SQUARESa" id="X_77_DIGITS_AND_SQUARESa"></a><a href="#X_77_DIGITS_AND_SQUARES"><b>77.&mdash;DIGITS AND SQUARES.&mdash;<i>solution</i></b></a></p>
+
+<p>The top row must be one of the four following numbers: 192, 219, 273,
+327. The first was the example given.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 156<a name="Page_156" id="Page_156"></a></span><a name="X_78_ODD_AND_EVEN_DIGITSa" id="X_78_ODD_AND_EVEN_DIGITSa"></a><a href="#X_78_ODD_AND_EVEN_DIGITS"><b>78.&mdash;ODD AND EVEN DIGITS.&mdash;<i>solution</i></b></a></p>
+
+<p>As we have to exclude complex and improper fractions and recurring
+decimals, the simplest solution is this: 79&nbsp;+&nbsp;5<sup>1</sup>/<sub>3</sub> and 84&nbsp;+&nbsp;<sup>2</sup>/<sub>6</sub>, both
+equal 84<sup>1</sup>/<sub>3</sub>. Without any use of fractions it is obviously impossible.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_79_THE_LOCKERS_PUZZLEa" id="X_79_THE_LOCKERS_PUZZLEa"></a><a href="#X_79_THE_LOCKERS_PUZZLE"><b>79.&mdash;THE LOCKERS PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The smallest possible total is 356&nbsp;=&nbsp;107&nbsp;+&nbsp;249, and the largest sum
+possible is 981&nbsp;=&nbsp;235&nbsp;+&nbsp;746, or 657&nbsp;+&nbsp;324. The middle sum may be either 720
+=134+586, or 702&nbsp;=&nbsp;134&nbsp;+&nbsp;568, or 407&nbsp;=&nbsp;138&nbsp;+&nbsp;269. The total in this case must
+be made up of three of the figures 0, 2, 4, 7, but no sum other than
+the three given can possibly be obtained. We have therefore no choice
+in the case of the first locker, an alternative in the case of the
+third, and any one of three arrangements in the case of the middle
+locker. Here is one solution:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>107</td><td align='center'>134</td><td align='center'>235</td></tr>
+<tr><td align='center' class='bb'>249</td><td align='center' class='bb'>586</td><td align='center' class='bb'>746</td></tr>
+<tr><td align='center'>356</td><td align='center'>720</td><td align='center'>981</td></tr>
+</table></div>
+
+<p>Of course, in each case figures in the first two lines may be
+exchanged vertically without altering the total, and as a result there
+are just 3,072 different ways in which the figures might be actually
+placed on the locker doors. I must content myself with showing one
+little principle involved in this puzzle. The sum of the digits in the
+total is always governed by the digit omitted. <sup>9</sup>/<sub>9</sub>&nbsp;-&nbsp;<sup>7</sup>/<sub>10</sub>&nbsp;-&nbsp;<sup>5</sup>/<sub>11</sub> -
+<sup>3</sup>/<sub>12</sub>&nbsp;-&nbsp;<sup>1</sup>/<sub>13</sub>&nbsp;-&nbsp;<sup>8</sup>/<sub>14</sub>&nbsp;-&nbsp;<sup>6</sup>/<sub>15</sub>&nbsp;-&nbsp;<sup>4</sup>/<sub>16</sub>&nbsp;-&nbsp;<sup>2</sup>/<sub>17</sub>&nbsp;-&nbsp;<sup>0</sup>/<sub>18</sub>. Whichever digit shown
+here in the upper line we omit, the sum of the digits in the total
+will be found beneath it. Thus in the case of locker A we omitted 8,
+and the figures in the total sum up to 14. If, therefore, we wanted to
+get 356, we may know at once to a certainty that it can only be
+obtained (if at all) by dropping the 8.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_80_THE_THREE_GROUPSa" id="X_80_THE_THREE_GROUPSa"></a><a href="#X_80_THE_THREE_GROUPS"><b>80.&mdash;THE THREE GROUPS.&mdash;<i>solution</i></b></a></p>
+
+<p>There are nine solutions to this puzzle, as follows, and no more:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>12&nbsp;&times;&nbsp;483&nbsp;=&nbsp;5,796</td></tr>
+<tr><td align='center'>27&nbsp;&times;&nbsp;198&nbsp;=&nbsp;5,346</td></tr>
+<tr><td align='center'>42&nbsp;&times;&nbsp;138&nbsp;=&nbsp;5,796</td></tr>
+<tr><td align='center'>39&nbsp;&times;&nbsp;186&nbsp;=&nbsp;7,254</td></tr>
+<tr><td align='center'>18&nbsp;&times;&nbsp;297&nbsp;=&nbsp;5,346</td></tr>
+<tr><td align='center'>48&nbsp;&times;&nbsp;159&nbsp;=&nbsp;7,632</td></tr>
+<tr><td align='center'>28&nbsp;&times;&nbsp;157&nbsp;=&nbsp;4,396</td></tr>
+<tr><td align='center'>4&nbsp;&times;&nbsp;1,738&nbsp;=&nbsp;6,952</td></tr>
+<tr><td align='center'>4&nbsp;&times;&nbsp;1,963&nbsp;=&nbsp;7,852</td></tr>
+</table></div>
+
+<p>The seventh answer is the one that is most likely to be overlooked by
+solvers of the puzzle.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_81_THE_NINE_COUNTERSa" id="X_81_THE_NINE_COUNTERSa"></a><a href="#X_81_THE_NINE_COUNTERS"><b>81.&mdash;THE NINE COUNTERS.&mdash;<i>solution</i></b></a></p>
+
+<p>In this case a certain amount of mere "trial" is unavoidable. But
+there are two kinds of "trials"&mdash;those that are purely haphazard, and
+those that are methodical. The true puzzle lover is never satisfied
+with mere haphazard trials. The reader will find that by just
+reversing the figures in 23 and 46 (making the multipliers 32 and 64)
+both products will be 5,056. This is an improvement, but it is not the
+correct answer. We can get as large a product as 5,568 if we multiply
+174 by 32 and 96 by 58, but this solution is not to be found without
+the exercise of some judgment and patience.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_82_THE_TEN_COUNTERSa" id="X_82_THE_TEN_COUNTERSa"></a><a href="#X_82_THE_TEN_COUNTERS"><b>82.&mdash;THE TEN COUNTERS.&mdash;<i>solution</i></b></a></p>
+
+<p>As I pointed out, it is quite easy so to arrange the counters that
+they shall form a pair of simple multiplication sums, each of which
+will give the same product&mdash;in fact, this can be done by anybody in
+five minutes with a little patience. But it is quite another matter to
+find that pair which gives the largest product and that which gives
+the smallest product.</p>
+
+<p>Now, in order to get the smallest product, it is necessary to select
+as multipliers the two smallest possible numbers. If, therefore, we
+place 1 and 2 as multipliers, all we have to do is to arrange the
+remaining eight counters in such a way that they shall form two
+numbers, one of which is just double the other; and in doing this we
+must, of course, try to make the smaller number as low as possible. Of
+course the lowest number we could get would be 3,045; but this will
+not work, neither will 3,405, 3,450, etc., and it may be ascertained
+that 3,485 is the lowest possible. One of the required answers is
+3,485&nbsp;&times;&nbsp;2&nbsp;=&nbsp;6,970, and 6,970&nbsp;&times;&nbsp;1&nbsp;=&nbsp;6,970.</p>
+
+<p>The other part of the puzzle (finding the pair with the highest
+product) is, however, the real knotty point, for it is not at all easy
+to discover whether we should let the multiplier consist of one or of
+two figures, though it is clear that we must keep, so far as we can,
+the largest figures to the left in both multiplier and multiplicand.
+It will be seen that by the following arrangement so high a number as
+58,560 may be obtained. Thus, 915&nbsp;&times;&nbsp;64&nbsp;=&nbsp;58,560, and 732&nbsp;&times;&nbsp;80&nbsp;=&nbsp;58,560.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_83_DIGITAL_MULTIPLICATIONa" id="X_83_DIGITAL_MULTIPLICATIONa"></a><a href="#X_83_DIGITAL_MULTIPLICATION"><b>83.&mdash;DIGITAL MULTIPLICATION.&mdash;<i>solution</i></b></a></p>
+
+<p>The solution that gives the smallest possible sum of digits in the
+common product is 23&nbsp;&times;&nbsp;174&nbsp;=&nbsp;58&nbsp;&times;&nbsp;69&nbsp;=&nbsp;4,002, and the solution that gives the
+largest possible sum of digits, 9&nbsp;&times;&nbsp;654&nbsp;=&nbsp;18&nbsp;&times;&nbsp;327&nbsp;=&nbsp;5,886. In the first case
+the digits sum to 6 and in the second case to 27. There is no way of
+obtaining the solution but by actual trial.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_84_THE_PIERROTS_PUZZLEa" id="X_84_THE_PIERROTS_PUZZLEa"></a><a href="#X_84_THE_PIERROTS_PUZZLE"><b>84.&mdash;THE PIERROT'S PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>There are just six different solutions to this puzzle, as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>8</td><td align='center'>multiplied by</td><td align='right'>473</td><td align='center'>equals</td><td align='right'>3784</td></tr>
+<tr><td align='right'>9</td><td align='center'>"</td><td align='right'>351</td><td align='center'>"</td><td align='right'>3159</td></tr>
+<tr><td align='right'>15</td><td align='center'>"</td><td align='right'>93</td><td align='center'>"</td><td align='right'>1395</td></tr>
+<tr><td align='right'>21</td><td align='center'>"</td><td align='right'>87</td><td align='center'>"</td><td align='right'>1287</td></tr>
+<tr><td align='right'>27</td><td align='center'>"</td><td align='right'>81</td><td align='center'>"</td><td align='right'>2187</td></tr>
+<tr><td align='right'>35</td><td align='center'>"</td><td align='right'>41</td><td align='center'>"</td><td align='right'>1435</td></tr>
+</table></div>
+
+<p>It will be seen that in every case the two multipliers contain exactly
+the same figures as the product.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 157<a name="Page_157" id="Page_157"></a></span><a name="X_85_THE_CAB_NUMBERSa" id="X_85_THE_CAB_NUMBERSa"></a><a href="#X_85_THE_CAB_NUMBERS"><b>85.&mdash;THE CAB NUMBERS.&mdash;<i>solution</i></b></a></p>
+
+<p>The highest product is, I think, obtained by multiplying 8,745,231 by
+96&mdash;namely, 839,542,176.</p>
+
+<p>Dealing here with the problem generally, I have shown in the last
+puzzle that with three digits there are only two possible solutions,
+and with four digits only six different solutions.</p>
+
+<p>These cases have all been given. With five digits there are just
+twenty-two solutions, as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>3</td><td align='center'>x</td><td align='right'>4128</td><td align='center'>=</td><td align='right'>12384</td></tr>
+<tr><td align='right'>3</td><td align='center'>x</td><td align='right'>4281</td><td align='center'>=</td><td align='right'>12843</td></tr>
+<tr><td align='right'>3</td><td align='center'>x</td><td align='right'>7125</td><td align='center'>=</td><td align='right'>21375</td></tr>
+<tr><td align='right'>3</td><td align='center'>x</td><td align='right'>7251</td><td align='center'>=</td><td align='right'>21753</td></tr>
+<tr><td align='right'>2541</td><td align='center'>x</td><td align='right'>6</td><td align='center'>=</td><td align='right'>15246</td></tr>
+<tr><td align='right'>651</td><td align='center'>x</td><td align='right'>24</td><td align='center'>=</td><td align='right'>15624</td></tr>
+<tr><td align='right'>678</td><td align='center'>x</td><td align='right'>42</td><td align='center'>=</td><td align='right'>28476</td></tr>
+<tr><td align='right'>246</td><td align='center'>x</td><td align='right'>51</td><td align='center'>=</td><td align='right'>12546</td></tr>
+<tr><td align='right'>57</td><td align='center'>x</td><td align='right'>834</td><td align='center'>=</td><td align='right'>47538</td></tr>
+<tr><td align='right'>75</td><td align='center'>x</td><td align='right'>231</td><td align='center'>=</td><td align='right'>17325</td></tr>
+<tr><td align='right'>624</td><td align='center'>x</td><td align='right'>78</td><td align='center'>=</td><td align='right'>48672</td></tr>
+<tr><td align='right'>435</td><td align='center'>x</td><td align='right'>87</td><td align='center'>=</td><td align='right'>37845</td></tr>
+<tr><td align='center' colspan='5'>&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;</td></tr>
+<tr><td align='right'>9</td><td align='center'>x</td><td align='right'>7461</td><td align='center'>=</td><td align='right'>67149</td></tr>
+<tr><td align='right'>72</td><td align='center'>x</td><td align='right'>936</td><td align='center'>=</td><td align='right'>67392</td></tr>
+<tr><td align='center' colspan='5'>&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;</td></tr>
+<tr><td align='right'>2</td><td align='center'>x</td><td align='right'>8714</td><td align='center'>=</td><td align='right'>17428</td></tr>
+<tr><td align='right'>2</td><td align='center'>x</td><td align='right'>8741</td><td align='center'>=</td><td align='right'>17482</td></tr>
+<tr><td align='right'>65</td><td align='center'>x</td><td align='right'>281</td><td align='center'>=</td><td align='right'>18265</td></tr>
+<tr><td align='right'>65</td><td align='center'>x</td><td align='right'>983</td><td align='center'>=</td><td align='right'>63985</td></tr>
+<tr><td align='center' colspan='5'>&mdash;&mdash;&mdash;&mdash;&mdash;&mdash;</td></tr>
+<tr><td align='right'>4973</td><td align='center'>x</td><td align='right'>8</td><td align='center'>=</td><td align='right'>39784</td></tr>
+<tr><td align='right'>6521</td><td align='center'>x</td><td align='right'>8</td><td align='center'>=</td><td align='right'>52168</td></tr>
+<tr><td align='right'>14</td><td align='center'>x</td><td align='right'>926</td><td align='center'>=</td><td align='right'>12964</td></tr>
+<tr><td align='right'>86</td><td align='center'>x</td><td align='right'>251</td><td align='center'>=</td><td align='right'>21586</td></tr>
+</table></div>
+
+<p>Now, if we took every possible combination and tested it by
+multiplication, we should need to make no fewer than 30,240 trials,
+or, if we at once rejected the number 1 as a multiplier, 28,560
+trials&mdash;a task that I think most people would be inclined to shirk.
+But let us consider whether there be no shorter way of getting at the
+results required. I have already explained that if you add together
+the digits of any number and then, as often as necessary, add the
+digits of the result, you must ultimately get a number composed of one
+figure. This last number I call the "digital root." It is necessary in
+every solution of our problem that the root of the sum of the digital
+roots of our multipliers shall be the same as the root of their
+product. There are only four ways in which this can happen: when the
+digital roots of the multipliers are 3 and 6, or 9 and 9, or 2 and 2,
+or 5 and 8. I have divided the twenty-two answers above into these
+four classes. It is thus evident that the digital root of any product
+in the first two classes must be 9, and in the second two classes 4.</p>
+
+<p>Owing to the fact that no number of five figures can have a digital
+sum less than 15 or more than 35, we find that the figures of our
+product must sum to either 18 or 27 to produce the root 9, and to
+either 22 or 31 to produce the root 4. There are 3 ways of selecting
+five different figures that add up to 18, there are 11 ways of
+selecting five figures that add up to 27, there are 9 ways of
+selecting five figures that add up to 22, and 5 ways of selecting five
+figures that add up to 31. There are, therefore, 28 different groups,
+and no more, from any one of which a product may be formed.</p>
+
+<p>We next write out in a column these 28 sets of five figures, and
+proceed to tabulate the possible factors, or multipliers, into which
+they may be split. Roughly speaking, there would now appear to be
+about 2,000 possible cases to be tried, instead of the 30,240
+mentioned above; but the process of elimination now begins, and if the
+reader has a quick eye and a clear head he can rapidly dispose of the
+large bulk of these cases, and there will be comparatively few test
+multiplications necessary. It would take far too much space to explain
+my own method in detail, but I will take the first set of figures in
+my table and show how easily it is done by the aid of little tricks
+and dodges that should occur to everybody as he goes along.</p>
+
+<p>My first product group of five figures is 84,321. Here, as we have
+seen, the root of each factor must be 3 or a multiple of 3. As there
+is no 6 or 9, the only single multiplier is 3. Now, the remaining four
+figures can be arranged in 24 different ways, but there is no need to
+make 24 multiplications. We see at a glance that, in order to get a
+five-figure product, either the 8 or the 4 must be the first figure to
+the left. But unless the 2 is preceded on the right by the 8, it will
+produce when multiplied either a 6 or a 7, which must not occur. We
+are, therefore, reduced at once to the two cases, 3&nbsp;&times;&nbsp;4,128 and 3 &times;
+4,281, both of which give correct solutions. Suppose next that we are
+trying the two-figure factor, 21. Here we see that if the number to be
+multiplied is under 500 the product will either have only four figures
+or begin with 10. Therefore we have only to examine the cases 21&nbsp;&times;&nbsp;843
+and 21&nbsp;&times;&nbsp;834. But we know that the first figure will be repeated, and
+that the second figure will be twice the first figure added to the
+second. Consequently, as twice 3 added to 4 produces a nought in our
+product, the first case is at once rejected. It only remains to try
+the remaining case by multiplication, when we find it does not give a
+correct answer. If we are next trying the factor 12, we see at the
+start that neither the 8 nor the 3 can be in the units place, because
+they would produce a 6, and so on. A sharp eye and an alert judgment
+will enable us thus to run through our table in a much shorter time
+than would be expected. The process took me a little more than three
+hours.</p>
+
+<p>I have not attempted to enumerate the solutions in the cases of six,
+seven, eight, and nine digits, but I have recorded nearly fifty
+examples with nine digits alone.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_86_QUEER_MULTIPLICATIONa" id="X_86_QUEER_MULTIPLICATIONa"></a><a href="#X_86_QUEER_MULTIPLICATION"><b>86.&mdash;QUEER MULTIPLICATION.&mdash;<i>solution</i></b></a></p>
+
+<p>If we multiply 32547891 by 6, we get the product, 195287346. In both
+cases all the nine digits are used once and once only.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 158<a name="Page_158" id="Page_158"></a></span><a name="X_87_THE_NUMBER_CHECKS_PUZZLEa" id="X_87_THE_NUMBER_CHECKS_PUZZLEa"></a><a href="#X_87_THE_NUMBER_CHECKS_PUZZLE"><b>87.&mdash;THE NUMBER CHECKS PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Divide the ten checks into the following three groups: 7 1 5&mdash;4 6&mdash;3 2
+8 9 0, and the first multiplied by the second produces the third.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_88_DIGITAL_DIVISIONa" id="X_88_DIGITAL_DIVISIONa"></a><a href="#X_88_DIGITAL_DIVISION"><b>88.&mdash;DIGITAL DIVISION.&mdash;<i>solution</i></b></a></p>
+
+<p>It is convenient to consider the digits as arranged to form fractions
+of the respective values, one-half, one-third, one-fourth, one-fifth,
+one-sixth, one-seventh, one-eighth, and one-ninth. I will first give
+the eight answers, as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'><sup>6729</sup>/<sub>13458</sub>&nbsp;=&nbsp;�</td></tr>
+<tr><td align='center'><sup>5823</sup>/<sub>17469</sub>&nbsp;=&nbsp;<sup>1</sup>/<sub>3</sub></td></tr>
+<tr><td align='center'><sup>3942</sup>/<sub>15768</sub>&nbsp;=&nbsp;�</td></tr>
+<tr><td align='center'><sup>2697</sup>/<sub>13485</sub>&nbsp;=&nbsp;<sup>1</sup>/<sub>5</sub></td></tr>
+<tr><td align='center'><sup>2943</sup>/<sub>17658</sub>&nbsp;=&nbsp;<sup>1</sup>/<sub>6</sub></td></tr>
+<tr><td align='center'><sup>2394</sup>/<sub>16758</sub>&nbsp;=&nbsp;<sup>1</sup>/<sub>7</sub></td></tr>
+<tr><td align='center'><sup>3187</sup>/<sub>25496</sub>&nbsp;=&nbsp;<sup>1</sup>/<sub>8</sub></td></tr>
+<tr><td align='center'><sup>6381</sup>/<sub>57429</sub>&nbsp;=&nbsp;<sup>1</sup>/<sub>9</sub></td></tr>
+</table></div>
+
+<p>The sum of the numerator digits and the denominator digits will, of
+course, always be 45, and the "digital root" is 9. Now, if we separate
+the nine digits into any two groups, the sum of the two digital roots
+will always be 9. In fact, the two digital roots must be either 9&mdash;9,
+8&mdash;1, 7&mdash;2, 6&mdash;3, or 5&mdash;4. In the first case the actual sum is 18, but
+then the digital root of this number is itself 9. The solutions in the
+cases of one-third, one-fourth, one-sixth, one-seventh, and one-ninth
+must be of the form 9&mdash;9; that is to say, the digital roots of both
+numerator and denominator will be 9. In the cases of one-half and
+one-fifth, however, the digital roots are 6&mdash;3, but of course the
+higher root may occur either in the numerator or in the denominator;
+thus <sup>2697</sup>/<sub>13485</sub>, <sup>2769</sup>/<sub>13845</sub>, <sup>2973</sup>/<sub>14865</sub>, <sup>3729</sup>/<sub>18645</sub>, where, in the
+first two arrangements, the roots of the numerator and denominator are
+respectively 6&mdash;3, and in the last two 3&mdash;6. The most curious case of
+all is, perhaps, one-eighth, for here the digital roots may be of any
+one of the five forms given above.</p>
+
+<p>The denominators of the fractions being regarded as the numerators
+multiplied by 2, 3, 4, 5, 6, 7, 8, and 9 respectively, we must pay
+attention to the "carryings over." In order to get five figures in the
+product there will, of course, always be a carry-over after
+multiplying the last figure to the left, and in every case higher than
+4 we must carry over at least three times. Consequently in cases from
+one-fifth to one-ninth we cannot produce different solutions by a mere
+change of position of pairs of figures, as, for example, we may with
+<sup>5832</sup>/<sub>17496</sub> and <sup>5823</sup>/<sub>17469</sub>, where the <sup>2</sup>/<sub>6</sub> and <sup>3</sup>/<sub>9</sub> change places. It is
+true that the same figures may often be differently arranged, as shown
+in the two pairs of values for one-fifth that I have given in the last
+paragraph, but here it will be found there is a general readjustment
+of figures and not a simple changing of the positions of pairs. There
+are other little points that would occur to every solver&mdash;such as that
+the figure 5 cannot ever appear to the extreme right of the numerator,
+as this would result in our getting either a nought or a second 5 in
+the denominator. Similarly 1 cannot ever appear in the same position,
+nor 6 in the fraction one-sixth, nor an even figure in the fraction
+one-fifth, and so on. The preliminary consideration of such points as
+I have touched upon will not only prevent our wasting a lot of time in
+trying to produce impossible forms, but will lead us more or less
+directly to the desired solutions.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_89_ADDING_THE_DIGITSa" id="X_89_ADDING_THE_DIGITSa"></a><a href="#X_89_ADDING_THE_DIGITS"><b>89.&mdash;ADDING THE DIGITS.&mdash;<i>solution</i></b></a></p>
+
+<p>The smallest possible sum of money is &pound;1, 8<i>s</i>. 9&frac34;d., the digits of
+which add to 25.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_90_THE_CENTURY_PUZZLEa" id="X_90_THE_CENTURY_PUZZLEa"></a><a href="#X_90_THE_CENTURY_PUZZLE"><b>90.&mdash;THE CENTURY PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The problem of expressing the number 100 as a mixed number or
+fraction, using all the nine digits once, and once only, has, like all
+these digital puzzles, a fascinating side to it. The merest tyro can
+by patient trial obtain correct results, and there is a singular
+pleasure in discovering and recording each new arrangement akin to the
+delight of the botanist in finding some long-sought plant. It is
+simply a matter of arranging those nine figures correctly, and yet
+with the thousands of possible combinations that confront us the task
+is not so easy as might at first appear, if we are to get a
+considerable number of results. Here are eleven answers, including the
+one I gave as a specimen:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>96<sup>2148</sup>/<sub>537</sub></td></tr>
+<tr><td align='center'>96<sup>1752</sup>/<sub>438</sub></td></tr>
+<tr><td align='center'>96<sup>1428</sup>/<sub>357</sub></td></tr>
+<tr><td align='center'>94<sup>1578</sup>/<sub>263</sub></td></tr>
+<tr><td align='center'>91<sup>7524</sup>/<sub>836</sub></td></tr>
+<tr><td align='center'>91<sup>5823</sup>/<sub>647</sub></td></tr>
+<tr><td align='center'>91<sup>5742</sup>/<sub>638</sub></td></tr>
+<tr><td align='center'>82<sup>3546</sup>/<sub>197</sub></td></tr>
+<tr><td align='center'>81<sup>7524</sup>/<sub>396</sub></td></tr>
+<tr><td align='center'>81<sup>5643</sup>/<sub>297</sub></td></tr>
+<tr><td align='center'>3<sup>69258</sup>/<sub>714</sub></td></tr>
+</table></div>
+
+<p>Now, as all the fractions necessarily represent whole numbers, it will
+be convenient to deal with them in the following form: 96&nbsp;+&nbsp;4, 94&nbsp;+&nbsp;6,
+91&nbsp;+&nbsp;9, 82&nbsp;+&nbsp;18, 81&nbsp;+&nbsp;19, and 3&nbsp;+&nbsp;97.</p>
+
+<p>With any whole number the digital roots of the fraction that brings it
+up to 100 will always be of one particular form. Thus, in the case of
+96&nbsp;+&nbsp;4, one can say at once that if any answers are obtainable, then
+the roots of both the numerator and the denominator of the fraction
+will be 6. Examine the first three arrangements given above, and you
+will find that this is so. In the case of 94&nbsp;+&nbsp;6 the roots of the
+numerator and denominator will be respectively 3&mdash;2, in the case of 91
++ 9 and of 82&nbsp;+&nbsp;18 they will be 9&mdash;8, in the case of 81&nbsp;+&nbsp;19 they will
+be 9&mdash;9, and in the case of 3&nbsp;+&nbsp;97 they will be 3&mdash;3. Every fraction
+that can be employed has, therefore, its particular digital root form,
+and you are only wasting your time in unconsciously attempting to
+break through this law.</p>
+
+<p>Every reader will have perceived that certain whole numbers are
+evidently impossible. Thus, if there is a 5 in the whole number, there
+will also be a nought or a second 5 in the fraction, which are barred
+by the conditions. Then multiples of 10, such as 90 and 80, cannot of
+course occur, nor can the whole number conclude with a 9, like 89 and
+79, because the fraction, equal to 11 or 21, will have 1 in the last
+place, and will therefore repeat a figure. Whole numbers that repeat a
+figure, such as 88 and 77, are also clearly useless. These cases, as I
+have said, are all obvious to every reader. But when I declare <span class='pagenum'>Pg 159<a name="Page_159" id="Page_159"></a></span>that
+such combinations as 98&nbsp;+&nbsp;2, 92&nbsp;+&nbsp;8, 86&nbsp;+&nbsp;14, 83&nbsp;+&nbsp;17, 74&nbsp;+&nbsp;26, etc.,
+etc., are to be at once dismissed as impossible, the reason is not so
+evident, and I unfortunately cannot spare space to explain it.</p>
+
+<p>But when all those combinations have been struck out that are known to
+be impossible, it does not follow that all the remaining "possible
+forms" will actually work. The elemental form may be right enough, but
+there are other and deeper considerations that creep in to defeat our
+attempts. For example, 98&nbsp;+&nbsp;2 is an impossible combination, because we
+are able to say at once that there is no possible form for the digital
+roots of the fraction equal to 2. But in the case of 97&nbsp;+&nbsp;3 there is a
+possible form for the digital roots of the fraction, namely, 6&mdash;5, and
+it is only on further investigation that we are able to determine that
+this form cannot in practice be obtained, owing to curious
+considerations. The working is greatly simplified by a process of
+elimination, based on such considerations as that certain
+multiplications produce a repetition of figures, and that the whole
+number cannot be from 12 to 23 inclusive, since in every such case
+sufficiently small denominators are not available for forming the
+fractional part.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_91_MORE_MIXED_FRACTIONSa" id="X_91_MORE_MIXED_FRACTIONSa"></a><a href="#X_91_MORE_MIXED_FRACTIONS"><b>91.&mdash;MORE MIXED FRACTIONS.&mdash;<i>solution</i></b></a></p>
+
+<p>The point of the present puzzle lies in the fact that the numbers 15
+and 18 are not capable of solution. There is no way of determining
+this without trial. Here are answers for the ten possible numbers:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>9<sup>5472</sup>/<sub>1368</sub></td><td align='center'>=</td><td align='right'>13;</td></tr>
+<tr><td align='right'>9<sup>6435</sup>/<sub>1287</sub></td><td align='center'>=</td><td align='right'>14;</td></tr>
+<tr><td align='right'>12<sup>3576</sup>/<sub>894</sub></td><td align='center'>=</td><td align='right'>16;</td></tr>
+<tr><td align='right'>6<sup>13258</sup>/<sub>947</sub></td><td align='center'>=</td><td align='right'>20;</td></tr>
+<tr><td align='right'>15<sup>9432</sup>/<sub>786</sub></td><td align='center'>=</td><td align='right'>27;</td></tr>
+<tr><td align='right'>24<sup>9756</sup>/<sub>813</sub></td><td align='center'>=</td><td align='right'>36;</td></tr>
+<tr><td align='right'>27<sup>5148</sup>/<sub>396</sub></td><td align='center'>=</td><td align='right'>40;</td></tr>
+<tr><td align='right'>65<sup>1892</sup>/<sub>473</sub></td><td align='center'>=</td><td align='right'>69;</td></tr>
+<tr><td align='right'>59<sup>3614</sup>/<sub>278</sub></td><td align='center'>=</td><td align='right'>72;</td></tr>
+<tr><td align='right'>75<sup>3648</sup>/<sub>192</sub></td><td align='center'>=</td><td align='right'>94.</td></tr>
+</table></div>
+
+<p>I have only found the one arrangement for each of the numbers 16, 20,
+and 27; but the other numbers are all capable of being solved in more
+than one way. As for 15 and 18, though these may be easily solved as a
+simple fraction, yet a "mixed fraction" assumes the presence of a
+whole number; and though my own idea for dodging the conditions is the
+following, where the fraction is both complex and mixed, it will be
+fairer to keep exactly to the form indicated:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>3<sup>8952</sup>/<span style="font-size: smaller;">746</span>/<sub>1</sub>&nbsp;=&nbsp;15</td></tr>
+<tr><td align='center'>9<sup>5742</sup>/<span style="font-size: smaller;">638</span>/<sub>1</sub>&nbsp;=&nbsp;18</td></tr>
+</table></div>
+
+<p>I have proved the possibility of solution for all numbers up to 100,
+except 1, 2, 3, 4, 15, and 18. The first three are easily shown to be
+impossible. I have also noticed that numbers whose digital root is
+8&mdash;such as 26, 35, 44, 53, etc.&mdash;seem to lend themselves to the
+greatest number of answers. For the number 26 alone I have recorded no
+fewer than twenty-five different arrangements, and I have no doubt
+that there are many more.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_92_DIGITAL_SQUARE_NUMBERSa" id="X_92_DIGITAL_SQUARE_NUMBERSa"></a><a href="#X_92_DIGITAL_SQUARE_NUMBERS"><b>92.&mdash;DIGITAL SQUARE NUMBERS.&mdash;<i>solution</i></b></a></p>
+
+<p>So far as I know, there are no published tables of square numbers that
+go sufficiently high to be available for the purposes of this puzzle.
+The lowest square number containing all the nine digits once, and once
+only, is 139,854,276, the square of 11,826. The highest square number
+under the same conditions is, 923,187,456, the square of 30,384.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_93_THE_MYSTIC_ELEVENa" id="X_93_THE_MYSTIC_ELEVENa"></a><a href="#X_93_THE_MYSTIC_ELEVEN"><b>93&mdash;THE MYSTIC ELEVEN.&mdash;<i>solution</i></b></a></p>
+
+<p>Most people know that if the sum of the digits in the odd places of
+any number is the same as the sum of the digits in the even places,
+then the number is divisible by 11 without remainder. Thus in
+896743012 the odd digits, 20468, add up 20, and the even digits, 1379,
+also add up 20. Therefore the number may be divided by 11. But few
+seem to know that if the difference between the sum of the odd and the
+even digits is 11, or a multiple of 11, the rule equally applies. This
+law enables us to find, with a very little trial, that the smallest
+number containing nine of the ten digits (calling nought a digit) that
+is divisible by 11 is 102,347,586, and the highest number possible,
+987,652,413.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_94_THE_DIGITAL_CENTURYa" id="X_94_THE_DIGITAL_CENTURYa"></a><a href="#X_94_THE_DIGITAL_CENTURY"><b>94.&mdash;THE DIGITAL CENTURY.&mdash;<i>solution</i></b></a></p>
+
+<p>There is a very large number of different ways in which arithmetical
+signs may be placed between the nine digits, arranged in numerical
+order, so as to give an expression equal to 100. In fact, unless the
+reader investigated the matter very closely, he might not suspect that
+so many ways are possible. It was for this reason that I added the
+condition that not only must the fewest possible signs be used, but
+also the fewest possible strokes. In this way we limit the problem to
+a single solution, and arrive at the simplest and therefore (in this
+case) the best result.</p>
+
+<p>Just as in the case of magic squares there are methods by which we may
+write down with the greatest ease a large number of solutions, but not
+all the solutions, so there are several ways in which we may quickly
+arrive at dozens of arrangements of the "Digital Century," without
+finding all the possible arrangements. There is, in fact, very little
+principle in the thing, and there is no certain way of demonstrating
+that we have got the best possible solution. All I can say is that the
+arrangement I shall give as the best is the best I have up to the
+present succeeded in discovering. I will give the reader a few
+interesting specimens, the first being the solution usually published,
+and the last the best solution that I know.</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'></td><td align='center'>Signs.</td><td align='center'></td><td align='center'>Strokes.</td></tr>
+<tr><td align='right'>1&nbsp;+&nbsp;2&nbsp;+&nbsp;3&nbsp;+&nbsp;4&nbsp;+&nbsp;5&nbsp;+&nbsp;6&nbsp;+&nbsp;7&nbsp;+&nbsp;(8&nbsp;&times;&nbsp;9)&nbsp;=&nbsp;100</td><td align='center'>( 9</td><td align='center'>..</td><td align='center'>18)</td></tr>
+<tr><td align='right'>(1&nbsp;&times;&nbsp;2)&nbsp;-&nbsp;3&nbsp;-&nbsp;4&nbsp;-&nbsp;5&nbsp;+&nbsp;(6&nbsp;&times;&nbsp;7)&nbsp;+&nbsp;(8&nbsp;&times;&nbsp;9)&nbsp;=&nbsp;100</td><td align='center'>(12</td><td align='center'>..</td><td align='center'>20)</td></tr>
+<tr><td align='right'>1&nbsp;+&nbsp;(2&nbsp;&times;&nbsp;3)&nbsp;+&nbsp;(4&nbsp;&times;&nbsp;5)&nbsp;-&nbsp;6&nbsp;+&nbsp;7&nbsp;+&nbsp;(8&nbsp;&times;&nbsp;9)&nbsp;=&nbsp;100</td><td align='center'>(11</td><td align='center'>..</td><td align='center'>21)</td></tr>
+<tr><td align='right'>(1&nbsp;+&nbsp;2&nbsp;-&nbsp;3&nbsp;-&nbsp;4)(5&nbsp;-&nbsp;6&nbsp;-&nbsp;7&nbsp;-&nbsp;8&nbsp;-&nbsp;9)&nbsp;=&nbsp;100</td><td align='center'>( 9</td><td align='center'>..</td><td align='center'>12)</td></tr>
+<tr><td align='right'>1&nbsp;+&nbsp;(2&nbsp;&times;&nbsp;3)&nbsp;+&nbsp;4&nbsp;+&nbsp;5&nbsp;+&nbsp;67&nbsp;+&nbsp;8&nbsp;+&nbsp;9&nbsp;=&nbsp;100</td><td align='center'>(8</td><td align='center'>..</td><td align='center'>16)</td></tr>
+<tr><td align='right'>(1&nbsp;&times;&nbsp;2)&nbsp;+&nbsp;34&nbsp;+&nbsp;56&nbsp;+&nbsp;7&nbsp;-&nbsp;8&nbsp;+&nbsp;9&nbsp;=&nbsp;100</td><td align='center'>(7</td><td align='center'>..</td><td align='center'>13)</td></tr>
+<tr><td align='right'>12&nbsp;+&nbsp;3&nbsp;-&nbsp;4&nbsp;+&nbsp;5&nbsp;+&nbsp;67&nbsp;+&nbsp;8&nbsp;+&nbsp;9&nbsp;=&nbsp;100</td><td align='center'>(6</td><td align='center'>..</td><td align='center'>11)</td></tr>
+<tr><td align='right'>123&nbsp;-&nbsp;4&nbsp;-&nbsp;5&nbsp;-&nbsp;6&nbsp;-&nbsp;7&nbsp;+&nbsp;8&nbsp;-&nbsp;9&nbsp;=&nbsp;100</td><td align='center'>(6</td><td align='center'>..</td><td align='center'>7)</td></tr>
+<tr><td align='right'>123&nbsp;+&nbsp;4&nbsp;-&nbsp;5&nbsp;+&nbsp;67&nbsp;-&nbsp;8&nbsp;-&nbsp;9&nbsp;=&nbsp;100</td><td align='center'>(4</td><td align='center'>..</td><td align='center'>6)</td></tr>
+<tr><td align='right'>123&nbsp;+&nbsp;45&nbsp;-&nbsp;67&nbsp;+&nbsp;8&nbsp;-&nbsp;9&nbsp;=&nbsp;100</td><td align='center'>(4</td><td align='center'>..</td><td align='center'>6)</td></tr>
+<tr><td align='right'>123&nbsp;-&nbsp;45&nbsp;-&nbsp;67&nbsp;+&nbsp;89&nbsp;=&nbsp;100</td><td align='center'>(3</td><td align='center'>..</td><td align='center'>4)</td></tr>
+</table></div>
+
+<p><span class='pagenum'>Pg 160<a name="Page_160" id="Page_160"></a></span></p>
+<p>It will be noticed that in the above I have counted the bracket as one
+sign and two strokes. The last solution is singularly simple, and I do
+not think it will ever be beaten.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_95_THE_FOUR_SEVENSa" id="X_95_THE_FOUR_SEVENSa"></a><a href="#X_95_THE_FOUR_SEVENS"><b>95.&mdash;THE FOUR SEVENS.&mdash;<i>solution</i></b></a></p>
+
+<p>The way to write four sevens with simple arithmetical signs so that
+they represent 100 is as follows:&mdash;</p>
+
+<div class='center'>
+<sup>7</sup>/<sub>.7</sub>&times;<sup>7</sup>/<sub>.7</sub>&nbsp;=&nbsp;100.
+</div>
+
+<p>Of course the fraction, 7 over decimal 7, equals 7 divided by <sup>7</sup>/<sub>10</sub>,
+which is the same as 70 divided by 7, or 10. Then 10 multiplied by 10
+is 100, and there you are! It will be seen that this solution applies
+equally to any number whatever that you may substitute for 7.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_96_THE_DICE_NUMBERSa" id="X_96_THE_DICE_NUMBERSa"></a><a href="#X_96_THE_DICE_NUMBERS"><b>96.&mdash;THE DICE NUMBERS.&mdash;<i>solution</i></b></a></p>
+
+<p>The sum of all the numbers that can be formed with any given set of
+four different figures is always 6,666 multiplied by the sum of the
+four figures. Thus, 1, 2, 3, 4 add up 10, and ten times 6,666 is
+66,660. Now, there are thirty-five different ways of selecting four
+figures from the seven on the dice&mdash;remembering the 6 and 9 trick. The
+figures of all these thirty-five groups add up to 600. Therefore 6,666
+multiplied by 600 gives us 3,999,600 as the correct answer.</p>
+
+<p>Let us discard the dice and deal with the problem generally, using the
+nine digits, but excluding nought. Now, if you were given simply the
+sum of the digits&mdash;that is, if the condition were that you could use
+any four figures so long as they summed to a given amount&mdash;then we
+have to remember that several combinations of four digits will, in
+many cases, make the same sum.</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'></td><td align='center'>10</td><td align='center'>11</td><td align='center'>12</td><td align='center'>13</td><td align='center'>14</td><td align='center'>15</td><td align='center'>16</td><td align='center'>17</td><td align='center'>18</td><td align='center'>19</td><td align='center'>20</td></tr>
+<tr><td align='center'></td><td align='center'>1</td><td align='center'>1</td><td align='center'>2</td><td align='center'>3</td><td align='center'>5</td><td align='center'>6</td><td align='center'>8</td><td align='center'>9</td><td align='center'>11</td><td align='center'>11</td><td align='center'>12</td></tr>
+<tr><td>&nbsp;</td></tr>
+<tr><td align='center'></td><td align='center'>21</td><td align='center'>22</td><td align='center'>23</td><td align='center'>24</td><td align='center'>25</td><td align='center'>26</td><td align='center'>27</td><td align='center'>28</td><td align='center'>29</td><td align='center'>30</td></tr>
+<tr><td align='center'></td><td align='center'>11</td><td align='center'>11</td><td align='center'>9</td><td align='center'>8</td><td align='center'>6</td><td align='center'>5</td><td align='center'>3</td><td align='center'>2</td><td align='center'>1</td><td align='center'>1</td></tr>
+</table></div>
+
+<p>Here the top row of numbers gives all the possible sums of four
+different figures, and the bottom row the number of different ways in
+which each sum may be made. For example 13 may be made in three ways:
+1237, 1246, and 1345. It will be found that the numbers in the bottom
+row add up to 126, which is the number of combinations of nine figures
+taken four at a time. From this table we may at once calculate the
+answer to such a question as this: What is the sum of all the numbers
+composed of our different digits (nought excluded) that add up to 14?
+Multiply 14 by the number beneath t in the table, 5, and multiply the
+result by 6,666, and you will have the answer. It follows that, to
+know the sum of all the numbers composed of four different digits, if
+you multiply all the pairs in the two rows and then add the results
+together, you will get 2,520, which, multiplied by 6,666, gives the
+answer 16,798,320.</p>
+
+<p>The following general solution for any number of digits will
+doubtless interest readers. Let n represent number of digits, then 5
+(10<sup>n</sup>&nbsp;-&nbsp;1) ) 8! divided by (9&nbsp;-&nbsp;n)! equals the required sum.
+Note that 0! equals 1. This may be reduced to the following practical
+rule: Multiply together 4&nbsp;&times;&nbsp;7&nbsp;&times;&nbsp;6&nbsp;&times;&nbsp;5 ... to (n&nbsp;-&nbsp;1)
+factors; now add (n&nbsp;+&nbsp;1) ciphers to the right, and from this result
+subtract the same set of figures with a single cipher to the right. Thus
+for n&nbsp;=&nbsp;4 (as in the case last mentioned), 4&nbsp;&times;&nbsp;7&nbsp;&times;&nbsp;6&nbsp;=&nbsp;168.
+Therefore 16,800,000 less 1,680 gives us 16,798,320 in another way.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_97_THE_SPOT_ON_THE_TABLEa" id="X_97_THE_SPOT_ON_THE_TABLEa"></a><a href="#X_97_THE_SPOT_ON_THE_TABLE"><b>97.&mdash;THE SPOT ON THE TABLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The ordinary schoolboy would correctly treat this as a quadratic
+equation. Here is the actual arithmetic. Double the product of the two
+distances from the walls. This gives us 144, which is the square of
+12. The sum of the two distances is 17. If we add these two numbers,
+12 and 17, together, and also subtract one from the other, we get the
+two answers that 29 or 5 was the radius, or half-diameter, of the
+table. Consequently, the full diameter was 58 in. or 10 in. But a
+table of the latter dimensions would be absurd, and not at all in
+accordance with the illustration. Therefore the table must have been
+58 in. in diameter. In this case the spot was on the edge nearest to
+the corner of the room&mdash;to which the boy was pointing. If the other
+answer were admissible, the spot would be on the edge farthest from
+the corner of the room.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_98_ACADEMIC_COURTESIESa" id="X_98_ACADEMIC_COURTESIESa"></a><a href="#X_98_ACADEMIC_COURTESIES"><b>98.&mdash;ACADEMIC COURTESIES.&mdash;<i>solution</i></b></a></p>
+
+<p>There must have been ten boys and twenty girls. The number of bows
+girl to girl was therefore 380, of boy to boy 90, of girl with boy
+400, and of boys and girls to teacher 30, making together 900, as
+stated. It will be remembered that it was not said that the teacher
+himself returned the bows of any child.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_99_THE_THIRTY-THREE_PEARLSa" id="X_99_THE_THIRTY-THREE_PEARLSa"></a><a href="#X_99_THE_THIRTY-THREE_PEARLS"><b>99.&mdash;THE THIRTY-THREE PEARLS.&mdash;<i>solution</i></b></a></p>
+
+<p>The value of the large central pearl must have been &pound;3,000. The pearl
+at one end (from which they increased in value by &pound;100) was &pound;1,400;
+the pearl at the other end, &pound;600.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_100_THE_LABOURERS_PUZZLEa" id="X_100_THE_LABOURERS_PUZZLEa"></a><a href="#X_100_THE_LABOURERS_PUZZLE"><b>100.&mdash;THE LABOURER'S PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The man said, "I am going twice as deep," not "as deep again." That is
+to say, he was still going twice as deep as he had gone already, so
+<span class='pagenum'>Pg 161<a name="Page_161" id="Page_161"></a></span>that when finished the hole would be three times its present depth.
+Then the answer is that at present the hole is 3 ft. 6 in. deep and
+the man 2 ft. 4 in. above ground. When completed the hole will be 10
+ft. 6 in. deep, and therefore the man will then be 4 ft. 8 in. below
+the surface, or twice the distance that he is now above ground.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_101_THE_TRUSSES_OF_HAYa" id="X_101_THE_TRUSSES_OF_HAYa"></a><a href="#X_101_THE_TRUSSES_OF_HAY"><b>101.&mdash;THE TRUSSES OF HAY.&mdash;<i>solution</i></b></a></p>
+
+<p>Add together the ten weights and divide by 4, and we get 289 lbs. as
+the weight of the five trusses together. If we call the five trusses
+in the order of weight A, B, C, D, and E, the lightest being A and the
+heaviest E, then the lightest, no lbs., must be the weight of A and B;
+and the next lightest, 112 lbs., must be the weight of A and C. Then
+the two heaviest, D and E, must weigh 121 lbs., and C and E must weigh
+120 lbs. We thus know that A, B, D, and E weigh together 231 lbs.,
+which, deducted from 289 lbs. (the weight of the five trusses), gives
+us the weight of C as 58 lbs. Now, by mere subtraction, we find the
+weight of each of the five trusses&mdash;54 lbs., 56 lbs., 58 lbs., 59
+lbs., and 62 lbs. respectively.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_102_MR_GUBBINS_IN_A_FOGa" id="X_102_MR_GUBBINS_IN_A_FOGa"></a><a href="#X_102_MR_GUBBINS_IN_A_FOG"><b>102.&mdash;MR. GUBBINS IN A FOG.&mdash;<i>solution</i></b></a></p>
+
+<p>The candles must have burnt for three hours and three-quarters. One
+candle had one-sixteenth of its total length left and the other
+four-sixteenths.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_103_PAINTING_THE_LAMP-POSTSa" id="X_103_PAINTING_THE_LAMP-POSTSa"></a><a href="#X_103_PAINTING_THE_LAMP-POSTS"><b>103.&mdash;PAINTING THE LAMP-POSTS.&mdash;<i>solution</i></b></a></p>
+
+<p>Pat must have painted six more posts than Tim, no matter how many
+lamp-posts there were. For example, suppose twelve on each side; then
+Pat painted fifteen and Tim nine. If a hundred on each side, Pat
+painted one hundred and three, and Tim only ninety-seven</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_104_CATCHING_THE_THIEFa" id="X_104_CATCHING_THE_THIEFa"></a><a href="#X_104_CATCHING_THE_THIEF"><b>104.&mdash;CATCHING THE THIEF.&mdash;<i>solution</i></b></a></p>
+
+<p>The constable took thirty steps. In the same time the thief would take
+forty-eight, which, added to his start of twenty-seven, carried him
+seventy-five steps. This distance would be exactly equal to thirty
+steps of the constable.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_105_THE_PARISH_COUNCIL_ELECTIONa" id="X_105_THE_PARISH_COUNCIL_ELECTIONa"></a><a href="#X_105_THE_PARISH_COUNCIL_ELECTION"><b>105.&mdash;THE PARISH COUNCIL ELECTION,&mdash;<i>solution</i></b></a></p>
+
+<p>The voter can vote for one candidate in 23 ways, for two in 253 ways,
+for three in 1,771, for four in 8,855, for five in 33,649, for six in
+100,947, for seven in 245,157, for eight in 490,314, and for nine
+candidates in 817,190 different ways. Add these together, and we get
+the total of 1,698,159 ways of voting.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_106_THE_MUDDLETOWN_ELECTIONa" id="X_106_THE_MUDDLETOWN_ELECTIONa"></a><a href="#X_106_THE_MUDDLETOWN_ELECTION"><b>106.&mdash;THE MUDDLETOWN ELECTION.&mdash;<i>solution</i></b></a></p>
+
+<p>The numbers of votes polled respectively by the Liberal, the
+Conservative, the Independent, and the Socialist were 1,553, 1,535,
+1,407, and 978 All that was necessary was to add the sum of the three
+majorities (739) to the total poll of 5,473 (making 6,212) and divide
+by 4, which gives us 1,553 as the poll of the Liberal. Then the polls
+of the other three candidates can, of course, be found by deducting
+the successive majorities from the last-mentioned number.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_107_THE_SUFFRAGISTS_MEETINGa" id="X_107_THE_SUFFRAGISTS_MEETINGa"></a><a href="#X_107_THE_SUFFRAGISTS_MEETING"><b>107.&mdash;THE SUFFRAGISTS' MEETING.&mdash;<i>solution</i></b></a></p>
+
+<p>Eighteen were present at the meeting and eleven left. If twelve had
+gone, two-thirds would have retired. If only nine had gone, the
+meeting would have lost half its members.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_108_THE_LEAP-YEAR_LADIESa" id="X_108_THE_LEAP-YEAR_LADIESa"></a><a href="#X_108_THE_LEAP-YEAR_LADIES"><b>108.&mdash;THE LEAP-YEAR LADIES.&mdash;<i>solution</i></b></a></p>
+
+<p>The correct and only answer is that 11,616 ladies made proposals of
+marriage. Here are all the details, which the reader can check for
+himself with the original statements. Of 10,164 spinsters, 8,085
+married bachelors, 627 married widowers, 1,221 were declined by
+bachelors, and 231 declined by widowers. Of the 1,452 widows, 1,155
+married bachelors, and 297 married widowers. No widows were declined.
+The problem is not difficult, by algebra, when once we have succeeded
+in correctly stating it.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_109_THE_GREAT_SCRAMBLEa" id="X_109_THE_GREAT_SCRAMBLEa"></a><a href="#X_109_THE_GREAT_SCRAMBLE"><b>109.&mdash;THE GREAT SCRAMBLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The smallest number of sugar plums that will fulfil the conditions is
+26,880. The five boys obtained respectively: Andrew, 2,863; Bob,
+6,335; Charlie, 2,438; David, 10,294; Edgar, 4,950. There is a
+little trap concealed in the words near the end, "one-fifth of the
+same," that seems at first sight to upset the whole account of the
+affair. But a little thought will show that the words could only mean
+"one-fifth of five-eighths", the fraction last mentioned&mdash;that is,
+one-eighth of the three-quarters that Bob and Andrew had last
+acquired.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_110_THE_ABBOTS_PUZZLEa" id="X_110_THE_ABBOTS_PUZZLEa"></a><a href="#X_110_THE_ABBOTS_PUZZLE"><b>110.&mdash;THE ABBOT'S PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The only answer is that there were 5 men, 25 women, and 70 children.
+There were thus 100 persons in all, 5 times as many women as men, and
+as the men would together receive 15 bushels, the women 50 bushels,
+and the children 35 bushels, exactly 100 bushels would be distributed.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_111_REAPING_THE_CORNa" id="X_111_REAPING_THE_CORNa"></a><a href="#X_111_REAPING_THE_CORN"><b>111.&mdash;REAPING THE CORN.&mdash;<i>solution</i></b></a></p>
+
+<p>The whole field must have contained 46.626 square rods. The side of
+the central square, left by the farmer, is 4.8284 rods, so it contains
+23.313 square rods. The area of the field was thus something more than
+a quarter of an acre and less than one-third; to be more precise,
+.2914 of an acre.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_112_A_PUZZLING_LEGACYa" id="X_112_A_PUZZLING_LEGACYa"></a><a href="#X_112_A_PUZZLING_LEGACY"><b>112.&mdash;A PUZZLING LEGACY.&mdash;<i>solution</i></b></a></p>
+
+<p>As the share of Charles falls in through his death, we have merely to
+divide the whole hundred acres between Alfred and Benjamin in the
+proportion of one-third to one-fourth&mdash;that is in the proportion of
+four-twelfths to three-<span class='pagenum'>Pg 162<a name="Page_162" id="Page_162"></a></span>twelfths, which is the same as four to three.
+Therefore Alfred takes four-sevenths of the hundred acres and Benjamin
+three-sevenths.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_113_THE_TORN_NUMBERa" id="X_113_THE_TORN_NUMBERa"></a><a href="#X_113_THE_TORN_NUMBER"><b>113.&mdash;THE TORN NUMBER.&mdash;<i>solution</i></b></a></p>
+
+<p>The other number that answers all the requirements of the puzzle is
+9,801. If we divide this in the middle into two numbers and add them
+together we get 99, which, multiplied by itself, produces 9,801. It is
+true that 2,025 may be treated in the same way, only this number is
+excluded by the condition which requires that no two figures should be
+alike.</p>
+
+<p>The general solution is curious. Call the number of figures in each
+half of the torn label n. Then, if we add 1 to each of the exponents
+of the prime factors (other than 3) of 10<sup>n</sup>&nbsp;-&nbsp;1 (1 being regarded as a
+factor with the constant exponent, 1), their product will be the
+number of solutions. Thus, for a label of six figures, n&nbsp;=&nbsp;3. The
+factors of 10<sup>n</sup>&nbsp;-&nbsp;1 are 1<sup>1</sup>&nbsp;&times;&nbsp;37<sup>1</sup> (not considering the 3<sup>3</sup>), and the
+product of 2&nbsp;&times;&nbsp;2&nbsp;=&nbsp;4, the number of solutions. This always includes
+the special cases 98&nbsp;-&nbsp;01, 00&nbsp;-&nbsp;01, 998&nbsp;-&nbsp;01, 000&nbsp;-&nbsp;001, etc. The
+solutions are obtained as follows:&mdash;Factorize 10<sup>3</sup>&nbsp;-&nbsp;1 in all possible
+ways, always keeping the powers of 3 together, thus, 37&nbsp;&times;&nbsp;27, 999&nbsp;&times;&nbsp;1.
+Then solve the equation 37x&nbsp;=&nbsp;27y&nbsp;+&nbsp;1. Here x&nbsp;=&nbsp;19 and y&nbsp;=&nbsp;26.
+Therefore, 19&nbsp;&times;&nbsp;37&nbsp;=&nbsp;703, the square of which gives one label,
+494,209. A complementary solution (through 27x&nbsp;=&nbsp;37x&nbsp;+&nbsp;1) can at once
+be found by 10<sup>n</sup>&nbsp;-&nbsp;703&nbsp;=&nbsp;297, the square of which gives 088,209 for
+second label. (These non-significant noughts to the left must be
+included, though they lead to peculiar cases like 00238&nbsp;-&nbsp;04641 =
+4879<sup>2</sup>, where 0238&nbsp;-&nbsp;4641 would not work.) The special case 999&nbsp;&times;&nbsp;1 we
+can write at once 998,001, according to the law shown above, by adding
+nines on one half and noughts on the other, and its complementary will
+be 1 preceded by five noughts, or 000001. Thus we get the squares of
+999 and 1. These are the four solutions.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_114_CURIOUS_NUMBERSa" id="X_114_CURIOUS_NUMBERSa"></a><a href="#X_114_CURIOUS_NUMBERS"><b>114.&mdash;CURIOUS NUMBERS.&mdash;<i>solution</i></b></a></p>
+
+<p>The three smallest numbers, in addition to 48, are 1,680, 57,120, and
+1,940,448. It will be found that 1,681 and 841, 57,121 and 28,561,
+1,940,449 and 970,225, are respectively the squares of 41 and 29, 239
+and 169, 1,393 and 985.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_115_A_PRINTERS_ERRORa" id="X_115_A_PRINTERS_ERRORa"></a><a href="#X_115_A_PRINTERS_ERROR"><b>115.&mdash;A PRINTER'S ERROR.&mdash;<i>solution</i></b></a></p>
+
+<p>The answer is that 2<sup>5</sup>&nbsp;&times;&nbsp;9<sup>2</sup> is the same as 2592, and this is the only
+possible solution to the puzzle.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_116_THE_CONVERTED_MISERa" id="X_116_THE_CONVERTED_MISERa"></a><a href="#X_116_THE_CONVERTED_MISER"><b>116.&mdash;THE CONVERTED MISER.&mdash;<i>solution</i></b></a></p>
+
+<p>As we are not told in what year Mr. Jasper Bullyon made the generous
+distribution of his accumulated wealth, but are required to find the
+lowest possible amount of money, it is clear that we must look for a
+year of the most favourable form.</p>
+
+<p>There are four cases to be considered&mdash;an ordinary year with fifty-two
+Sundays and with fifty-three Sundays, and a leap-year with fifty-two
+and fifty-three Sundays respectively. Here are the lowest possible
+amounts in each case:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>313</td><td align='center'>weekdays,</td><td align='center'>52 Sundays</td><td align='center'>&pound;112,055</td></tr>
+<tr><td align='center'>312</td><td align='center'>weekdays,</td><td align='center'>53 Sundays</td><td align='center'>19,345</td></tr>
+<tr><td align='center'>314</td><td align='center'>weekdays,</td><td align='center'>52 Sundays</td><td align='center'>No solution possible.</td></tr>
+<tr><td align='center'>313</td><td align='center'>weekdays,</td><td align='center'>53 Sundays</td><td align='center'>&pound;69,174</td></tr>
+</table></div>
+
+<p>The lowest possible amount, and therefore the correct answer, is
+&pound;19,345, distributed in an ordinary year that began on a Sunday. The
+last year of this kind was 1911. He would have paid &pound;53 on every day
+of the year, or &pound;62 on every weekday, with &pound;1 left over, as required,
+in the latter event.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_117_A_FENCE_PROBLEMa" id="X_117_A_FENCE_PROBLEMa"></a><a href="#X_117_A_FENCE_PROBLEM"><b>117.&mdash;A FENCE PROBLEM.&mdash;<i>solution</i></b></a></p>
+
+<p>Though this puzzle presents no great difficulty to any one possessing
+a knowledge of algebra, it has perhaps rather interesting features.</p>
+
+<p>Seeing, as one does in the illustration, just one corner of the
+proposed square, one is scarcely prepared for the fact that the field,
+in order to comply with the conditions, must contain exactly 501,760
+acres, the fence requiring the same number of rails. Yet this is the
+correct answer, and the only answer, and if that gentleman in Iowa
+carries out his intention, his field will be twenty-eight miles long
+on each side, and a little larger than the county of Westmorland. I am
+not aware that any limit has ever been fixed to the size of a "field,"
+though they do not run so large as this in Great Britain. Still, out
+in Iowa, where my correspondent resides, they do these things on a
+very big scale. I have, however, reason to believe that when he finds
+the sort of task he has set himself, he will decide to abandon it; for
+if that cow decides to roam to fresh woods and pastures new, the
+milkmaid may have to start out a week in advance in order to obtain
+the morning's milk.</p>
+
+<p>Here is a little rule that will always apply where the length of the
+rail is half a pole. Multiply the number of rails in a hurdle by four,
+and the result is the exact number of miles in the side of a square
+field containing the same number of acres as there are rails in the
+complete fence. Thus, with a one-rail fence the field is four miles
+square; a two-rail fence gives eight miles square; a three-rail fence,
+twelve miles square; and so on, until we find that a seven-rail fence
+multiplied by four gives a field of twenty-eight miles square. In the
+case of our present problem, if the field be made smaller, then the
+number of rails will exceed the number of acres; while if the field be
+made larger, the number of rails will be less than the acres of the
+field.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_118_CIRCLING_THE_SQUARESa" id="X_118_CIRCLING_THE_SQUARESa"></a><a href="#X_118_CIRCLING_THE_SQUARES"><b>118.&mdash;CIRCLING THE SQUARES.&mdash;<i>solution</i></b></a></p>
+
+<p>Though this problem might strike the novice as being rather difficult,
+it is, as a matter of fact, quite easy, and is made still easier by
+inserting four out of the ten numbers.</p>
+
+<p><span class='pagenum'>Pg 163<a name="Page_163" id="Page_163"></a></span>First, it will be found that squares that are diametrically opposite
+have a common difference. For example, the difference between the
+square of 14 and the square of 2, in the diagram, is 192; and the
+difference between the square of 16 and the square of 8 is also 192.
+This must be so in every case. Then it should be remembered that the
+difference between squares of two consecutive numbers is always twice
+the smaller number plus 1, and that the difference between the squares
+of any two numbers can always be expressed as the difference of the
+numbers multiplied by their sum. Thus the square of 5 (25) less the
+square of 4 (16) equals (2&nbsp;&times;&nbsp;4)&nbsp;+&nbsp;1, or 9; also, the square of 7 (49)
+less the square of 3 (9) equals (7&nbsp;+&nbsp;3)&nbsp;&times;&nbsp;(7&nbsp;-&nbsp;3), or 40.</p>
+
+<p>Now, the number 192, referred to above, may be divided into five
+different pairs of even factors: 2&nbsp;&times;&nbsp;96, 4&nbsp;&times;&nbsp;48, 6&nbsp;&times;&nbsp;32, 8&nbsp;&times;&nbsp;24, and
+12&nbsp;&times;&nbsp;16, and these divided by 2 give us, 1&nbsp;&times;&nbsp;48, 2&nbsp;&times;&nbsp;24, 3&nbsp;&times;&nbsp;16, 4 &times;
+12, and 6&nbsp;&times;&nbsp;8. The difference and sum respectively of each of these
+pairs in turn produce 47, 49; 22, 26; 13, 19; 8, 16; and 2, 14. These
+are the required numbers, four of which are already placed. The six
+numbers that have to be added may be placed in just six different
+ways, one of which is as follows, reading round the circle clockwise:
+16, 2, 49, 22, 19, 8, 14, 47, 26, 13.</p>
+
+<p>I will just draw the reader's attention to one other little point. In
+all circles of this kind, the difference between diametrically
+opposite numbers increases by a certain ratio, the first numbers (with
+the exception of a circle of 6) being 4 and 6, and the others formed
+by doubling the next preceding but one. Thus, in the above case, the
+first difference is 2, and then the numbers increase by 4, 6, 8, and
+12. Of course, an infinite number of solutions may be found if we
+admit fractions. The number of squares in a circle of this kind must,
+however, be of the form 4n&nbsp;+&nbsp;6; that is, it must be a number composed
+of 6 plus a multiple of 4.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_119_RACKBRANES_LITTLE_LOSSa" id="X_119_RACKBRANES_LITTLE_LOSSa"></a><a href="#X_119_RACKBRANES_LITTLE_LOSS"><b>119.&mdash;RACKBRANE'S LITTLE LOSS.&mdash;<i>solution</i></b></a></p>
+
+<p>The professor must have started the game with thirteen shillings, Mr.
+Potts with four shillings, and Mrs. Potts with seven shillings.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_120_THE_FARMER_AND_HIS_SHEEPa" id="X_120_THE_FARMER_AND_HIS_SHEEPa"></a><a href="#X_120_THE_FARMER_AND_HIS_SHEEP"><b>120.&mdash;THE FARMER AND HIS SHEEP.&mdash;<i>solution</i></b></a></p>
+
+<p>The farmer had one sheep only! If he divided this sheep (which is best
+done by weight) into two parts, making one part two-thirds and the
+other part one-third, then the difference between these two numbers is
+the same as the difference between their squares&mdash;that is, one-third.
+Any two fractions will do if the denominator equals the sum of the two
+numerators.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_121_HEADS_OR_TAILSa" id="X_121_HEADS_OR_TAILSa"></a><a href="#X_121_HEADS_OR_TAILS"><b>121.&mdash;HEADS OR TAILS.&mdash;<i>solution</i></b></a></p>
+
+<p>Crooks must have lost, and the longer he went on the more he would
+lose. In two tosses he would be left with three-quarters of his money,
+in four tosses with nine-sixteenths of his money, in six tosses with
+twenty-seven sixty-fourths of his money, and so on. The order of the
+wins and losses makes no difference, so long as their number is in the
+end equal.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_122_THE_SEE-SAW_PUZZLEa" id="X_122_THE_SEE-SAW_PUZZLEa"></a><a href="#X_122_THE_SEE-SAW_PUZZLE"><b>122.&mdash;THE SEE-SAW PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The boy's weight must have been about 39.79 lbs. A brick weighed 3
+lbs. Therefore 16 bricks weighed 48 lbs. and 11 bricks 33 lbs.
+Multiply 48 by 33 and take the square root.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_123_A_LEGAL_DIFFICULTYa" id="X_123_A_LEGAL_DIFFICULTYa"></a><a href="#X_123_A_LEGAL_DIFFICULTY"><b>123.&mdash;A LEGAL DIFFICULTY.&mdash;<i>solution</i></b></a></p>
+
+<p>It was clearly the intention of the deceased to give the son twice as
+much as the mother, or the daughter half as much as the mother.
+Therefore the most equitable division would be that the mother should
+take two-sevenths, the son four-sevenths, and the daughter
+one-seventh.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_124_A_QUESTION_OF_DEFINITIONa" id="X_124_A_QUESTION_OF_DEFINITIONa"></a><a href="#X_124_A_QUESTION_OF_DEFINITION"><b>124.&mdash;A QUESTION OF DEFINITION.&mdash;<i>solution</i></b></a></p>
+
+<p>There is, of course, no difference in <i>area</i> between a mile square and
+a square mile. But there may be considerable difference in <i>shape</i>. A
+mile square can be no other shape than square; the expression
+describes a surface of a certain specific size and shape. A square
+mile may be of any shape; the expression names a unit of area, but
+does not prescribe any particular shape.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_125_THE_MINERS_HOLIDAYa" id="X_125_THE_MINERS_HOLIDAYa"></a><a href="#X_125_THE_MINERS_HOLIDAY"><b>125.&mdash;THE MINERS' HOLIDAY.&mdash;<i>solution</i></b></a></p>
+
+<p>Bill Harris must have spent thirteen shillings and sixpence, which
+would be three shillings more than the average for the seven men&mdash;half
+a guinea.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_126_SIMPLE_MULTIPLICATIONa" id="X_126_SIMPLE_MULTIPLICATIONa"></a><a href="#X_126_SIMPLE_MULTIPLICATION"><b>126.&mdash;SIMPLE MULTIPLICATION.&mdash;<i>solution</i></b></a></p>
+
+<p>The number required is 3,529,411,764,705,882, which may be multiplied
+by 3 and divided by 2, by the simple expedient of removing the 3 from
+one end of the row to the other. If you want a longer number, you can
+increase this one to any extent by repeating the sixteen figures in
+the same order.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_127_SIMPLE_DIVISIONa" id="X_127_SIMPLE_DIVISIONa"></a><a href="#X_127_SIMPLE_DIVISION"><b>127.&mdash;SIMPLE DIVISION.&mdash;<i>solution</i></b></a></p>
+
+<p>Subtract every number in turn from every other number, and we get 358
+(twice), 716, 1,611, 1,253, and 895. Now, we see at a glance that, as
+358 equals 2&nbsp;&times;&nbsp;179, the only number that can divide in every case
+without a remainder will be 179. On trial we find that this is such a
+divisor. Therefore, 179 is the divisor we want, which always leaves a
+remainder 164 in the case of the original numbers given.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_128_A_PROBLEM_IN_SQUARESa" id="X_128_A_PROBLEM_IN_SQUARESa"></a><a href="#X_128_A_PROBLEM_IN_SQUARES"><b>128.&mdash;A PROBLEM IN SQUARES.&mdash;<i>solution</i></b></a></p>
+
+<p>The sides of the three boards measure 31 in., 41 in., and 49 in. The
+common difference of area is exactly five square feet. Three numbers
+whose squares are in A.P., with a common difference of 7, are <sup>113</sup>/<sub>120</sub>,
+<sup>337</sup>/<sub>120</sub>, <sup>463</sup>/<sub>120</sub>; and with <span class='pagenum'>Pg 164<a name="Page_164" id="Page_164"></a></span>a common difference of 13 are <sup>80929</sup>/<sub>19380</sub>,
+<sup>106921</sup>/<sub>19380</sub>, and <sup>127729</sup>/<sub>19380</sub>. In the case of whole square numbers
+the common difference will always be divisible by 24, so it is obvious
+that our squares must be fractional. Readers should now try to solve
+the case where the common difference is 23. It is rather a hard nut.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_129_THE_BATTLE_OF_HASTINGSa" id="X_129_THE_BATTLE_OF_HASTINGSa"></a><a href="#X_129_THE_BATTLE_OF_HASTINGS"><b>129.&mdash;THE BATTLE OF HASTINGS.&mdash;<i>solution</i></b></a></p>
+
+<p>Any number (not itself a square number) may be multiplied by a square
+that will give a product 1 less than another square. The given number
+must not itself be a square, because a square multiplied by a square
+produces a square, and no square plus 1 can be a square. My remarks
+throughout must be understood to apply to whole numbers, because
+fractional soldiers are not of much use in war.</p>
+
+<p>Now, of all the numbers from 2 to 99 inclusive, 61 happens to be the
+most awkward one to work, and the lowest possible answer to our puzzle
+is that Harold's army consisted of 3,119,882,982,860,264,400 men. That
+is, there would be 51,145,622,669,840,400 men (the square of
+226,153,980) in each of the sixty-one squares. Add one man (Harold),
+and they could then form one large square with 1,766,319,049 men on
+every side. The general problem, of which this is a particular case,
+is known as the "Pellian Equation"&mdash;apparently because Pell neither
+first propounded the question nor first solved it! It was issued as a
+challenge by Fermat to the English mathematicians of his day. It is
+readily solved by the use of continued fractions.</p>
+
+<p>Next to 61, the most difficult number under 100 is 97, where
+97&nbsp;&times;&nbsp;6,377,352<sup>2</sup>&nbsp;+&nbsp;1&nbsp;=&nbsp;a square.</p>
+
+<p>The reason why I assumed that there must be something wrong with the
+figures in the chronicle is that we can confidently say that Harold's
+army did not contain over three trillion men! If this army (not to
+mention the Normans) had had the whole surface of the earth (sea
+included) on which to encamp, each man would have had slightly more
+than a quarter of a square inch of space in which to move about! Put
+another way: Allowing one square foot of standing-room per man, each
+small square would have required all the space allowed by a globe
+three times the diameter of the earth.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_130_THE_SCULPTORS_PROBLEMa" id="X_130_THE_SCULPTORS_PROBLEMa"></a><a href="#X_130_THE_SCULPTORS_PROBLEM"><b>130.&mdash;THE SCULPTOR'S PROBLEM.&mdash;<i>solution</i></b></a></p>
+
+<p>A little thought will make it clear that the answer must be
+fractional, and that in one case the numerator will be greater and in
+the other case less than the denominator. As a matter of fact, the
+height of the larger cube must be <sup>8</sup>/<sub>7</sub> ft., and of the smaller <sup>3</sup>/<sub>7</sub> ft.,
+if we are to have the answer in the smallest possible figures. Here
+the lineal measurement is <sup>11</sup>/<sub>7</sub> ft.&mdash;that is, 1<sup>4</sup>/<sub>7</sub> ft. What are the
+cubic contents of the two cubes? First <sup>8</sup>/<sub>7</sub>&nbsp;&times;&nbsp;<sup>3</sup>/<sub>7</sub>&nbsp;&times;&nbsp;<sup>8</sup>/<sub>7</sub>&nbsp;=&nbsp;<sup>512</sup>/<sub>343</sub>, and
+secondly <sup>3</sup>/<sub>7</sub>&nbsp;&times;&nbsp;<sup>3</sup>/<sub>7</sub>&nbsp;&times;&nbsp;<sup>3</sup>/<sub>7</sub>&nbsp;=&nbsp;<sup>27</sup>/<sub>343</sub>. Add these together and the result
+is <sup>539</sup>/<sub>343</sub>, which reduces to <sup>11</sup>/<sub>7</sub> or 1<sup>4</sup>/<sub>7</sub> ft. We thus see that the
+answers in cubic feet and lineal feet are precisely the same.</p>
+
+<p>The germ of the idea is to be found in the works of Diophantus of
+Alexandria, who wrote about the beginning of the fourth century. These
+fractional numbers appear in triads, and are obtained from three
+generators, <i>a</i>, <i>b</i>, <i>c</i>, where <i>a</i> is the largest and <i>c</i> the
+smallest.</p>
+
+<p>Then <i>ab</i>+<i>c</i><sup>2</sup>=denominator, and <i>a</i><sup>2</sup>-<i>c</i><sup>2</sup>, <i>b</i><sup>2</sup>-<i>c</i><sup>2</sup>,
+and <i>a</i><sup>2</sup>-<i>b</i><sup>2</sup> will be the three numerators. Thus, using the
+generators 3, 2, 1, we get <sup>8</sup>/<sub>7</sub>, <sup>3</sup>/<sub>7</sub>, <sup>5</sup>/<sub>7</sub> and we can pair the first and
+second, as in the above solution, or the first and third for a second
+solution. The denominator must always be a prime number of the form
+6<i>n</i>+1, or composed of such primes. Thus you can have 13, 19, etc., as
+denominators, but not 25, 55, 187, etc.</p>
+
+<p>When the principle is understood there is no difficulty in writing
+down the dimensions of as many sets of cubes as the most exacting
+collector may require. If the reader would like one, for example, with
+plenty of nines, perhaps the following would satisfy him:
+<sup>99999999</sup>/<sub>99990001</sub> and <sup>19999</sup>/<sub>99990001</sub>.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_131_THE_SPANISH_MISERa" id="X_131_THE_SPANISH_MISERa"></a><a href="#X_131_THE_SPANISH_MISER"><b>131.&mdash;THE SPANISH MISER.&mdash;<i>solution</i></b></a></p>
+
+<p>There must have been 386 doubloons in one box, 8,450 in another, and
+16,514 in the third, because 386 is the smallest number that can
+occur. If I had asked for the smallest aggregate number of coins, the
+answer would have been 482, 3,362, and 6,242. It will be found in
+either case that if the contents of any two of the three boxes be
+combined, they form a square number of coins. It is a curious
+coincidence (nothing more, for it will not always happen) that in the
+first solution the digits of the three numbers add to 17 in every
+case, and in the second solution to 14. It should be noted that the
+middle one of the three numbers will always be half a square.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_132_THE_NINE_TREASURE_BOXESa" id="X_132_THE_NINE_TREASURE_BOXESa"></a><a href="#X_132_THE_NINE_TREASURE_BOXES"><b>132.&mdash;THE NINE TREASURE BOXES.&mdash;<i>solution</i></b></a></p>
+
+<p>Here is the answer that fulfils the conditions:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='left'>A&nbsp;=&nbsp;4</td><td align='left'>B&nbsp;=&nbsp;3,364</td><td align='left'>C&nbsp;=&nbsp;6,724</td></tr>
+<tr><td align='left'>D&nbsp;=&nbsp;2,116</td><td align='left'>E&nbsp;=&nbsp;5,476</td><td align='left'>F&nbsp;=&nbsp;8,836</td></tr>
+<tr><td align='left'>G&nbsp;=&nbsp;9,409</td><td align='left'>H&nbsp;=&nbsp;12,769</td><td align='left'>I&nbsp;=&nbsp;16,129</td></tr>
+</table></div>
+
+
+<p>Each of these is a square number, the roots, taken in alphabetical
+order, being 2, 58, 82, 46, 74, 94, 97, 113, and 127, while the
+required difference between A and B, B and C, D and E. etc., is in
+every case 3,360.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_133_THE_FIVE_BRIGANDSa" id="X_133_THE_FIVE_BRIGANDSa"></a><a href="#X_133_THE_FIVE_BRIGANDS"><b>133.&mdash;THE FIVE BRIGANDS.&mdash;<i>solution</i></b></a></p>
+
+<p>The sum of 200 doubloons might have been held by the five brigands in
+any one of 6,627 different ways. Alfonso may have held any number from
+1 to 11. If he held 1 doubloon, there are 1,005 different ways of
+distributing the remainder; if he held 2, there are 985 ways; if 3,
+there are 977 ways; if 4, there are 903 ways; if 5 doubloons, 832
+ways; if 6 doubloons, 704 ways; if 7 doubloons, 570 ways; if 8
+doubloons, 388 ways; if 9 doubloons, 200 ways; if 10 doubloons, 60
+ways; and if Alfonso held 11 doubloons, the remainder could be
+distri<span class='pagenum'>Pg 165<a name="Page_165" id="Page_165"></a></span>buted in 3 different ways. More than 11 doubloons he could not
+possibly have had. It will scarcely be expected that I shall give all
+these 6,627 ways at length. What I propose to do is to enable the
+reader, if he should feel so disposed, to write out all the answers
+where Alfonso has one and the same amount. Let us take the cases where
+Alfonso has 6 doubloons, and see how we may obtain all the 704
+different ways indicated above. Here are two tables that will serve as
+keys to all these answers:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' colspan='3'>Table I.</td></tr>
+<tr><td align='center'>A</td><td align='center'>=</td><td align='left'>6.</td></tr>
+<tr><td align='center'>B</td><td align='center'>=</td><td align='left'>n.</td></tr>
+<tr><td align='center'>C</td><td align='center'>=</td><td align='left'>(63&nbsp;-&nbsp;5n)&nbsp;+&nbsp;m.</td></tr>
+<tr><td align='center'>D</td><td align='center'>=</td><td align='left'>(128&nbsp;+&nbsp;4n)&nbsp;-&nbsp;4m.</td></tr>
+<tr><td align='center'>E</td><td align='center'>=</td><td align='left'>3&nbsp;+&nbsp;3m.</td></tr>
+</table><br /><br /></div>
+
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' colspan='3'>Table II.</td></tr>
+<tr><td align='center'>A</td><td align='center'>=</td><td align='left'>6.</td></tr>
+<tr><td align='center'>B</td><td align='center'>=</td><td align='left'>n.</td></tr>
+<tr><td align='center'>C</td><td align='center'>=</td><td align='left'>1&nbsp;+&nbsp;m.</td></tr>
+<tr><td align='center'>D</td><td align='center'>=</td><td align='left'>(376&nbsp;-&nbsp;16n)&nbsp;-&nbsp;4m.</td></tr>
+<tr><td align='center'>E</td><td align='center'>=</td><td align='left'>(15n&nbsp;-&nbsp;183)&nbsp;+&nbsp;3m.</td></tr>
+</table></div>
+
+<p>In the first table we may substitute for n any whole number from 1 to
+12 inclusive, and m may be nought or any whole number from 1 to (31 +
+n) inclusive. In the second table n may have the value of any whole
+number from 13 to 23 inclusive, and m may be nought or any whole
+number from 1 to (93&nbsp;-&nbsp;4n) inclusive. The first table thus gives (32 +
+n) answers for every value of n; and the second table gives (94&nbsp;-&nbsp;4n)
+answers for every value of n. The former, therefore, produces 462 and
+the latter 242 answers, which together make 704, as already stated.</p>
+
+<p>Let us take Table I., and say n&nbsp;=&nbsp;5 and m&nbsp;=&nbsp;2; also in Table II. take
+n&nbsp;=&nbsp;13 and m&nbsp;=&nbsp;0. Then we at once get these two answers:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' colspan='3'>Table I.</td></tr>
+<tr><td align='center'>A</td><td align='center'>=</td><td align='right'>6</td></tr>
+<tr><td align='center'>B</td><td align='center'>=</td><td align='right'>5</td></tr>
+<tr><td align='center'>C</td><td align='center'>=</td><td align='right'>40</td></tr>
+<tr><td align='center'>D</td><td align='center'>=</td><td align='right'>140</td></tr>
+<tr><td align='center'>E</td><td align='center'>=</td><td align='right' class='bb'>9</td></tr>
+<tr><td align='center'></td><td></td><td align='right'>200 <br />doubloons</td></tr>
+</table><br /><br /></div>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' colspan='3'>Table II.</td></tr>
+<tr><td align='center'>A</td><td align='center'>=</td><td align='right'>6</td></tr>
+<tr><td align='center'>B</td><td align='center'>=</td><td align='right'>13</td></tr>
+<tr><td align='center'>C</td><td align='center'>=</td><td align='right'>1</td></tr>
+<tr><td align='center'>D</td><td align='center'>=</td><td align='right'>168</td></tr>
+<tr><td align='center'>E</td><td align='center'>=</td><td align='right' class='bb'>12</td></tr>
+<tr><td align='center'></td><td></td><td align='right'>200 <br />doubloons.</td></tr>
+</table></div>
+
+<p>These will be found to work correctly. All the rest of the 704
+answers, where Alfonso always holds six doubloons, may be obtained in
+this way from the two tables by substituting the different numbers for
+the letters m and n.</p>
+
+<p>Put in another way, for every holding of Alfonso the number of answers
+is the sum of two arithmetical progressions, the common difference in
+one case being 1 and in the other -4. Thus in the case where Alfonso
+holds 6 doubloons one progression is 33&nbsp;+&nbsp;34&nbsp;+&nbsp;35&nbsp;+&nbsp;36&nbsp;+&nbsp;...&nbsp;+&nbsp;43 +
+44, and the other 42&nbsp;+&nbsp;38&nbsp;+&nbsp;34&nbsp;+&nbsp;30&nbsp;+&nbsp;...&nbsp;+&nbsp;6&nbsp;+&nbsp;2. The sum of the
+first series is 462, and of the second 242&mdash;results which again agree
+with the figures already given. The problem may be said to consist in
+finding the first and last terms of these progressions. I should
+remark that where Alfonso holds 9, 10, or 11 there is only one
+progression, of the second form.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_134_THE_BANKERS_PUZZLEa" id="X_134_THE_BANKERS_PUZZLEa"></a><a href="#X_134_THE_BANKERS_PUZZLE"><b>134.&mdash;THE BANKER'S PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>In order that a number of sixpences may not be divisible into a number
+of equal piles, it is necessary that the number should be a prime. If
+the banker can bring about a prime number, he will win; and I will
+show how he can always do this, whatever the customer may put in the
+box, and that therefore the banker will win to a certainty. The banker
+must first deposit forty sixpences, and then, no matter how many the
+customer may add, he will desire the latter to transfer from the
+counter the square of the number next below what the customer put in.
+Thus, banker puts 40, customer, we will say, adds 6, then transfers
+from the counter 25 (the square of 5), which leaves 71 in all, a prime
+number. Try again. Banker puts 40, customer adds 12, then transfers
+121 (the square of 11), as desired, which leaves 173, a prime number.
+The key to the puzzle is the curious fact that any number up to 39, if
+added to its square and the sum increased by 41, makes a prime number.
+This was first discovered by Euler, the great mathematician. It has
+been suggested that the banker might desire the customer to transfer
+sufficient to raise the contents of the box to a given number; but
+this would not only make the thing an absurdity, but breaks the rule
+that neither knows what the other puts in.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_135_THE_STONEMASONS_PROBLEMa" id="X_135_THE_STONEMASONS_PROBLEMa"></a><a href="#X_135_THE_STONEMASONS_PROBLEM"><b>135.&mdash;THE STONEMASON'S PROBLEM.&mdash;<i>solution</i></b></a></p>
+
+<p>The puzzle amounts to this. Find the smallest square number that may
+be expressed as the sum of more than three consecutive cubes, the cube
+1 being barred. As more than three heaps were to be supplied, this
+condition shuts out the otherwise smallest answer, 23<sup>3</sup>&nbsp;+&nbsp;24<sup>3</sup>&nbsp;+&nbsp;25<sup>3</sup>
+= 204<sup>2</sup>. But it admits the answer, 25<sup>3</sup>&nbsp;+&nbsp;26<sup>3</sup>&nbsp;+&nbsp;27<sup>3</sup>&nbsp;+&nbsp;28<sup>3</sup>&nbsp;+&nbsp;29<sup>3</sup> =
+315<sup>2</sup>. The correct answer, however, requires more heaps, but a smaller
+aggregate number of blocks. Here it is: 14<sup>3</sup>&nbsp;+&nbsp;15<sup>3</sup>&nbsp;+&nbsp;... up to 25<sup>3</sup>
+inclusive, or twelve heaps in all, which, added together, make 97,344
+blocks of stone that may be laid out to form a square 312&nbsp;&times;&nbsp;312. I
+will just remark that one key to the solution lies in what are called
+triangular numbers. (See pp. <a href="#Page_13">13</a>, <a href="#Page_25">25</a>, and <a href="#Page_166">166</a>.)</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_136_THE_SULTANS_ARMYa" id="X_136_THE_SULTANS_ARMYa"></a><a href="#X_136_THE_SULTANS_ARMY"><b>136.&mdash;THE SULTAN'S ARMY.&mdash;<i>solution</i></b></a></p>
+
+<p>The smallest primes of the form 4n&nbsp;+&nbsp;1 are 5, 13, 17, 29, and 37, and
+the smallest of the form 4n&nbsp;-&nbsp;1 are 3, 7, 11, 19, and 23. Now, primes
+of the first form can always be expressed as the sum of two squares,
+and in only one way. Thus, 5&nbsp;=&nbsp;4&nbsp;+&nbsp;1; 13&nbsp;=&nbsp;9&nbsp;+&nbsp;4; 17&nbsp;=&nbsp;16&nbsp;+&nbsp;1; 29&nbsp;=&nbsp;25
++ 4; 37&nbsp;=&nbsp;36&nbsp;+&nbsp;1. But primes of the second form can never be
+expressed as the sum of two squares in any way whatever.</p>
+
+<p>In order that a number may be expressed as the sum of two squares in
+several different ways, it is necessary that it shall be a composite
+number containing a certain number of primes of our first form. Thus,
+5 or 13 alone can only be so expressed in one way; but 65, (5&nbsp;&times;&nbsp;13),
+can be expressed in two ways, 1,105, (5&nbsp;&times;&nbsp;13&nbsp;&times;&nbsp;17), in four ways,
+32,045, (5&nbsp;&times;&nbsp;13&nbsp;&times;&nbsp;17&nbsp;&times;&nbsp;29), in eight ways. We thus get double as many
+ways for every new factor of this form that we introduce. Note,
+however, that I say <i>new</i> <span class='pagenum'>Pg 166<a name="Page_166" id="Page_166"></a></span>factor, for the <i>repetition</i> of factors is
+subject to another law. We cannot express 25, (5&nbsp;&times;&nbsp;5), in two ways,
+but only in one; yet 125, (5&nbsp;&times;&nbsp;5&nbsp;&times;&nbsp;5), can be given in two ways, and
+so can 625, (5&nbsp;&times;&nbsp;5&nbsp;&times;&nbsp;5&nbsp;&times;&nbsp;5); while if we take in yet another 5 we can
+express the number as the sum of two squares in three different ways.</p>
+
+<p>If a prime of the second form gets into your composite number, then
+that number cannot be the sum of two squares. Thus 15, (3&nbsp;&times;&nbsp;5), will
+not work, nor will 135, (3&nbsp;&times;&nbsp;3&nbsp;&times;&nbsp;3&nbsp;&times;&nbsp;5); but if we take in an even
+number of 3's it will work, because these 3's will themselves form a
+square number, but you will only get one solution. Thus, 45, (3&nbsp;&times;&nbsp;3 &times;
+5, or 9&nbsp;&times;&nbsp;5)&nbsp;=&nbsp;36&nbsp;+&nbsp;9. Similarly, the factor 2 may always occur, or
+any power of 2, such as 4, 8, 16, 32; but its introduction or omission
+will never affect the number of your solutions, except in such a case
+as 50, where it doubles a square and therefore gives you the two
+answers, 49&nbsp;+&nbsp;1 and 25&nbsp;+&nbsp;25.</p>
+
+<p>Now, directly a number is decomposed into its prime factors, it is
+possible to tell at a glance whether or not it can be split into two
+squares; and if it can be, the process of discovery in how many ways
+is so simple that it can be done in the head without any effort. The
+number I gave was 130. I at once saw that this was 2&nbsp;&times;&nbsp;5&nbsp;&times;&nbsp;13, and
+consequently that, as 65 can be expressed in two ways (64&nbsp;+&nbsp;1 and 49 +
+16), 130 can also be expressed in two ways, the factor 2 not affecting
+the question.</p>
+
+<p>The smallest number that can be expressed as the sum of two squares in
+twelve different ways is 160,225, and this is therefore the smallest
+army that would answer the Sultan's purpose. The number is composed of
+the factors 5&nbsp;&times;&nbsp;5&nbsp;&times;&nbsp;13&nbsp;&times;&nbsp;17&nbsp;&times;&nbsp;29, each of which is of the required
+form. If they were all different factors, there would be sixteen ways;
+but as one of the factors is repeated, there are just twelve ways.
+Here are the sides of the twelve pairs of squares: (400 and 15), (399
+and 32), (393 and 76), (392 and 81), (384 and 113), (375 and 140),
+(360 and 175), (356 and 183), (337 and 216), (329 and 228), (311 and
+252), (265 and 300). Square the two numbers in each pair, add them
+together, and their sum will in every case be 160,225.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_137_A_STUDY_IN_THRIFTa" id="X_137_A_STUDY_IN_THRIFTa"></a><a href="#X_137_A_STUDY_IN_THRIFT"><b>137.&mdash;A STUDY IN THRIFT.&mdash;<i>solution</i></b></a></p>
+
+<p>Mrs. Sandy McAllister will have to save a tremendous sum out of her
+housekeeping allowance if she is to win that sixth present that her
+canny husband promised her. And the allowance must be a very liberal
+one if it is to admit of such savings. The problem required that we
+should find five numbers higher than 36 the units of which may be
+displayed so as to form a square, a triangle, two triangles, and three
+triangles, using the complete number in every one of the four cases.</p>
+
+<p>Every triangular number is such that if we multiply it by 8 and add 1
+the result is an odd square number. For example, multiply 1, 3, 6, 10,
+15 respectively by 8 and add 1, and we get 9, 25, 49, 81, 121, which
+are the squares of the odd numbers 3, 5, 7, 9, 11. Therefore in every
+case where 8x<sup>2</sup>&nbsp;+&nbsp;1&nbsp;=&nbsp;a square number, x<sup>2</sup> is also a triangular. This
+point is dealt with in our puzzle, "The Battle of Hastings." I will
+now merely show again how, when the first solution is found, the
+others may be discovered without any difficulty. First of all, here
+are the figures:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>8</td><td align='center'>&times;</td><td align='right'>1<sup>2</sup></td><td align='center'>+ 1 =</td><td align='right'>3<sup>2</sup></td></tr>
+<tr><td align='center'>8</td><td align='center'>&times;</td><td align='right'>6<sup>2</sup></td><td align='center'>+ 1 =</td><td align='right'>17<sup>2</sup></td></tr>
+<tr><td align='center'>8</td><td align='center'>&times;</td><td align='right'>35<sup>2</sup></td><td align='center'>+ 1 =</td><td align='right'>99<sup>2</sup></td></tr>
+<tr><td align='center'>8</td><td align='center'>&times;</td><td align='right'>204<sup>2</sup></td><td align='center'>+ 1 =</td><td align='right'>577<sup>2</sup></td></tr>
+<tr><td align='center'>8</td><td align='center'>&times;</td><td align='right'>1189<sup>2</sup></td><td align='center'>+ 1 =</td><td align='right'>3363<sup>2</sup></td></tr>
+<tr><td align='center'>8</td><td align='center'>&times;</td><td align='right'>6930<sup>2</sup></td><td align='center'>+ 1 =</td><td align='right'>19601<sup>2</sup></td></tr>
+<tr><td align='center'>8</td><td align='center'>&times;</td><td align='right'>40391<sup>2</sup></td><td align='center'>+ 1 =</td><td align='right'>114243<sup>2</sup></td></tr>
+</table></div>
+
+<p>The successive pairs of numbers are found in this way:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>(1&nbsp;&times;&nbsp;3)&nbsp;+&nbsp;(3&nbsp;&times;&nbsp;1)</td><td align='center'>=</td><td align='right'>6</td><td>&nbsp;</td><td align='center'>(8&nbsp;&times;&nbsp;1)&nbsp;+&nbsp;(3&nbsp;&times;&nbsp;3)</td><td align='right'>=</td><td align='right'>17</td></tr>
+<tr><td align='right'>(1&nbsp;&times;&nbsp;17)&nbsp;+&nbsp;(3&nbsp;&times;&nbsp;6)</td><td align='center'>=</td><td align='right'>35</td><td>&nbsp;</td><td align='center'>(8&nbsp;&times;&nbsp;6)&nbsp;+&nbsp;(3&nbsp;&times;&nbsp;17)</td><td align='right'>=</td><td align='right'>99</td></tr>
+<tr><td align='right'>(1&nbsp;&times;&nbsp;99)&nbsp;+&nbsp;(3&nbsp;&times;&nbsp;35)</td><td align='center'>=</td><td align='right'>204</td><td>&nbsp;</td><td align='center'>(8&nbsp;&times;&nbsp;35)&nbsp;+&nbsp;(3&nbsp;&times;&nbsp;99)</td><td align='right'>=</td><td align='right'>577</td></tr>
+</table></div>
+
+<p>and so on. Look for the numbers in the table above, and the method
+will explain itself.</p>
+
+<p>Thus we find that the numbers 36, 1225, 41616, 1413721, 48024900, and
+1631432881 will form squares with sides of 6, 35, 204, 1189, 6930, and
+40391; and they will also form single triangles with sides of 8, 49,
+288, 1681, 9800, and 57121. These numbers may be obtained from the
+last column in the first table above in this way: simply divide the
+numbers by 2 and reject the remainder. Thus the integral halves of 17,
+99, and 577 are 8, 49, and 288.</p>
+
+<p>All the numbers we have found will form either two or three triangles
+at will. The following little diagram will show you graphically at a
+glance that every square number must necessarily be the sum of two
+triangulars, and that the side of one triangle will be the same as the
+side of the corresponding square, while the other will be just 1 less.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a137.png" width="400" height="206" alt="" title="" />
+</div>
+
+<p>Thus a square may always be divided easily into two triangles, and the
+sum of two consecutive triangulars will always make a square. In
+numbers it is equally clear, for if we examine the first
+triangulars&mdash;1, 3, 6, 10, 15, 21, 28&mdash;we find that by adding all the
+consecutive pairs in turn we get the series of square numbers&mdash;9, 16,
+25, 36, 49, etc.</p>
+
+<p>The method of forming three triangles from our numbers is equally
+direct, and not at all a matter of trial. But I must content myself
+with giving actual figures, and just stating that every triangular
+higher than 6 will form three triangulars. I give the sides of the
+triangles, and readers will know from my remarks when stat<span class='pagenum'>Pg 167<a name="Page_167" id="Page_167"></a></span>ing the
+puzzle how to find from these sides the number of counters or coins in
+each, and so check the results if they so wish.</p>
+
+<div class='center'>
+<table border="1" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>Number</td><td align='right'>Side of <br />Square.</td><td align='center'>Side of <br />Triangle.</td><td align='center'>Sides of <br />Two Triangles.</td><td align='center'>Sides of <br />Three Triangles.</td></tr>
+<tr><td align='right'>36</td><td align='right'>6</td><td align='center'>8</td><td align='center'>6&nbsp;+&nbsp;5</td><td align='center'>5&nbsp;+&nbsp;5&nbsp;+&nbsp;3</td></tr>
+<tr><td align='right'>1225</td><td align='right'>35</td><td align='center'>49</td><td align='center'>36&nbsp;+&nbsp;34</td><td align='center'>33&nbsp;+&nbsp;32&nbsp;+&nbsp;16</td></tr>
+<tr><td align='right'>41616</td><td align='right'>204</td><td align='center'>288</td><td align='center'>204&nbsp;+&nbsp;203</td><td align='center'>192&nbsp;+&nbsp;192&nbsp;+&nbsp;95</td></tr>
+<tr><td align='right'>1413721</td><td align='right'>1189</td><td align='center'>1681</td><td align='center'>1189&nbsp;+&nbsp;1188</td><td align='center'>1121&nbsp;+&nbsp;1120&nbsp;+&nbsp;560</td></tr>
+<tr><td align='right'>48024900</td><td align='right'>6930</td><td align='center'>9800</td><td align='center'>6930&nbsp;+&nbsp;6929</td><td align='center'>6533&nbsp;+&nbsp;6533&nbsp;+&nbsp;3267</td></tr>
+<tr><td align='right'>1631432881</td><td align='right'>40391</td><td align='center'>57121</td><td align='center'>40391&nbsp;+&nbsp;40390</td><td align='center'>38081&nbsp;+&nbsp;38080&nbsp;+&nbsp;19040</td></tr>
+</table></div>
+
+<p>I should perhaps explain that the arrangements given in the last two
+columns are not the only ways of forming two and three triangles.
+There are others, but one set of figures will fully serve our purpose.
+We thus see that before Mrs. McAllister can claim her sixth &pound;5 present
+she must save the respectable sum of &pound;1,631,432,881.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_138_THE_ARTILLERYMENS_DILEMMAa" id="X_138_THE_ARTILLERYMENS_DILEMMAa"></a><a href="#X_138_THE_ARTILLERYMENS_DILEMMA"><b>138.&mdash;THE ARTILLERYMEN'S DILEMMA.&mdash;<i>solution</i></b></a></p>
+
+<p>We were required to find the smallest number of cannon balls that we
+could lay on the ground to form a perfect square, and could pile into
+a square pyramid. I will try to make the matter clear to the merest
+novice.</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>1</td><td align='right'>2</td><td align='right'>3</td><td align='right'>4</td><td align='right'>5</td><td align='right'>6</td><td align='right'>7</td></tr>
+<tr><td align='right'>1</td><td align='right'>3</td><td align='right'>6</td><td align='right'>10</td><td align='right'>15</td><td align='right'>21</td><td align='right'>28</td></tr>
+<tr><td align='right'>1</td><td align='right'>4</td><td align='right'>10</td><td align='right'>20</td><td align='right'>35</td><td align='right'>56</td><td align='right'>84</td></tr>
+<tr><td align='right'>1</td><td align='right'>5</td><td align='right'>14</td><td align='right'>30</td><td align='right'>55</td><td align='right'>91</td><td align='right'>140</td></tr>
+</table></div>
+
+<p>Here in the first row we place in regular order the natural numbers.
+Each number in the second row represents the sum of the numbers in the
+row above, from the beginning to the number just over it. Thus 1, 2,
+3, 4, added together, make 10. The third row is formed in exactly the
+same way as the second. In the fourth row every number is formed by
+adding together the number just above it and the preceding number.
+Thus 4 and 10 make 14, 20 and 35 make 55. Now, all the numbers in the
+second row are triangular numbers, which means that these numbers of
+cannon balls may be laid out on the ground so as to form equilateral
+triangles. The numbers in the third row will all form our triangular
+pyramids, while the numbers in the fourth row will all form square
+pyramids.</p>
+
+<p>Thus the very process of forming the above numbers shows us that every
+square pyramid is the sum of two triangular pyramids, one of which has
+the same number of balls in the side at the base, and the other one
+ball fewer. If we continue the above table to twenty-four places, we
+shall reach the number 4,900 in the fourth row. As this number is the
+square of 70, we can lay out the balls in a square, and can form a
+square pyramid with them. This manner of writing out the series until
+we come to a square number does not appeal to the mathematical mind,
+but it serves to show how the answer to the particular puzzle may be
+easily arrived at by anybody. As a matter of fact, I confess my
+failure to discover any number other than 4,900 that fulfils the
+conditions, nor have I found any rigid proof that this is the only
+answer. The problem is a difficult one, and the second answer, if it
+exists (which I do not believe), certainly runs into big figures.</p>
+
+<p>For the benefit of more advanced mathematicians I will add that the
+general expression for square pyramid numbers is (2n<sup>3</sup>&nbsp;+&nbsp;3n<sup>2</sup>&nbsp;+&nbsp;n)/6.
+For this expression to be also a square number (the special case of 1
+excepted) it is necessary that n&nbsp;=&nbsp;p<sup>2</sup>&nbsp;-&nbsp;1&nbsp;=&nbsp;6t<sup>2</sup>, where 2p<sup>2</sup>&nbsp;-&nbsp;1 =
+q<sup>2</sup> (the "Pellian Equation"). In the case of our solution above, n =
+24, p&nbsp;=&nbsp;5, t&nbsp;=&nbsp;2, q&nbsp;=&nbsp;7.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_139_THE_DUTCHMENS_WIVESa" id="X_139_THE_DUTCHMENS_WIVESa"></a><a href="#X_139_THE_DUTCHMENS_WIVES"><b>139.&mdash;THE DUTCHMEN'S WIVES.&mdash;<i>solution</i></b></a></p>
+
+<p>The money paid in every case was a square number of shillings, because
+they bought 1 at 1<i>s</i>., 2 at 2<i>s</i>., 3 at 3<i>s</i>., and so on. But every husband
+pays altogether 63<i>s</i>. more than his wife, so we have to find in how
+many ways 63 may be the difference between two square numbers. These
+are the three only possible ways: the square of 8 less the square of
+1, the square of 12 less the square of 9, and the square of 32 less
+the square of 31. Here 1, 9, and 31 represent the number of pigs
+bought and the number of shillings per pig paid by each woman, and 8,
+12, and 32 the same in the case of their respective husbands. From the
+further information given as to their purchases, we can now pair them
+off as follows: Cornelius and Gurtr&uuml;n bought 8 and 1; Elas and Katr&uuml;n
+bought 12 and 9; Hendrick and Anna bought 32 and 31. And these pairs
+represent correctly the three married couples.</p>
+
+<p>The reader may here desire to know how we may determine the maximum
+number of ways in which a number may be expressed as the difference
+between two squares, and how we are to find the actual squares. Any
+integer except 1, 4, and twice any odd number, may be expressed as the
+difference of two integral squares in as many ways as it can be split
+up into pairs of factors, counting 1 as a factor. Suppose the number
+to be 5,940. The factors are <span class='pagenum'>Pg 168<a name="Page_168" id="Page_168"></a></span>2<sup>2</sup>.3<sup>3</sup>.5.11. Here the exponents are 2,
+3, 1, 1. Always deduct 1 from the exponents of 2 and add 1 to all the
+other exponents; then we get 1, 4, 2, 2, and half the product of these
+four numbers will be the required number of ways in which 5,940 may be
+the difference of two squares&mdash;that is, 8. To find these eight
+squares, as it is an <i>even</i> number, we first divide by 4 and get 1485,
+the eight pairs of factors of which are 1&nbsp;&times;&nbsp;1485, 3&nbsp;&times;&nbsp;495, 5&nbsp;&times;&nbsp;297, 9
+&times; 165, 11&nbsp;&times;&nbsp;135, 15&nbsp;&times;&nbsp;99, 27&nbsp;&times;&nbsp;55, and 33&nbsp;&times;&nbsp;45. The sum and difference
+of any one of these pairs will give the required numbers. Thus, the
+square of 1,486 less the square of 1,484 is 5,940, the square of 498
+less the square of 492 is the same, and so on. In the case of 63
+above, the number is <i>odd</i>; so we factorize at once, 1&nbsp;&times;&nbsp;63, 3&nbsp;&times;&nbsp;21, 7
+&times; 9. Then we find that <i>half</i> the sum and difference will give us the
+numbers 32 and 31, 12 and 9, and 8 and 1, as shown in the solution to
+the puzzle.</p>
+
+<p>The reverse problem, to find the factors of a number when you have
+expressed it as the difference of two squares, is obvious. For
+example, the sum and difference of any pair of numbers in the last
+sentence will give us the factors of 63. Every prime number (except 1
+and 2) may be expressed as the difference of two squares in one way,
+and in one way only. If a number can be expressed as the difference of
+two squares in more than one way, it is composite; and having so
+expressed it, we may at once obtain the factors, as we have seen.
+Fermat showed in a letter to Mersenne or Fr&eacute;nicle, in 1643, how we may
+discover whether a number may be expressed as the difference of two
+squares in more than one way, or proved to be a prime. But the method,
+when dealing with large numbers, is necessarily tedious, though in
+practice it may be considerably shortened. In many cases it is the
+shortest method known for factorizing large numbers, and I have always
+held the opinion that Fermat used it in performing a certain feat in
+factorizing that is historical and wrapped in mystery.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_140_FIND_ADAS_SURNAMEa" id="X_140_FIND_ADAS_SURNAMEa"></a><a href="#X_140_FIND_ADAS_SURNAME"><b>140.&mdash;FIND ADA'S SURNAME.&mdash;<i>solution</i></b></a></p>
+
+<p>The girls' names were Ada Smith, Annie Brown, Emily Jones, Mary
+Robinson, and Bessie Evans.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_141_SATURDAY_MARKETINGa" id="X_141_SATURDAY_MARKETINGa"></a><a href="#X_141_SATURDAY_MARKETING"><b>141.&mdash;SATURDAY MARKETING.&mdash;<i>solution</i></b></a></p>
+
+<p>As every person's purchase was of the value of an exact number of
+shillings, and as the party possessed when they started out forty
+shilling coins altogether, there was no necessity for any lady to have
+any smaller change, or any evidence that they actually had such
+change. This being so, the only answer possible is that the women were
+named respectively Anne Jones, Mary Robinson, Jane Smith, and Kate
+Brown. It will now be found that there would be exactly eight
+shillings left, which may be divided equally among the eight persons
+in coin without any change being required.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_142_THE_SILK_PATCHWORKa" id="X_142_THE_SILK_PATCHWORKa"></a><a href="#X_142_THE_SILK_PATCHWORK"><b>142.&mdash;THE SILK PATCHWORK.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a142.png" width="400" height="397" alt="" title="" />
+</div>
+
+<p>Our illustration will show how to cut the stitches of the patchwork so
+as to get the square F entire, and four equal pieces, G, H, I, K, that
+will form a perfect Greek cross. The reader will know how to assemble
+these four pieces from Fig. 13 in the article.</p>
+
+
+<hr style="width: 30%;" />
+<p><a name="X_143_TWO_CROSSES_FROM_ONEa" id="X_143_TWO_CROSSES_FROM_ONEa"></a><a href="#X_143_TWO_CROSSES_FROM_ONE"><b>143.&mdash;TWO CROSSES FROM ONE.&mdash;<i>solution</i></b></a></p>
+
+<p>It will be seen that one cross is cut out entire, as A in Fig. 1,
+while the four pieces marked <span class='pagenum'>Pg 169<a name="Page_169" id="Page_169"></a></span>B, C, D and E form the second cross, as
+in Fig. 2, which will be of exactly the same size as the other. I will
+leave the reader the pleasant task of discovering for himself the best
+way of finding the direction of the cuts. Note that the Swastika again
+appears.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a143.png" width="600" height="335" alt="" title="" />
+</div>
+
+<p>The difficult question now presents itself: How are we to cut three
+Greek crosses from one in the fewest possible pieces? As a matter of
+fact, this problem may be solved in as few as thirteen pieces; but as
+I know many of my readers, advanced geometricians, will be glad to
+have something to work on of which they are not shown the solution, I
+leave the mystery for the present undisclosed.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_144_THE_CROSS_AND_THE_TRIANGLEa" id="X_144_THE_CROSS_AND_THE_TRIANGLEa"></a><a href="#X_144_THE_CROSS_AND_THE_TRIANGLE"><b>144.&mdash;THE CROSS AND THE TRIANGLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The line A B in the following diagram represents the side of a square
+having the same area as the cross. I have shown elsewhere, as stated,
+how to make a square and equilateral triangle of equal area. I need
+not go, therefore, into the preliminary question of finding the
+dimensions of the triangle that is to equal our cross. We will assume
+that we have already found this, and the question then becomes, How
+are we to cut up one of these into pieces that will form the other?</p>
+
+<p>First draw the line A B where A and B are midway between the
+extremities of the two side arms. Next make the lines D C and E F
+equal in length to half the side of the triangle. Now from E and F
+describe with the same radius the intersecting arcs at G and draw F G.
+Finally make I K equal to H C and L B equal to A D. If we now draw I
+L, it should be parallel to F G, and all the six pieces are marked
+out. These fit together and form a perfect equilateral triangle, as
+shown in the second diagram. Or we might have first found the
+direction of the line M N in our triangle, then placed the point O
+over the point E in the cross and turned round the triangle over the
+cross until the line M N was parallel to A B. The piece 5 can then be
+marked off and the other pieces in succession.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a144.png" width="600" height="331" alt="" title="" />
+</div>
+
+<p>I have seen many attempts at a solution involving the assumption that
+the height of the triangle is exactly the same as the height of the
+cross. This is a fallacy: the cross will always be higher than the
+triangle of equal area.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_145_THE_FOLDED_CROSSa" id="X_145_THE_FOLDED_CROSSa"></a><a href="#X_145_THE_FOLDED_CROSS"><b>145.&mdash;THE FOLDED CROSS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a145a.png" width="400" height="264" alt="" title="" />
+</div>
+
+<p>First fold the cross along the dotted line A B in Fig. 1. You then
+have it in the form shown in Fig. 2. Next fold it along the dotted
+line C D (where D is, of course, the centre of the cross), and you get
+the form shown in Fig. 3. Now take your scissors and cut from G to F,
+and the four pieces, all of the same size and shape, will fit together
+and form a square, as shown in Fig. 4.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a145b.png" width="400" height="251" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 170<a name="Page_170" id="Page_170"></a></span><a name="X_146_AN_EASY_DISSECTION_PUZZLEa" id="X_146_AN_EASY_DISSECTION_PUZZLEa"></a><a href="#X_146_AN_EASY_DISSECTION_PUZZLE"><b>146.&mdash;AN EASY DISSECTION PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a146.png" width="400" height="205" alt="" title="" />
+</div>
+
+<p>The solution to this puzzle is shown in the illustration. Divide the
+figure up into twelve equal triangles, and it is easy to discover the
+directions of the cuts, as indicated by the dark lines.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_147_AN_EASY_SQUARE_PUZZLEa" id="X_147_AN_EASY_SQUARE_PUZZLEa"></a><a href="#X_147_AN_EASY_SQUARE_PUZZLE"><b>147.&mdash;AN EASY SQUARE PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a147.png" width="400" height="386" alt="" title="" />
+</div>
+
+<p>The diagram explains itself, one of the five pieces having been cut in
+two to form a square.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_148_THE_BUN_PUZZLEa" id="X_148_THE_BUN_PUZZLEa"></a><a href="#X_148_THE_BUN_PUZZLE"><b>148.&mdash;THE BUN PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a148.png" width="600" height="272" alt="" title="" />
+</div>
+
+<p>The secret of the bun puzzle lies in the fact that, with the relative
+dimensions of the circles as given, the three diameters will form a
+right-angled triangle, as shown by A, B, C. It follows that the two
+smaller buns are exactly equal to the large bun. Therefore, if we give
+David and Edgar the two halves marked D and E, they will have their
+fair shares&mdash;one quarter of the confectionery each. Then if we place
+the small bun, H, on the top of the remaining one and trace its
+circumference in the manner shown, Fred's piece, F, will exactly equal
+Harry's small bun, H, with the addition of the piece marked G&mdash;half
+the rim of the other. Thus each boy gets an exactly equal share, and
+there are only five pieces necessary.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_149_THE_CHOCOLATE_SQUARESa" id="X_149_THE_CHOCOLATE_SQUARESa"></a><a href="#X_149_THE_CHOCOLATE_SQUARES"><b>149.&mdash;THE CHOCOLATE SQUARES.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a149.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p>Square A is left entire; the two pieces marked B fit together and make
+a second square; the two pieces C make a third square; and the four
+pieces marked D will form the fourth square.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_150_DISSECTING_A_MITREa" id="X_150_DISSECTING_A_MITREa"></a><a href="#X_150_DISSECTING_A_MITRE"><b>150.&mdash;DISSECTING A MITRE.&mdash;<i>solution</i></b></a></p>
+
+<p>The diagram on the next page shows how to cut into five pieces to form
+a square. The dotted lines are intended to show how to find the points
+C and F&mdash;the only difficulty. A B is half B D, and A E is parallel to
+B H. With the point of the compasses at B describe the arc H E, and A
+E will be the distance of C from B. Then F G equals B C less A B.</p>
+
+<p>This puzzle&mdash;with the added condition that it shall be cut into four
+parts of the same size and shape&mdash;I have not been able to trace to an
+earlier date than 1835. Strictly speaking, <span class='pagenum'>Pg 171<a name="Page_171" id="Page_171"></a></span>it is, in that form,
+impossible of solution; but I give the answer that is always
+presented, and that seems to satisfy most people.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a150a.png" width="600" height="282" alt="" title="" />
+</div>
+
+<p>We are asked to assume that the two portions containing the same
+letter&mdash;AA, BB, CC, DD&mdash;are joined by "a mere hair," and are,
+therefore, only one piece. To the geometrician this is absurd, and the
+four shares are not equal in area unless they consist of two pieces
+each. If you make them equal in area, they will not be exactly alike
+in shape.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a150b.png" width="400" height="388" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_151_THE_JOINERS_PROBLEMa" id="X_151_THE_JOINERS_PROBLEMa"></a><a href="#X_151_THE_JOINERS_PROBLEM"><b>151.&mdash;THE JOINER'S PROBLEM.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a151.png" width="400" height="254" alt="" title="" />
+</div>
+
+<p>Nothing could be easier than the solution of this puzzle&mdash;when you
+know how to do it. And yet it is apt to perplex the novice a good deal
+if he wants to do it in the fewest possible pieces&mdash;three. All you
+have to do is to find the point A, midway between B and C, and then
+cut from A to D and from A to E. The three pieces then form a square
+in the manner shown. Of course, the proportions of the original figure
+must be correct; thus the triangle BEF is just a quarter of the square
+BCDF. Draw lines from B to D and from C to F and this will be clear.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_152_ANOTHER_JOINERS_PROBLEMa" id="X_152_ANOTHER_JOINERS_PROBLEMa"></a><a href="#X_152_ANOTHER_JOINERS_PROBLEM"><b>152.&mdash;ANOTHER JOINER'S PROBLEM.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a152.png" width="400" height="389" alt="" title="" />
+</div>
+
+<p>The point was to find a general rule for forming a perfect square out
+of another square combined with a "right-angled isosceles triangle."
+The triangle to which geometricians give this high-sounding name is,
+of course, nothing more or less than half a square that has been
+divided from corner to corner.</p>
+
+<p>The precise relative proportions of the square and triangle are of no
+consequence whatever. <span class='pagenum'>Pg 172<a name="Page_172" id="Page_172"></a></span>It is only necessary to cut the wood or
+material into five pieces.</p>
+
+<p>Suppose our original square to be ACLF in the above diagram and our
+triangle to be the shaded portion CED. Now, we first find half the
+length of the long side of the triangle (CD) and measure off this
+length at AB. Then we place the triangle in its present position
+against the square and make two cuts&mdash;one from B to F, and the other
+from B to E. Strange as it may seem, that is all that is necessary! If
+we now remove the pieces G, H, and M to their new places, as shown in
+the diagram, we get the perfect square BEKF.</p>
+
+<p>Take any two square pieces of paper, of different sizes but perfect
+squares, and cut the smaller one in half from corner to corner. Now
+proceed in the manner shown, and you will find that the two pieces may
+be combined to form a larger square by making these two simple cuts,
+and that no piece will be required to be turned over.</p>
+
+<p>The remark that the triangle might be "a little larger or a good deal
+smaller in proportion" was intended to bar cases where area of
+triangle is greater than area of square. In such cases six pieces are
+necessary, and if triangle and square are of equal area there is an
+obvious solution in three pieces, by simply cutting the square in half
+diagonally.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_153_A_CUTTING-OUT_PUZZLEa" id="X_153_A_CUTTING-OUT_PUZZLEa"></a><a href="#X_153_A_CUTTING-OUT_PUZZLE"><b>153.&mdash;A CUTTING-OUT PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a153.png" width="400" height="300" alt="" title="" />
+</div>
+
+<p>The illustration shows how to cut the four pieces and form with them a
+square. First find the side of the square (the mean proportional
+between the length and height of the rectangle), and the method is
+obvious. If our strip is exactly in the proportions 9x1, or 16x1, or
+25x1, we can clearly cut it in 3, 4, or 5 rectangular pieces
+respectively to form a square. Excluding these special cases, the
+general law is that for a strip in length more than n&sup2; times the
+breadth, and not more than (n+1)&sup2; times the breadth, it may be cut in
+n+2 pieces to form a square, and there will be n-1 rectangular pieces
+like piece 4 in the diagram. Thus, for example, with a strip 24x1, the
+length is more than 16 and less than 25 times the breadth. Therefore
+it can be done in 6 pieces (n here being 4), 3 of which will be
+rectangular. In the case where n equals 1, the rectangle disappears
+and we get a solution in three pieces. Within these limits, of course,
+the sides need not be rational: the solution is purely geometrical.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_154_MRS_HOBSONS_HEARTHRUGa" id="X_154_MRS_HOBSONS_HEARTHRUGa"></a><a href="#X_154_MRS_HOBSONS_HEARTHRUG"><b>154.&mdash;MRS. HOBSON'S HEARTHRUG.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a154.png" width="400" height="297" alt="" title="" />
+</div>
+
+<p>As I gave full measurements of the mutilated rug, it was quite an easy
+matter to find the precise dimensions for the square. The two pieces
+cut off would, if placed together, make an oblong piece 12x6, giving
+an area of 72 (inches or yards, as we please), and as the original
+complete rug measured 36x27, it had an area of 972. If, therefore, we
+deduct the pieces that have been cut away, we find that our new rug
+will contain 972 less 72, or 900; and as 900 is the square of 30, we
+know that the new rug must measure 30x30 to be a perfect square. This
+is a great help towards the solution, because we may safely conclude
+that the two horizontal sides measuring 30 each may be left intact.</p>
+
+<p>There is a very easy way of solving the puzzle in four pieces, and
+also a way in three pieces that can scarcely be called difficult, but
+the correct answer is in only two pieces.</p>
+
+<p>It will be seen that if, after the cuts are made, we insert the teeth
+of the piece B one tooth lower down, the two portions will fit
+together and form a square.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_155_THE_PENTAGON_AND_SQUAREa" id="X_155_THE_PENTAGON_AND_SQUAREa"></a><a href="#X_155_THE_PENTAGON_AND_SQUARE"><b>155.&mdash;THE PENTAGON AND SQUARE.&mdash;<i>solution</i></b></a></p>
+
+<p>A regular pentagon may be cut into as few as six pieces that will fit
+together without any turning over and form a square, as I shall show
+below. Hitherto the best answer has been in seven pieces&mdash;the solution
+produced some years ago by a foreign mathematician, Paul Busschop. We
+first form a parallelogram, and from that the square. The process will
+be seen in the diagram on the next page.</p>
+
+<p>The pentagon is ABCDE. By the cut AC and the cut FM (F being the
+middle point between A and C, and M being the same distance from A as
+F) we get two pieces that may be placed in position at GHEA and form
+the parallelogram GHDC. We then find the mean proportional between the
+length HD and the <i>height</i> of the parallelogram. This distance we mark
+off from C at K, then draw CK, <span class='pagenum'>Pg 173<a name="Page_173" id="Page_173"></a></span>and from G drop the line GL,
+perpendicular to KC. The rest is easy and rather obvious. It will be
+seen that the six pieces will form either the pentagon or the square.</p>
+
+<p>I have received what purported to be a solution in five pieces, but
+the method was based on the rather subtle fallacy that half the
+diagonal plus half the side of a pentagon equals the side of a square
+of the same area. I say subtle, because it is an extremely close
+approximation that will deceive the eye, and is quite difficult to
+prove inexact. I am not aware that attention has before been drawn to
+this curious approximation.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a155.png" width="400" height="367" alt="" title="" />
+</div>
+
+<p>Another correspondent made the side of his square 1� of the side of
+the pentagon. As a matter of fact, the ratio is irrational. I
+calculate that if the side of the pentagon is 1&mdash;inch, foot, or
+anything else&mdash;the side of the square of equal area is 1.3117 nearly,
+or say roughly 1<sup>3</sup>/<sub>10</sub>. So we can only hope to solve the puzzle by
+geometrical methods.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_156_THE_DISSECTED_TRIANGLEa" id="X_156_THE_DISSECTED_TRIANGLEa"></a><a href="#X_156_THE_DISSECTED_TRIANGLE"><b>156.&mdash;THE DISSECTED TRIANGLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Diagram A is our original triangle. We will say it measures 5 inches
+(or 5 feet) on each side. If we take off a slice at the bottom of any
+equilateral triangle by a cut parallel with the base, the portion that
+remains will always be an equilateral triangle; so we first cut off
+piece 1 and get a triangle 3 inches on every side. The manner of
+finding directions of the other cuts in A is obvious from the diagram.</p>
+
+<p>Now, if we want two triangles, 1 will be one of them, and 2, 3, 4, and
+5 will fit together, as in B, to form the other. If we want three
+equilateral triangles, 1 will be one, 4 and 5 will form the second, as
+in C, and 2 and 3 will form the third, as in D. In B and C the piece 5
+is turned over; but there can be no objection to this, as it is not
+forbidden, and is in no way opposed to the nature of the puzzle.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a156.png" width="400" height="492" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_157_THE_TABLE-TOP_AND_STOOLSa" id="X_157_THE_TABLE-TOP_AND_STOOLSa"></a><a href="#X_157_THE_TABLE-TOP_AND_STOOLS"><b>157.&mdash;THE TABLE-TOP AND STOOLS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a157a.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p>One object that I had in view when presenting this little puzzle was
+to point out the uncertainty of the meaning conveyed by the word
+"oval." Though originally derived from the Latin word <i>ovum</i>, an egg,
+yet what we understand as the egg-shape (with one end smaller than the
+other) is only one of many forms of the oval; while some eggs are
+spherical in shape, and a sphere or circle is most certainly not an
+oval. If we speak of an ellipse&mdash;a conical ellipse&mdash;we are on safer
+ground, but here we must be careful of error. I recollect a Liverpool
+town councillor, many years ago, whose ignorance of the poultry-yard
+led him to substitute the word "hen" for "fowl," remarking, "We must
+remember, gentlemen, that although every cock is a hen, every hen is
+not a cock!" Similarly, we must always note that although every
+ellipse is an oval, every oval is not an <span class='pagenum'>Pg 174<a name="Page_174" id="Page_174"></a></span>ellipse. It is correct to
+say that an oval is an oblong curvilinear figure, having two unequal
+diameters, and bounded by a curve line returning into itself; and this
+includes the ellipse, but all other figures which in any way approach
+towards the form of an oval without necessarily having the properties
+above described are included in the term "oval." Thus the following
+solution that I give to our puzzle involves the pointed "oval," known
+among architects as the "vesica piscis."</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a157b.png" width="400" height="298" alt="" title="" />
+</div>
+
+<p>The dotted lines in the table are given for greater clearness, the
+cuts being made along the other lines. It will be seen that the eight
+pieces form two stools of exactly the same size and shape with similar
+hand-holes. These holes are a trifle longer than those in the
+schoolmaster's stools, but they are much narrower and of considerably
+smaller area. Of course 5 and 6 can be cut out in one piece&mdash;also 7
+and 8&mdash;making only <i>six pieces</i> in all. But I wished to keep the same
+number as in the original story.</p>
+
+<p>When I first gave the above puzzle in a London newspaper, in
+competition, no correct solution was received, but an ingenious and
+neatly executed attempt by a man lying in a London infirmary was
+accompanied by the following note: "Having no compasses here, I was
+compelled to improvise a pair with the aid of a small penknife, a bit
+of firewood from a bundle, a piece of tin from a toy engine, a tin
+tack, and two portions of a hairpin, for points. They are a fairly
+serviceable pair of compasses, and I shall keep them as a memento of
+your puzzle."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_158_THE_GREAT_MONADa" id="X_158_THE_GREAT_MONADa"></a><a href="#X_158_THE_GREAT_MONAD"><b>158.&mdash;THE GREAT MONAD.&mdash;<i>solution</i></b></a></p>
+
+<p>The areas of circles are to each other as the squares of their
+diameters. If you have a circle 2 in. in diameter and another 4 in. in
+diameter, then one circle will be four times as great in area as the
+other, because the square of 4 is four times as great as the square of
+2. Now, if we refer to Diagram 1, we see how two equal squares may be
+cut into four pieces that will form one larger square; from which it
+is self-evident that any square has just half the area of the square
+of its diagonal. In Diagram 2 I have introduced a square as it often
+occurs in ancient drawings of the Monad; which was my reason for
+believing that the symbol had mathematical meanings, since it will be
+found to demonstrate the fact that the area of the outer ring or
+annulus is exactly equal to the area of the inner circle. Compare
+Diagram 2 with Diagram 1, and you will see that as the square of the
+diameter CD is double the square of the diameter of the inner circle,
+or CE, therefore the area of the larger circle is double the area of
+the smaller one, and consequently the area of the annulus is exactly
+equal to that of the inner circle. This answers our first question.</p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/a158.png" width="500" height="485" alt="" title="" />
+</div>
+
+<p>In Diagram 3 I show the simple solution to the second question. It is
+obviously correct, and may be proved by the cutting and superposition
+of parts. The dotted lines will also serve to make it evident. The
+third question is solved by the cut CD in Diagram 2, but it remains to
+be proved that the piece F is really one-half of the Yin or the Yan.
+This we will <span class='pagenum'>Pg 175<a name="Page_175" id="Page_175"></a></span>do in Diagram 4. The circle K has one-quarter the area
+of the circle containing Yin and Yan, because its diameter is just
+one-half the length. Also L in Diagram 3 is, we know, one-quarter the
+area. It is therefore evident that G is exactly equal to H, and
+therefore half G is equal to half H. So that what F loses from L it
+gains from K, and F must be half of Yin or Yan.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_159_THE_SQUARE_OF_VENEERa" id="X_159_THE_SQUARE_OF_VENEERa"></a><a href="#X_159_THE_SQUARE_OF_VENEER"><b>159.&mdash;THE SQUARE OF VENEER.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a159.png" width="400" height="394" alt="" title="" />
+</div>
+
+<p>Any square number may be expressed as the sum of two squares in an
+infinite number of different ways. The solution of the present puzzle
+forms a simple demonstration of this rule. It is a condition that we
+give actual dimensions.</p>
+
+<p>In this puzzle I ignore the known dimensions of our square and work on
+the assumption that it is 13n by 13n. The value of n we can afterwards
+determine. Divide the square as shown (where the dotted lines indicate
+the original markings) into 169 squares. As 169 is the sum of the two
+squares 144 and 25, we will proceed to divide the veneer into two
+squares, measuring respectively 12x12 and 5x5; and as we know that two
+squares may be formed from one square by dissection in four pieces, we
+seek a solution in this number. The dark lines in the diagram show
+where the cuts are to be made. The square 5x5 is cut out whole, and
+the larger square is formed from the remaining three pieces, B, C, and
+D, which the reader can easily fit together.</p>
+
+<p>Now, n is clearly <sup>5</sup>/<sub>13</sub> of an inch. Consequently our larger square must
+be <sup>60</sup>/<sub>13</sub> in.&nbsp;&times;&nbsp;<sup>60</sup>/<sub>13</sub> in., and our smaller square <sup>25</sup>/<sub>13</sub> in.&nbsp;&times;&nbsp;<sup>25</sup>/<sub>13</sub> in.
+The square of <sup>60</sup>/<sub>13</sub> added to the square of <sup>25</sup>/<sub>13</sub> is 25. The square is
+thus divided into as few as four pieces that form two squares of known
+dimensions, and all the sixteen nails are avoided.</p>
+
+<p>Here is a general formula for finding two squares whose sum shall
+equal a given square, say a&sup2;. In the case of the solution of our
+puzzle p&nbsp;=&nbsp;3, q&nbsp;=&nbsp;2, and a&nbsp;=&nbsp;5.</p>
+
+<div style="font-size: larger;">
+<pre>
+                                 _________________________
+                2pqa           \/ a�( p�&nbsp;+&nbsp;q�)�&nbsp;-&nbsp;(2pqa)�
+              ---------&nbsp;=&nbsp;x;   ---------------------------&nbsp;=&nbsp;y
+               p�&nbsp;+&nbsp;q�                  p�&nbsp;+&nbsp;q�
+
+             Here x�&nbsp;+&nbsp;y�&nbsp;=&nbsp;a�.
+</pre>
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_160_THE_TWO_HORSESHOESa" id="X_160_THE_TWO_HORSESHOESa"></a><a href="#X_160_THE_TWO_HORSESHOES"><b>160.&mdash;THE TWO HORSESHOES.&mdash;<i>solution</i></b></a></p>
+
+<p>The puzzle was to cut the two shoes (including the hoof contained
+within the outlines) into four pieces, two pieces each, that would fit
+together and form a perfect circle. It was also stipulated that all
+four pieces should be different in shape. As a matter of fact, it is a
+puzzle based on the principle contained in that curious Chinese symbol
+the Monad. (See No. <a href="#X_158_THE_GREAT_MONAD">158</a>.)</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a160.png" width="600" height="262" alt="" title="" />
+</div>
+
+<p>The above diagrams give the correct solution to the problem. It will
+be noticed that 1 and 2 are cut into the required four pieces, all
+differ<span class='pagenum'>Pg 176<a name="Page_176" id="Page_176"></a></span>ent in shape, that fit together and form the perfect circle
+shown in Diagram 3. It will further be observed that the two pieces A
+and B of one shoe and the two pieces C and D of the other form two
+exactly similar halves of the circle&mdash;the Yin and the Yan of the great
+Monad. It will be seen that the shape of the horseshoe is more easily
+determined from the circle than the dimensions of the circle from the
+horseshoe, though the latter presents no difficulty when you know that
+the curve of the long side of the shoe is part of the circumference of
+your circle. The difference between B and D is instructive, and the
+idea is useful in all such cases where it is a condition that the
+pieces must be different in shape. In forming D we simply add on a
+symmetrical piece, a curvilinear square, to the piece B. Therefore, in
+giving either B or D a quarter turn before placing in the new
+position, a precisely similar effect must be produced.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_161_THE_BETSY_ROSS_PUZZLEa" id="X_161_THE_BETSY_ROSS_PUZZLEa"></a><a href="#X_161_THE_BETSY_ROSS_PUZZLE"><b>161.&mdash;THE BETSY ROSS PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Fold the circular piece of paper in half along the dotted line shown
+in Fig. 1, and divide the upper half into five equal parts as
+indicated. Now fold the paper along the lines, and it will have the
+appearance shown in Fig. 2. If you want a star like Fig. 3, cut from A
+to B; if you wish one like Fig. 4, cut from A to C. Thus, the nearer
+you cut to the point at the bottom the longer will be the points of
+the star, and the farther off from the point that you cut the shorter
+will be the points of the star.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a161.png" width="600" height="447" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_162_THE_CARDBOARD_CHAINa" id="X_162_THE_CARDBOARD_CHAINa"></a><a href="#X_162_THE_CARDBOARD_CHAIN"><b>162.&mdash;THE CARDBOARD CHAIN.&mdash;<i>solution</i></b></a></p>
+
+<p>The reader will probably feel rewarded for any care and patience that
+he may bestow on cutting out the cardboard chain. We will suppose that
+he has a piece of cardboard measuring 8 in. by 2� in., though the
+dimensions are of no importance. Yet if you want a long chain you must,
+of course, take a long strip of cardboard. First rule pencil lines B B
+and C C, half an inch from the edges, and also the short perpendicular
+lines half an inch apart. Rule lines on the other side in just the same
+way, and in order that they shall coincide it is well to prick through
+the card with a needle the points where the short lines end. Now take
+your penknife and split the card from A A down to B B, and from D D up
+to C C. Then cut right through the card along all the short
+perpendicular lines, and half through the card along the short portions
+of B B and C C that are not dotted. Next turn the card over and cut half
+through along the short lines on B B and C C at the places that are
+immediately beneath the dotted lines on the upper side. With a little
+careful separation of the parts with the penknife, the cardboard may now
+be divided into two interlacing ladder-like portions, as shown in Fig.
+2; and if you cut away all the shaded parts you will get the chain, cut
+solidly out of the cardboard, without any join, as shown in the
+illustrations on page 40.</p>
+
+<p>It is an interesting variant of the puzzle to cut out two keys on a
+ring&mdash;in the same manner without join.</p>
+
+<p><span class='pagenum'>Pg 177<a name="Page_177" id="Page_177"></a></span></p><div class="figcenter" style="width: 600px;">
+<img src="images/a162.png" width="600" height="422" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_164_THE_POTATO_PUZZLEa" id="X_164_THE_POTATO_PUZZLEa"></a><a href="#X_164_THE_POTATO_PUZZLE"><b>164.&mdash;THE POTATO PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>As many as twenty-two pieces may be obtained by the six cuts. The
+illustration shows a pretty symmetrical solution. The rule in such
+cases is that every cut shall intersect every other cut and no two
+intersections coincide; that is to say, every line passes through
+every other line, but more than two lines do not cross at the same
+point anywhere. There are other ways of making the cuts, but this rule
+must always be observed if we are to get the full number of pieces.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a164.png" width="400" height="394" alt="" title="" />
+</div>
+
+<p>The general formula is that with <i>n</i> cuts we can always produce <sup>(<i>n</i>(<i>n</i>&nbsp;+&nbsp;1)&nbsp;+&nbsp;1)</sup>/<sub>2</sub> . One of the problems proposed by the late Sam Loyd
+was to produce the maximum number of pieces by <i>n</i> straight cuts
+through a solid cheese. Of course, again, the pieces cut off may not
+be moved or piled. Here we have to deal with the intersection of
+planes (instead of lines), and the general formula is that with <i>n</i>
+cuts we may produce <sup>((<i>n</i>&nbsp;-&nbsp;1)<i>n</i>(<i>n</i>&nbsp;+&nbsp;1))</sup>/<sub>6</sub>&nbsp;+&nbsp;<i>n</i>&nbsp;+&nbsp;1 pieces.
+It is extremely difficult to "see" the direction and effects of the
+successive cuts for more than a few of the lowest values of <i>n</i>.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_165_THE_SEVEN_PIGSa" id="X_165_THE_SEVEN_PIGSa"></a><a href="#X_165_THE_SEVEN_PIGS"><b>165.&mdash;THE SEVEN PIGS.&mdash;<i>solution</i></b></a></p>
+
+<p>The illustration shows the direction for placing the three fences so
+as to enclose every pig in a separate sty. The greatest number of
+spaces that can be enclosed with three straight lines in a square is
+seven, as shown in the last puzzle. Bearing this fact in mind, the
+puzzle must be solved by trial.</p>
+
+<p><span class='pagenum'>Pg 178<a name="Page_178" id="Page_178"></a></span></p><div class="figcenter" style="width: 400px;">
+<img src="images/a165.png" width="400" height="394" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_166_THE_LANDOWNERS_FENCESa" id="X_166_THE_LANDOWNERS_FENCESa"></a><a href="#X_166_THE_LANDOWNERS_FENCES"><b>166.&mdash;THE LANDOWNER'S FENCES.&mdash;<i>solution</i></b></a></p>
+
+<p>Four fences only are necessary, as follows:&mdash;</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a166.png" width="400" height="392" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_167_THE_WIZARDS_CATSa" id="X_167_THE_WIZARDS_CATSa"></a><a href="#X_167_THE_WIZARDS_CATS"><b>167.&mdash;THE WIZARD'S CATS.&mdash;<i>solution</i></b></a></p>
+
+<p>The illustration requires no explanation. It shows clearly how the
+three circles may be drawn so that every cat has a separate enclosure,
+and cannot approach another cat without crossing a line.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a167.png" width="400" height="400" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_168_THE_CHRISTMAS_PUDDINGa" id="X_168_THE_CHRISTMAS_PUDDINGa"></a><a href="#X_168_THE_CHRISTMAS_PUDDING"><b>168.&mdash;THE CHRISTMAS PUDDING.&mdash;<i>solution</i></b></a></p>
+
+<p>The illustration shows how the pudding may be cut into two parts of
+exactly the same size and shape. The lines must necessarily pass
+through the points A, B, C, D, and E. But, subject to this condition,
+they may be varied in an infinite number of ways. For example, at a
+point midway between A and the edge, the line may be completed in an
+unlimited number of ways (straight or crooked), provided it be exactly
+reflected from E to the opposite edge. And similar variations may be
+introduced at other places.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a168.png" width="400" height="400" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_169_A_TANGRAM_PARADOXa" id="X_169_A_TANGRAM_PARADOXa"></a><a href="#X_169_A_TANGRAM_PARADOX"><b>169.&mdash;A TANGRAM PARADOX.&mdash;<i>solution</i></b></a></p>
+
+<p>The diagrams will show how the figures are constructed&mdash;each with the
+seven Tangrams. It will be noticed that in both cases the head, hat,
+and arm are precisely alike, and the width at the base of the body the
+same. But this body contains four pieces in the first case, and in the
+second design only three. The first is larger than the second by
+exactly that narrow strip indicated by the dotted line between A and
+B. <span class='pagenum'>Pg 179<a name="Page_179" id="Page_179"></a></span>This strip is therefore exactly equal in area to the piece forming
+the foot in the other design, though when thus distributed along the
+side of the body the increased dimension is not easily apparent to the
+eye.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a169.png" width="400" height="368" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_170_THE_CUSHION_COVERSa" id="X_170_THE_CUSHION_COVERSa"></a><a href="#X_170_THE_CUSHION_COVERS"><b>170.&mdash;THE CUSHION COVERS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a170.png" width="400" height="419" alt="" title="" />
+</div>
+
+<p>The two pieces of brocade marked A will fit together and form one
+perfect square cushion top, and the two pieces marked B will form the
+other.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_171_THE_BANNER_PUZZLEa" id="X_171_THE_BANNER_PUZZLEa"></a><a href="#X_171_THE_BANNER_PUZZLE"><b>171.&mdash;THE BANNER PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The illustration explains itself. Divide the bunting into 25 squares
+(because this number is the sum of two other squares&mdash;16 and 9), and
+then cut along the thick lines. The two pieces marked A form one
+square, and the two pieces marked B form the other.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a171.png" width="400" height="400" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_172_MRS_SMILEYS_CHRISTMAS_PRESENTa" id="X_172_MRS_SMILEYS_CHRISTMAS_PRESENTa"></a><a href="#X_172_MRS_SMILEYS_CHRISTMAS_PRESENT"><b>172.&mdash;MRS. SMILEY'S CHRISTMAS PRESENT.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a172.png" width="400" height="794" alt="" title="" />
+</div>
+
+<p><span class='pagenum'>Pg 180<a name="Page_180" id="Page_180"></a></span>The first step is to find six different square numbers that sum to
+196. For example, 1&nbsp;+&nbsp;4&nbsp;+&nbsp;25&nbsp;+&nbsp;36&nbsp;+&nbsp;49&nbsp;+&nbsp;81&nbsp;=&nbsp;196; 1&nbsp;+&nbsp;4&nbsp;+&nbsp;9&nbsp;+&nbsp;25&nbsp;+&nbsp;36
++ 121&nbsp;=&nbsp;196; 1&nbsp;+&nbsp;9&nbsp;+&nbsp;16&nbsp;+&nbsp;25&nbsp;+&nbsp;64&nbsp;+&nbsp;81&nbsp;=&nbsp;196. The rest calls for
+individual judgment and ingenuity, and no definite rules can be given
+for procedure. The annexed diagrams will show solutions for the first
+two cases stated. Of course the three pieces marked A and those marked
+B will fit together and form a square in each case. The assembling of
+the parts may be slightly varied, and the reader may be interested in
+finding a solution for the third set of squares I have given.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_173_MRS_PERKINSS_QUILTa" id="X_173_MRS_PERKINSS_QUILTa"></a><a href="#X_173_MRS_PERKINSS_QUILT"><b>173.&mdash;MRS. PERKINS'S QUILT.&mdash;<i>solution</i></b></a></p>
+
+<p>The following diagram shows how the quilt should be constructed.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a173.png" width="400" height="392" alt="" title="" />
+</div>
+
+<p>There is, I believe, practically only one solution to this puzzle. The
+fewest separate squares must be eleven. The portions must be of the
+sizes given, the three largest pieces must be arranged as shown, and
+the remaining group of eight squares may be "reflected," but cannot be
+differently arranged.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_174_THE_SQUARES_OF_BROCADEa" id="X_174_THE_SQUARES_OF_BROCADEa"></a><a href="#X_174_THE_SQUARES_OF_BROCADE"><b>174.&mdash;THE SQUARES OF BROCADE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a174a.png" width="400" height="415" alt="" title="" />
+</div>
+
+<p>So far as I have been able to discover, there is only one possible
+solution to fulfil the conditions. The pieces fit together as in
+Diagram 1, Diagrams 2 and 3 showing how the two original squares are
+to be cut. It will be seen that the pieces A and C have each twenty
+chequers, and are therefore of equal area. Diagram 4 (built up with
+the dissected square No. 5) solves the puzzle, except for the small
+condition contained in the words, "I cut the <i>two</i> squares in the
+manner desired." In this case the smaller square is preserved intact.
+Still I give it as an illustration of a feature of the puzzle. It is
+impossible in a problem of this kind to give a <i>quarter-turn</i> to any
+of the pieces if the pattern is to properly match, but (as in the case
+of F, in Diagram 4) we may give a symmetrical piece a
+<i>half-turn</i>&mdash;that is, turn it upside down. Whether or not a piece may
+be given a quarter-turn, a half-turn, or no turn at all in these
+chequered problems, depends on the character of the design, on the
+material employed, and also on the form of the piece itself.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a174b.png" width="400" height="225" alt="" title="" />
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a174c.png" width="400" height="431" alt="" title="" />
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a174d.png" width="400" height="442" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_175_ANOTHER_PATCHWORK_PUZZLEa" id="X_175_ANOTHER_PATCHWORK_PUZZLEa"></a><a href="#X_175_ANOTHER_PATCHWORK_PUZZLE"><b>175.&mdash;ANOTHER PATCHWORK PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The lady need only unpick the stitches along the dark lines in the
+larger portion of patchwork, <span class='pagenum'>Pg 181<a name="Page_181" id="Page_181"></a></span>when the four pieces will fit together
+and form a square, as shown in our illustration.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a175.png" width="400" height="173" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_176_LINOLEUM_CUTTINGa" id="X_176_LINOLEUM_CUTTINGa"></a><a href="#X_176_LINOLEUM_CUTTING"><b>176.&mdash;LINOLEUM CUTTING.&mdash;<i>solution</i></b></a></p>
+
+<p>There is only one solution that will enable us to retain the larger of
+the two pieces with as little as possible cut from it. Fig. 1 in the
+following diagram shows how the smaller piece is to be cut, and Fig. 2
+how we should dissect the larger piece, while in Fig. 3 we have the
+new square 10&nbsp;&times;&nbsp;10 formed by the four pieces with all the chequers
+properly matched. It will be seen that the piece D contains fifty-two
+chequers, and this is the largest piece that it is possible to
+preserve under the conditions.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a176.png" width="400" height="543" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_177_ANOTHER_LINOLEUM_PUZZLEa" id="X_177_ANOTHER_LINOLEUM_PUZZLEa"></a><a href="#X_177_ANOTHER_LINOLEUM_PUZZLE"><b>177.&mdash;ANOTHER LINOLEUM PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Cut along the thick lines, and the four pieces will fit together and
+form a perfect square in the manner shown in the smaller diagram.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a177.png" width="400" height="236" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_178_THE_CARDBOARD_BOXa" id="X_178_THE_CARDBOARD_BOXa"></a><a href="#X_178_THE_CARDBOARD_BOX"><b>178.&mdash;THE CARDBOARD BOX.&mdash;<i>solution</i></b></a></p>
+
+<p>The areas of the top and side multiplied together and divided by the
+area of the end give the square of the length. Similarly, the product
+of top and end divided by side gives the square of the breadth; and
+the product of side and end divided by the top gives the square of the
+depth. But we only need one of these operations. Let us take the
+first. Thus, 120&nbsp;&times;&nbsp;96 divided by 80 equals 144, the square of 12.
+Therefore the length is 12 inches, from which we can, of course, at
+once get the breadth and depth&mdash;10 in. and 8 in. respectively.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_179_STEALING_THE_BELL-ROPESa" id="X_179_STEALING_THE_BELL-ROPESa"></a><a href="#X_179_STEALING_THE_BELL-ROPES"><b>179.&mdash;STEALING THE BELL-ROPES.&mdash;<i>solution</i></b></a></p>
+
+<p>Whenever we have one side (<i>a</i>) of a right-angled triangle, and know the
+difference between the second side and the hypotenuse (which
+difference we will call <i>b</i>), then the length of the hypotenuse will be</p>
+
+<p class='center'><span class='su2'><i>a</i><sup>2</sup></span>/<sub>2<i>b</i></sub>&nbsp;+&nbsp;<sup><i>b</i></sup>/<sub>2</sub></p>
+
+<p>In the case of our puzzle this will be</p>
+
+<p class='center'><sup>(48&nbsp;&times;&nbsp;48)</sup>/<sub>6</sub> &nbsp;+&nbsp;1� in.&nbsp;=&nbsp;32 ft. 1� in.,</p>
+
+<p>which is the length of the rope.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_180_THE_FOUR_SONSa" id="X_180_THE_FOUR_SONSa"></a><a href="#X_180_THE_FOUR_SONS"><b>180.&mdash;THE FOUR SONS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a180.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p>The diagram shows the most equitable division of the land possible,
+"so that each son shall receive land of exactly the same area and
+exactly similar in shape," and so that each shall have access to the
+well in the centre without trespass on another's land. The conditions
+<span class='pagenum'>Pg 182<a name="Page_182" id="Page_182"></a></span>do not require that each son's land shall be in one piece, but it is
+necessary that the two portions assigned to an individual should be
+kept apart, or two adjoining portions might be held to be one piece,
+in which case the condition as to shape would have to be broken. At
+present there is only one shape for each piece of land&mdash;half a square
+divided diagonally. And A, B, C, and D can each reach their land from
+the outside, and have each equal access to the well in the centre.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_181_THE_THREE_RAILWAY_STATIONSa" id="X_181_THE_THREE_RAILWAY_STATIONSa"></a><a href="#X_181_THE_THREE_RAILWAY_STATIONS"><b>181.&mdash;THE THREE RAILWAY STATIONS.&mdash;<i>solution</i></b></a></p>
+
+<p>The three stations form a triangle, with sides 13, 14, and 15 miles.
+Make the 14 side the base; then the height of the triangle is 12 and
+the area 84. Multiply the three sides together and divide by four
+times the area. The result is eight miles and one-eighth, the distance
+required.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_182_THE_GARDEN_PUZZLEa" id="X_182_THE_GARDEN_PUZZLEa"></a><a href="#X_182_THE_GARDEN_PUZZLE"><b>182.&mdash;THE GARDEN PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Half the sum of the four sides is 144. From this deduct in turn the
+four sides, and we get 64, 99, 44, and 81. Multiply these together,
+and we have as the result the square of 4,752. Therefore the garden
+contained 4,752 square yards. Of course the tree being equidistant
+from the four corners shows that the garden is a quadrilateral that
+may be inscribed in a circle.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_183_DRAWING_A_SPIRALa" id="X_183_DRAWING_A_SPIRALa"></a><a href="#X_183_DRAWING_A_SPIRAL"><b>183.&mdash;DRAWING A SPIRAL.&mdash;<i>solution</i></b></a></p>
+
+<p>Make a fold in the paper, as shown by the dotted line in the
+illustration. Then, taking any two points, as A and B, describe
+semicircles on the line alternately from the centres B and A, being
+careful to make the ends join, and the thing is done. Of course this
+is not a <i>true</i> spiral, but the puzzle was to produce the <i>particular</i>
+spiral that was shown, and that was drawn in this simple manner.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a183.png" width="600" height="600" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_184_HOW_TO_DRAW_AN_OVALa" id="X_184_HOW_TO_DRAW_AN_OVALa"></a><a href="#X_184_HOW_TO_DRAW_AN_OVAL"><b>184.&mdash;HOW TO DRAW AN OVAL.&mdash;<i>solution</i></b></a></p>
+
+<p>If you place your sheet of paper round the surface of a cylindrical
+bottle or canister, the oval can be drawn with one sweep of the
+compasses.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_185_ST_GEORGES_BANNERa" id="X_185_ST_GEORGES_BANNERa"></a><a href="#X_185_ST_GEORGES_BANNER"><b>185.&mdash;ST. GEORGE'S BANNER.&mdash;<i>solution</i></b></a></p>
+
+<p>As the flag measures 4 ft. by 3 ft., the length of the diagonal (from
+corner to corner) is 5 ft. All you need do is to deduct half the
+length of this diagonal (2� ft.) from a quarter of the distance all
+round the edge of the flag (3� ft.)&mdash;a quarter of 14 ft. The
+difference (1 ft.) is the required width of the arm of the red cross.
+The area of the cross will then be the same as that of the white
+ground.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_186_THE_CLOTHES_LINE_PUZZLEa" id="X_186_THE_CLOTHES_LINE_PUZZLEa"></a><a href="#X_186_THE_CLOTHES_LINE_PUZZLE"><b>186.&mdash;THE CLOTHES LINE PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Multiply together, and also add together, the heights of the two poles
+and divide one result <span class='pagenum'>Pg 183<a name="Page_183" id="Page_183"></a></span>by the other. That is, if the two heights are <i>a</i>
+and <i>b</i> respectively, then <sup><i>ab</i></sup>/<sub>(<i>a</i>&nbsp;+&nbsp;<i>b</i>)</sub> will give the height of the
+intersection. In the particular case of our puzzle, the intersection
+was therefore 2 ft. 11 in. from the ground. The distance that the
+poles are apart does not affect the answer. The reader who may have
+imagined that this was an accidental omission will perhaps be
+interested in discovering the reason why the distance between the
+poles may be ignored.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_187_THE_MILKMAID_PUZZLEa" id="X_187_THE_MILKMAID_PUZZLEa"></a><a href="#X_187_THE_MILKMAID_PUZZLE"><b>187.&mdash;THE MILKMAID PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a187.png" width="400" height="442" alt="" title="" />
+</div>
+
+<p>Draw a straight line, as shown in the diagram, from the milking-stool
+perpendicular to the near bank of the river, and continue it to the
+point A, which is the same distance from that bank as the stool. If
+you now draw the straight line from A to the door of the dairy, it
+will cut the river at B. Then the shortest route will be from the
+stool to B and thence to the door. Obviously the shortest distance
+from A to the door is the straight line, and as the distance from the
+stool to any point of the river is the same as from A to that point,
+the correctness of the solution will probably appeal to every reader
+without any acquaintance with geometry.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_188_THE_BALL_PROBLEMa" id="X_188_THE_BALL_PROBLEMa"></a><a href="#X_188_THE_BALL_PROBLEM"><b>188.&mdash;THE BALL PROBLEM.&mdash;<i>solution</i></b></a></p>
+
+<p>If a round ball is placed on the level ground, six similar balls may
+be placed round it (all on the ground), so that they shall all touch
+the central ball.</p>
+
+<p>As for the second question, the ratio of the diameter of a circle to
+its circumference we call <i>pi</i>; and though we cannot express this
+ratio in exact numbers, we can get sufficiently near to it for all
+practical purposes. However, in this case it is not necessary to know
+the value of <i>pi</i> at all. Because, to find the area of the surface of
+a sphere we multiply the square of the diameter by <i>pi</i>; to find the
+volume of a sphere we multiply the cube of the diameter by one-sixth
+of <i>pi</i>. Therefore we may ignore <i>pi</i>, and have merely to seek a
+number whose square shall equal one-sixth of its cube. This number is
+obviously 6. Therefore the ball was 6 ft. in diameter, for the area of
+its surface will be 36 times <i>pi</i> in square feet, and its volume also
+36 times <i>pi</i> in cubic feet.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_189_THE_YORKSHIRE_ESTATESa" id="X_189_THE_YORKSHIRE_ESTATESa"></a><a href="#X_189_THE_YORKSHIRE_ESTATES"><b>189.&mdash;THE YORKSHIRE ESTATES.&mdash;<i>solution</i></b></a></p>
+
+<p>The triangular piece of land that was not for sale contains exactly
+eleven acres. Of course it is not difficult to find the answer if we
+follow the eccentric and tricky tracks of intricate trigonometry; or I
+might say that the application of a well-known formula reduces the
+problem to finding one-quarter of the square root of (4&nbsp;&times;&nbsp;370&nbsp;&times;&nbsp;116) -
+(370&nbsp;+&nbsp;116&nbsp;-&nbsp;74)&sup2;&mdash;that is a quarter of the square root of 1936, which
+is one-quarter of 44, or 11 acres. But all that the reader really
+requires to know is the Pythagorean law on which many puzzles have
+been built, that in any right-angled triangle the square of the
+hypotenuse is equal to the sum of the squares of the other two sides.
+I shall dispense with all "surds" and similar absurdities,
+notwithstanding the fact that the sides of our triangle are clearly
+incommensurate, since we cannot exactly extract the square roots of
+the three square areas.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a189.png" width="400" height="235" alt="" title="" />
+</div>
+
+<p>In the above diagram ABC represents our triangle. ADB is a
+right-angled triangle, AD measuring 9 and BD measuring 17, because the
+square of 9 added to the square of 17 equals 370, the known area of
+the square on AB. Also AEC is a right-angled triangle, and the square
+of 5 added to the square of 7 equals 74, the square estate on A C.
+Similarly, CFB is a right-angled triangle, for the square of 4 added
+to the square of 10 equals 116, the square estate on BC. Now, although
+the sides of our triangular estate are incommensurate, we have in this
+diagram all the exact figures that we need to discover the area with
+precision.</p>
+
+<p>The area of our triangle ADB is clearly half of 9&nbsp;&times;&nbsp;17, or 76&frac12; acres.
+The area of AEC is half of 5&nbsp;&times;&nbsp;7, or 17&frac12; acres; the area of CFB is
+half of 4&nbsp;&times;&nbsp;10, or 20 acres; and the area of the oblong EDFC is
+obviously 4&nbsp;&times;&nbsp;7, or 28 acres. Now, if we add together 17&frac12;, 20, and <span class='pagenum'>Pg 184<a name="Page_184" id="Page_184"></a></span>28
+= 65&frac12;, and deduct this sum from the area of the large triangle ADB
+(which we have found to be 76&frac12; acres), what remains must clearly be
+the area of ABC. That is to say, the area we want must be 76&frac12;&nbsp;-&nbsp;65&frac12; =
+11 acres exactly.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_190_FARMER_WURZELS_ESTATEa" id="X_190_FARMER_WURZELS_ESTATEa"></a><a href="#X_190_FARMER_WURZELS_ESTATE"><b>190.&mdash;FARMER WURZEL'S ESTATE.&mdash;<i>solution</i></b></a></p>
+
+<p>The area of the complete estate is exactly one hundred acres. To find
+this answer I use the following little formula,</p>
+
+<pre>
+                                  __________________
+                                \/4ab&nbsp;-&nbsp;(a&nbsp;+&nbsp;b&nbsp;-&nbsp;c)�;
+                                _____________________
+                                          4
+</pre>
+
+<p>where a, b, c represent the three square areas, in any order. The
+expression gives the area of the triangle A. This will be found to be
+9 acres. It can be easily proved that A, B, C, and D are all equal in
+area; so the answer is 26&nbsp;+&nbsp;20&nbsp;+&nbsp;18&nbsp;+&nbsp;9&nbsp;+&nbsp;9&nbsp;+&nbsp;9&nbsp;+&nbsp;9&nbsp;=&nbsp;100 acres.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a190.png" width="400" height="396" alt="" title="" />
+</div>
+
+<p>Here is the proof. If every little dotted square in the diagram
+represents an acre, this must be a correct plan of the estate, for the
+squares of 5 and 1 together equal 26; the squares of 4 and 2 equal 20;
+and the squares of 3 and 3 added together equal 18. Now we see at once
+that the area of the triangle E is 2&frac12;, F is 4&frac12;, and G is 4. These
+added together make 11 acres, which we deduct from the area of the
+rectangle, 20 acres, and we find that the field A contains exactly 9
+acres. If you want to prove that B, C, and D are equal in size to A,
+divide them in two by a line from the middle of the longest side to
+the opposite angle, and you will find that the two pieces in every
+case, if cut out, will exactly fit together and form A.</p>
+
+<p>Or we can get our proof in a still easier way. The complete area of
+the squared diagram is 12&nbsp;&times;&nbsp;12&nbsp;=&nbsp;144 acres, and the portions 1, 2, 3,
+4, not included in the estate, have the respective areas of 12&frac12;, 17&frac12;,
+9&frac12;, and 4&frac12;. These added together make 44, which, deducted from 144,
+leaves 100 as the required area of the complete estate.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_191_THE_CRESCENT_PUZZLEa" id="X_191_THE_CRESCENT_PUZZLEa"></a><a href="#X_191_THE_CRESCENT_PUZZLE"><b>191.&mdash;THE CRESCENT PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Referring to the original diagram, let AC be <i>x</i>, let CD be <i>x</i>&nbsp;-&nbsp;9, and
+let EC be <i>x</i>&nbsp;-&nbsp;5. Then <i>x</i>&nbsp;-&nbsp;5 is a mean proportional between <i>x</i>&nbsp;-&nbsp;9 and
+<i>x</i>, from which we find that <i>x</i> equals 25. Therefore the diameters are 50
+in. and 41 in. respectively.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_192_THE_PUZZLE_WALLa" id="X_192_THE_PUZZLE_WALLa"></a><a href="#X_192_THE_PUZZLE_WALL"><b>192.&mdash;THE PUZZLE WALL.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a192.png" width="400" height="300" alt="" title="" />
+</div>
+
+<p>The answer given in all the old books is that shown in Fig. 1, where
+the curved wall shuts out the cottages from access to the lake. But in
+seeking the direction for the "shortest possible" wall most readers
+to-day, remembering that the shortest distance between two points is a
+straight line, will adopt the method shown in Fig. 2. This is
+certainly an improvement, yet the correct answer is really that
+indicated in Fig. 3. A measurement of the lines will show that there
+is a considerable saving of length in this wall.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_193_THE_SHEEP_FOLDa" id="X_193_THE_SHEEP_FOLDa"></a><a href="#X_193_THE_SHEEP_FOLD"><b>193.&mdash;THE SHEEP-FOLD.&mdash;<i>solution</i></b></a></p>
+
+<p>This is the answer that is always given and accepted as correct: Two
+more hurdles would be necessary, for the pen was twenty-four by one
+(as in Fig. A on next page), and by moving one of the sides and
+placing an extra hurdle at each end (as in Fig. B) the area would be
+doubled. The diagrams are not to scale. Now there is no condition in
+the puzzle that requires the sheep-fold to be of any particular form.
+But even if we accept the point that the pen was twenty-four by one,
+the answer utterly fails, for two extra hurdles are certainly not at
+all necessary. For example, I arrange the fifty hurdles as in Fig. C,
+and as the area is increased from twenty-four "square hurdles" to 156,
+there is now accommodation for 650 sheep. If it be held that the area
+must be exactly double that of the original pen, then I construct it
+(as in Fig. D) with twenty-eight hurdles only, and have twenty-two in
+hand for other purposes on the farm. Even if it were insisted that all
+the original hurdles must be used, then I should construct it as in
+Fig. E, where I can get the area as exact as any farmer could possibly
+require, even if we have to allow for the fact that the sheep might
+not be able to graze at the extreme ends. Thus we see that, from any
+<span class='pagenum'>Pg 185<a name="Page_185" id="Page_185"></a></span>point of view, the accepted answer to this ancient little puzzle
+breaks down. And yet attention has never before been drawn to the
+absurdity.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a193.png" width="400" height="473" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_194_THE_GARDEN_WALLSa" id="X_194_THE_GARDEN_WALLSa"></a><a href="#X_194_THE_GARDEN_WALLS"><b>194.&mdash;THE GARDEN WALLS.&mdash;<i>solution</i></b></a></p>
+
+<p>The puzzle was to divide the circular field into four equal parts by
+three walls, each wall being of exactly the same length. There are two
+essential difficulties in this problem. These are: (1) the thickness
+of the walls, and (2) the condition that these walls are three in
+number. As to the first point, since we are told that the walls are
+brick walls, we clearly cannot ignore their thickness, while we have
+to find a solution that will equally work, whether the walls be of a
+thickness of one, two, three, or more bricks.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a194.png" width="400" height="412" alt="" title="" />
+</div>
+
+<p>The second point requires a little more consideration. How are we to
+distinguish between a wall and walls? A straight wall without any bend
+in it, no matter how long, cannot ever become "walls," if it is
+neither broken nor intersected in any way. Also our circular field is
+clearly enclosed by one wall. But if it had happened to be a square or
+a triangular enclosure, would there be respectively four and three
+walls or only one enclosing wall in each case? It is true that we
+speak of "the four walls" of a square building or garden, but this is
+only a conventional way of saying "the four sides." If you were
+speaking of the actual brickwork, you would say, "I am going to
+enclose this square garden with a wall." Angles clearly do not affect
+the question, for we may have a zigzag wall just as well as a straight
+one, and the Great Wall of China is a good example of a wall with
+plenty of angles. Now, if you look at Diagrams 1, 2, and 3, you may be
+puzzled to declare whether there are in each case two or four new
+walls; but you cannot call them three, as required in our puzzle. The
+intersection either affects the question or it does not affect it.</p>
+
+<p>If you tie two pieces of string firmly together, or splice them in a
+nautical manner, they become "one piece of string." If you simply let
+them lie across one another or overlap, they remain "two pieces of
+string." It is all a question of joining and welding. It may similarly
+be held that if two walls be built into one another&mdash;I might almost
+say, if they be made homogeneous&mdash;they become one wall, in which case
+Diagrams 1, 2, and 3 might each be said to show one wall or two, if it
+be indicated that the four ends only touch, and are not really built
+into, the outer circular wall.</p>
+
+<p>The objection to Diagram 4 is that although it shows the three
+required walls (assuming the ends are not built into the outer
+circular wall), yet it is only absolutely correct when we assume the
+walls to have no thickness. A brick has thickness, and therefore the
+fact throws the whole method out and renders it only approximately
+correct.</p>
+
+<p>Diagram 5 shows, perhaps, the only correct and perfectly satisfactory
+solution. It will be noticed that, in addition to the circular wall,
+there are three new walls, which touch (and so enclose) but are not
+built into one another. This solution may be adapted to any desired
+thickness of wall, and its correctness as to area and length of wall
+space is so obvious that it is unnecessary to explain it. I will,
+however, just say that the semicircular piece of ground that each
+tenant gives to his neighbour is exactly equal to the semicircular
+piece that his neighbour gives to him, while any section of wall space
+found in one garden is precisely repeated in all the others. Of course
+there is an infinite number of ways in which this solution may be
+correctly varied.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 186<a name="Page_186" id="Page_186"></a></span><a name="X_195_LADY_BELINDAS_GARDENa" id="X_195_LADY_BELINDAS_GARDENa"></a><a href="#X_195_LADY_BELINDAS_GARDEN"><b>195.&mdash;LADY BELINDA'S GARDEN.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a195.png" width="400" height="193" alt="" title="" />
+</div>
+
+<p>All that Lady Belinda need do was this: She should measure from A to
+B, fold her tape in four and mark off the point E, which is thus one
+quarter of the side. Then, in the same way, mark off the point F,
+one-fourth of the side AD Now, if she makes EG equal to AF, and GH
+equal to EF, then AH is the required width for the path in order that
+the bed shall be exactly half the area of the garden. An exact
+numerical measurement can only be obtained when the sum of the squares
+of the two sides is a square number. Thus, if the garden measured 12
+poles by 5 poles (where the squares of 12 and 5, 144 and 25, sum to
+169, the square of 13), then 12 added to 5, less 13, would equal four,
+and a quarter of this, 1 pole, would be the width of the path.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_196_THE_TETHERED_GOATa" id="X_196_THE_TETHERED_GOATa"></a><a href="#X_196_THE_TETHERED_GOAT"><b>196.&mdash;THE TETHERED GOAT.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a196.png" width="400" height="370" alt="" title="" />
+</div>
+
+<p>This problem is quite simple if properly attacked. Let us suppose the
+triangle ABC to represent our half-acre field, and the shaded portion
+to be the quarter-acre over which the goat will graze when tethered to
+the corner C. Now, as six equal equilateral triangles placed together
+will form a regular hexagon, as shown, it is evident that the shaded
+pasture is just one-sixth of the complete area of a circle. Therefore
+all we require is the radius (CD) of a circle containing six
+quarter-acres or 1� acres, which is equal to 9,408,960 square
+inches. As we only want our answer "to the nearest inch," it is
+sufficiently exact for our purpose if we assume that as 1 is to
+3.1416, so is the diameter of a circle to its circumference. If,
+therefore, we divide the last number I gave by 3.1416, and extract the
+square root, we find that 1,731 inches, or 48 yards 3 inches, is the
+required length of the tether "to the nearest inch."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_197_THE_COMPASSES_PUZZLEa" id="X_197_THE_COMPASSES_PUZZLEa"></a><a href="#X_197_THE_COMPASSES_PUZZLE"><b>197.&mdash;THE COMPASSES PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Let AB in the following diagram be the given straight line. With the
+centres A and B and radius AB describe the two circles. Mark off DE
+and EF equal to AD. With the centres A and F and radius DF describe
+arcs intersecting at G. With the centres A and B and distance BG
+describe arcs GHK and N. Make HK equal to AB and HL equal to HB. Then
+with centres K and L and radius AB describe arcs intersecting at I.
+Make BM equal to BI. Finally, with the centre M and radius MB cut the
+line in C, and the point C is the required middle of the line AB. For
+greater exactitude you can mark off R from A (as you did M from B),
+and from R describe another arc at C. This also solves the problem, to
+find a point midway between two given points without the straight
+line.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a197.png" width="400" height="326" alt="" title="" />
+</div>
+
+<p>I will put the young geometer in the way of a rigid proof. First prove
+that twice the square of the line AB equals the square of the distance
+BG, from which it follows that HABN are the four corners of a square.
+To prove that I is the centre of this square, draw a line from H to P
+through QIB and continue the arc HK to P. Then, conceiving the
+necessary lines to be drawn, the angle HKP, being in a semicircle, is
+a right angle. Let fall the perpendicular KQ, and by similar
+triangles, and from the fact that HKI is an isosceles triangle by the
+construction, it can be proved that HI is half of HB. We can similarly
+prove that C is the centre of the square of which AIB are three
+corners.</p>
+
+<p>I am aware that this is not the simplest possible solution.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_198_THE_EIGHT_STICKSa" id="X_198_THE_EIGHT_STICKSa"></a><a href="#X_198_THE_EIGHT_STICKS"><b>198.&mdash;THE EIGHT STICKS.&mdash;<i>solution</i></b></a></p>
+
+<p>The first diagram is the answer that nearly every one will give to
+this puzzle, and at first sight it seems quite satisfactory. But
+consider the conditions. We have to lay "every one of the sticks on
+the table." Now, if a ladder be <span class='pagenum'>Pg 187<a name="Page_187" id="Page_187"></a></span>placed against a wall with only one
+end on the ground, it can hardly be said that it is "laid on the
+ground." And if we place the sticks in the above manner, it is only
+possible to make one end of two of them touch the table: to say that
+every one lies on the table would not be correct. To obtain a solution
+it is only necessary to have our sticks of proper dimensions. Say the
+long sticks are each 2 ft. in length and the short ones 1 ft. Then the
+sticks must be 3 in. thick, when the three equal squares may be
+enclosed, as shown in the second diagram. If I had said "matches"
+instead of "sticks," the puzzle would be impossible, because an
+ordinary match is about twenty-one times as long as it is broad, and
+the enclosed rectangles would not be squares.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a198.png" width="400" height="311" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_199_PAPAS_PUZZLEa" id="X_199_PAPAS_PUZZLEa"></a><a href="#X_199_PAPAS_PUZZLE"><b>199.&mdash;PAPA'S PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>I have found that a large number of people imagine that the following
+is a correct solution of the problem. Using the letters in the diagram
+below, they argue that if you make the distance BA one-third of BC,
+and therefore the area of the rectangle ABE equal to that of the
+triangular remainder, the card must hang with the long side
+horizontal. Readers will remember the jest of Charles II., who induced
+the Royal Society to meet and discuss the reason why the water in a
+vessel will not rise if you put a live fish in it; but in the middle
+of the proceedings one of the least distinguished among them quietly
+slipped out and made the experiment, when he found that the water
+<i>did</i> rise! If my correspondents had similarly made the experiment
+with a piece of cardboard, they would have found at once their error.
+Area is one thing, but gravitation is quite another. The fact of that
+triangle sticking its leg out to D has to be compensated for by
+additional area in the rectangle. As a matter of fact, the ratio of BA
+to AC is as 1 is to the square root of 3, which latter cannot be given
+in an exact numerical measure, but is approximately 1.732. Now let us
+look at the correct general solution. There are many ways of arriving
+at the desired result, but the one I give is, I think, the simplest
+for beginners.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a199.png" width="400" height="342" alt="" title="" />
+</div>
+
+<p>Fix your card on a piece of paper and draw the equilateral triangle
+BCF, BF and CF being equal to BC. Also mark off the point G so that DG
+shall equal DC. Draw the line CG and produce it until it cuts the line
+BF in H. If we now make HA parallel to BE, then A is the point from
+which our cut must be made to the corner D, as indicated by the dotted
+line.</p>
+
+<p>A curious point in connection with this problem is the fact that the
+position of the point A is independent of the side CD. The reason for
+this is more obvious in the solution I have given than in any other
+method that I have seen, and (although the problem may be solved with
+all the working on the cardboard) that is partly why I have preferred
+it. It will be seen at once that however much you may reduce the width
+of the card by bringing E nearer to B and D nearer to C, the line CG,
+being the diagonal of a square, will always lie in the same direction,
+and will cut BF in H. Finally, if you wish to get an approximate
+measure for the distance BA, all you have to do is to multiply the
+length of the card by the decimal .366. Thus, if the card were 7
+inches long, we get 7&nbsp;&times;&nbsp;.366&nbsp;=&nbsp;2.562, or a little more than 2&frac12; inches, for
+the distance from B to A.</p>
+
+<p>But the real joke of the puzzle is this: We have seen that the
+position of the point A is independent of the width of the card, and
+depends entirely on the length. Now, in the illustration it will be
+found that both cards have the same length; consequently all the
+little maid had to do was to lay the clipped card on top of the other
+one and mark off the point A at precisely the same distance from the
+top left-hand corner! So, after all, Pappus' puzzle, as he presented
+it to his little maid, was quite an infantile problem, when he was
+able to show her how to perform the feat without first introducing her
+to the elements of statics and geometry.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_200_A_KITE-FLYING_PUZZLEa" id="X_200_A_KITE-FLYING_PUZZLEa"></a><a href="#X_200_A_KITE-FLYING_PUZZLE"><b>200.&mdash;A KITE-FLYING PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Solvers of this little puzzle, I have generally found, may be roughly
+divided into two classes: those who get within a mile of the correct
+answer by means of more or less complex calcu<span class='pagenum'>Pg 188<a name="Page_188" id="Page_188"></a></span>lations, involving
+"<i>pi</i>," and those whose arithmetical kites fly hundreds and thousands
+of miles away from the truth. The comparatively easy method that I
+shall show does not involve any consideration of the ratio that the
+diameter of a circle bears to its circumference. I call it the
+"hat-box method."</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a200.png" width="400" height="469" alt="" title="" />
+</div>
+
+<p>Supposing we place our ball of wire, A, in a cylindrical hat-box, B,
+that exactly fits it, so that it touches the side all round and
+exactly touches the top and bottom, as shown in the illustration.
+Then, by an invariable law that should be known by everybody, that box
+contains exactly half as much again as the ball. Therefore, as the
+ball is 24 in. in diameter, a hat-box of the same circumference but
+two-thirds of the height (that is, 16 in. high) will have exactly the
+same contents as the ball.</p>
+
+<p>Now let us consider that this reduced hat-box is a cylinder of metal
+made up of an immense number of little wire cylinders close together
+like the hairs in a painter's brush. By the conditions of the puzzle
+we are allowed to consider that there are no spaces between the wires.
+How many of these cylinders one one-hundredth of an inch thick are
+equal to the large cylinder, which is 24 in. thick? Circles are to one
+another as the squares of their diameters. The square of <sup>1</sup>/<sub>100</sub> is
+<sup>1</sup>/<sub>100000</sub>, and the square of 24 is 576; therefore the large cylinder
+contains 5,760,000 of the little wire cylinders. But we have seen that
+each of these wires is 16 in. long; hence 16&nbsp;&times;&nbsp;5,760,000 = 92,160,000
+inches as the complete length of the wire. Reduce this to miles, and
+we get 1,454 miles 2,880 ft. as the length of the wire attached to the
+professor's kite.</p>
+
+<p>Whether a kite would fly at such a height, or support such a weight,
+are questions that do not enter into the problem.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_201_HOW_TO_MAKE_CISTERNSa" id="X_201_HOW_TO_MAKE_CISTERNSa"></a><a href="#X_201_HOW_TO_MAKE_CISTERNS"><b>201.&mdash;HOW TO MAKE CISTERNS.&mdash;<i>solution</i></b></a></p>
+
+<p>Here is a general formula for solving this problem. Call the two sides
+of the rectangle <i>a</i> and <i>b</i>. Then</p>
+
+<div style="font-size: larger;">
+<p class='center'><span class='su2'>( <i>a</i>&nbsp;+&nbsp;<i>b</i>&nbsp;-&nbsp;(<i>a</i><sup>2</sup>&nbsp;+&nbsp;<i>b</i><sup>2</sup>&nbsp;-&nbsp;<i>ab</i>)<sup>&frac12;</sup> )</span>/<sub>6</sub></p>
+</div>
+
+<p>equals the side of the little square pieces to cut away. The measurements given
+were 8 ft. by 3 ft., and the above rule gives 8 in. as the side of the
+square pieces that have to be cut away. Of course it will not always
+come out exact, as in this case (on account of that square root), but
+you can get as near as you like with decimals.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_202_THE_CONE_PUZZLEa" id="X_202_THE_CONE_PUZZLEa"></a><a href="#X_202_THE_CONE_PUZZLE"><b>202.&mdash;THE CONE PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The simple rule is that the cone must be cut at one-third of its
+altitude.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_203_CONCERNING_WHEELSa" id="X_203_CONCERNING_WHEELSa"></a><a href="#X_203_CONCERNING_WHEELS"><b>203.&mdash;CONCERNING WHEELS.&mdash;<i>solution</i></b></a></p>
+
+<p>If you mark a point A on the circumference of a wheel that runs on the
+surface of a level road, like an ordinary cart-wheel, the curve
+described by that point will be a common cycloid, as in Fig. 1. But if
+you mark a point B on the circumference of the flange of a
+locomotive-wheel, the curve will be a curtate cycloid, as in Fig. 2,
+terminating in nodes. Now, if we consider one of these nodes or loops,
+we shall see that "at any given moment" certain points at the bottom
+of the loop must be moving in the opposite direction to the train. As
+there is an infinite number of such points on the flange's
+circumference, there must be an infinite number of these loops being
+described while the train is in motion. In fact, at any given moment
+certain points on the flanges are always moving in a direction
+opposite to that in which the train is going.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a203.png" width="400" height="305" alt="" title="" />
+</div>
+
+<p>In the case of the two wheels, the wheel that runs round the
+stationary one makes two revolutions round its own centre. As both
+wheels are of the same size, it is obvious that if at the start we
+mark a point on the circumference of the upper wheel, at the very top,
+this point will be in contact with the lower wheel at its lowest part
+when half the journey has been made. Therefore this point is again at
+the top of the moving wheel, and one revolution has been made.
+Consequently there are two such revolutions in the complete journey.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_204_A_NEW_MATCH_PUZZLEa" id="X_204_A_NEW_MATCH_PUZZLEa"></a><a href="#X_204_A_NEW_MATCH_PUZZLE"><b>204.&mdash;A NEW MATCH PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>1. The easiest way is to arrange the eighteen matches as in Diagrams 1
+and 2, making the length of the perpendicular AB equal to a match and
+a half. Then, if the matches are an inch in <span class='pagenum'>Pg 189<a name="Page_189" id="Page_189"></a></span>length, Fig. 1 contains
+two square inches and Fig. 2 contains six square inches&mdash;4&nbsp;&times;&nbsp;1�. The
+second case (2) is a little more difficult to solve. The solution is
+given in Figs. 3 and 4. For the purpose of construction, place matches
+temporarily on the dotted lines. Then it will be seen that as 3
+contains five equal equilateral triangles and 4 contains fifteen
+similar triangles, one figure is three times as large as the other,
+and exactly eighteen matches are used.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a204.png" width="400" height="339" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_205_THE_SIX_SHEEP-PENSa" id="X_205_THE_SIX_SHEEP-PENSa"></a><a href="#X_205_THE_SIX_SHEEP-PENS"><b>205.&mdash;THE SIX SHEEP-PENS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 300px;">
+<img src="images/a205.png" width="300" height="274" alt="" title="" />
+</div>
+
+<p>Place the twelve matches in the manner shown in the
+illustration, and you will have six pens of equal size.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_206_THE_KING_AND_THE_CASTLESa" id="X_206_THE_KING_AND_THE_CASTLESa"></a><a href="#X_206_THE_KING_AND_THE_CASTLES"><b>206.&mdash;THE KING AND THE CASTLES.&mdash;<i>solution</i></b></a></p>
+
+<p>There are various ways of building the ten castles so that they shall
+form five rows with four castles in every row, but the arrangement in
+the next column is the only one that also provides that two castles
+(the greatest number possible) shall not be approachable from the
+outside. It will be seen that you must cross the walls to reach these
+two.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a206.png" width="400" height="565" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_207_CHERRIES_AND_PLUMSa" id="X_207_CHERRIES_AND_PLUMSa"></a><a href="#X_207_CHERRIES_AND_PLUMS"><b>207.&mdash;CHERRIES AND PLUMS.&mdash;<i>solution</i></b></a></p>
+
+<p>There are several ways in which this problem might be solved were it
+not for the condition that as few cherries and plums as possible shall
+be planted on the north and east sides of the orchard. The best
+possible arrangement is that shown in the diagram, where the cherries,
+plums, and apples are indicated respectively by the letters C, P, and
+A. The dotted lines connect the cherries, and the other lines the
+plums. It will be seen that the ten cherry trees and the ten plum
+trees are so planted that each fruit forms five lines with four trees
+of its kind in line. This is the only arrangement that allows of so
+few as two cherries or plums being planted on the north and east
+outside rows.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a207.png" width="400" height="388" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_208_A_PLANTATION_PUZZLEa" id="X_208_A_PLANTATION_PUZZLEa"></a><a href="#X_208_A_PLANTATION_PUZZLE"><b>208.&mdash;A PLANTATION PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The illustration shows the ten trees that must be left to form five
+rows with four trees in every <span class='pagenum'>Pg 190<a name="Page_190" id="Page_190"></a></span>row. The dots represent the positions
+of the trees that have been cut down.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a208.png" width="400" height="402" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_209_THE_TWENTY-ONE_TREESa" id="X_209_THE_TWENTY-ONE_TREESa"></a><a href="#X_209_THE_TWENTY-ONE_TREES"><b>209.&mdash;THE TWENTY-ONE TREES.&mdash;<i>solution</i></b></a></p>
+
+<p>I give two pleasing arrangements of the trees. In each case there are
+twelve straight rows with five trees in every row.</p>
+
+<div class="figcenter" style="width: 300px;">
+<img src="images/a209a.png" width="300" height="302" alt="" title="" />
+</div>
+
+<div class="figcenter" style="width: 300px;">
+<img src="images/a209b.png" width="300" height="291" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_210_THE_TEN_COINSa" id="X_210_THE_TEN_COINSa"></a><a href="#X_210_THE_TEN_COINS"><b>210.&mdash;THE TEN COINS.&mdash;<i>solution</i></b></a></p>
+
+<p>The answer is that there are just 2,400 different ways. Any three
+coins may be taken from one side to combine with one coin taken from
+the other side. I give four examples on this and the next page. We may
+thus select three from the top in ten ways and one from the bottom in
+five ways, making fifty. But we may also select three from the bottom
+and one from the top in fifty ways. We may thus select the four coins
+in one hundred ways, and the four removed may be arranged by
+permutation in twenty-four ways. Thus there are 24&nbsp;&times;&nbsp;100&nbsp;=&nbsp;2,400
+different solutions.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a210a.png" width="400" height="390" alt="" title="" />
+</div>
+
+
+<p>As all the points and lines puzzles that I have given so far,
+excepting the last, are variations of the case of ten points arranged
+to form five lines of four, it will be well to consider this
+particular case generally. There are six fundamental solutions, and no
+more, as shown in the six diagrams. These, for the sake of
+convenience, I named some years ago the Star, the Dart, the Compasses,
+the Funnel, the Scissors, and the Nail. (See next page.) Readers will
+understand that any one of these forms may be distorted in an infinite
+number of different ways without destroying its real character.</p>
+
+<p>In "The King and the Castles" we have the Star, and its solution gives
+the Compasses. In the "Cherries and Plums" solution we find that the
+Cherries represent the Funnel and the Plums the Dart. The solution of
+the "Plantation Puzzle" is an example of the Dart distorted. Any
+solution to the "Ten Coins" will represent the Scissors. Thus examples
+of all have been given except the Nail.</p>
+
+<p><span class='pagenum'>Pg 191<a name="Page_191" id="Page_191"></a></span></p><div class="figcenter" style="width: 600px;">
+<img src="images/a210b.png" width="600" height="164" alt="" title="" />
+</div>
+
+<p>On a reduced chessboard, 7 by 7, we may place the ten pawns in just
+three different ways, but they must all represent the Dart. The
+"Plantation" shows one way, the Plums show a second way, and the
+reader may like to find the third way for himself. On an ordinary
+chessboard, 8 by 8, we can also get in a beautiful example of the
+Funnel&mdash;symmetrical in relation to the diagonal of the board. The
+smallest board that will take a Star is one 9 by 7. The Nail requires
+a board 11 by 7, the Scissors 11 by 9, and the Compasses 17 by 12. At least these are the best
+results recorded in my note-book. They may be beaten, but I do not
+think so. If you divide a chessboard into two parts by a diagonal
+zigzag line, so that the larger part contains 36 squares and the
+smaller part 28 squares, you can place three separate schemes on the
+larger part and one on the smaller part (all Darts) without their
+conflicting&mdash;that is, they occupy forty different squares. They can be
+placed in other ways without a division of the board. The smallest
+square board that will contain six different schemes (not
+fundamentally different), without any line of one scheme crossing the
+line of another, is 14 by 14; and the smallest board that will contain
+one scheme entirely enclosed within the lines of a second scheme,
+without any of the lines of the one, when drawn from point to point,
+crossing a line of the other, is 14 by 12.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a210c.png" width="600" height="131" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_211_THE_TWELVE_MINCE-PIESa" id="X_211_THE_TWELVE_MINCE-PIESa"></a><a href="#X_211_THE_TWELVE_MINCE-PIES"><b>211.&mdash;THE TWELVE MINCE-PIES.&mdash;<i>solution</i></b></a></p>
+
+<p>If you ignore the four black pies in our illustration, the remaining
+twelve are in their original positions. Now remove the four detached
+pies to the places occupied by the black ones, and you will have your
+seven straight rows of four, as shown by the dotted lines.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a211.png" width="400" height="502" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_212_THE_BURMESE_PLANTATIONa" id="X_212_THE_BURMESE_PLANTATIONa"></a><a href="#X_212_THE_BURMESE_PLANTATION"><b>212.&mdash;THE BURMESE PLANTATION.&mdash;<i>solution</i></b></a></p>
+
+<p>The arrangement on the next page is the most symmetrical answer that
+can probably be found for twenty-one rows, which is, I believe, the
+greatest number of rows possible. There are several ways of doing it.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a212.png" width="400" height="413" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_213_TURKS_AND_RUSSIANSa" id="X_213_TURKS_AND_RUSSIANSa"></a><a href="#X_213_TURKS_AND_RUSSIANS"><b>213.&mdash;TURKS AND RUSSIANS.&mdash;<i>solution</i></b></a></p>
+
+<p>The main point is to discover the smallest possible number of Russians
+that there could have been. As the enemy opened fire from all
+directions, it is clearly necessary to find what is the smallest
+number of heads that could form sixteen lines with three heads in
+every line. Note that I say sixteen, and not thirty-two, because every
+line taken by a bullet may be also taken by another bullet fired in
+exactly the opposite direction. Now, as few as eleven points, or
+heads, may be arranged to form the required sixteen lines of three,
+but the discovery of this arrangement is a hard nut. The diagram<span class='pagenum'>Pg 192<a name="Page_192" id="Page_192"></a></span>
+at the foot of this page will show exactly how the thing is to be
+done.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a213.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>If, therefore, eleven Russians were in the positions shown by the
+stars, and the thirty-two Turks in the positions indicated by the
+black dots, it will be seen, by the lines shown, that each Turk may
+fire exactly over the heads of three Russians. But as each bullet
+kills a man, it is essential that every Turk shall shoot one of his
+comrades and be shot by him in turn; otherwise we should have to
+provide extra Russians to be shot, which would be destructive of the
+correct solution of our problem. As the firing was simultaneous, this
+point presents no difficulties. The answer we thus see is that there
+were at least eleven Russians amongst whom there was no casualty, and
+that all the thirty-two Turks were shot by one another. It was not
+stated whether the Russians fired any shots, but it will be evident
+that even if they did their firing could not have been effective: for
+if one of their bullets killed a Turk, then we have immediately to
+provide another man for one of the Turkish bullets to kill; and as the
+Turks were known to be thirty-two in number, this <span class='pagenum'>Pg 193<a name="Page_193" id="Page_193"></a></span>would necessitate
+our introducing another Russian soldier and, of course, destroying the
+solution. I repeat that the difficulty of the puzzle consists in
+finding how to arrange eleven points so that they shall form sixteen
+lines of three. I am told that the possibility of doing this was first
+discovered by the Rev. Mr. Wilkinson some twenty years ago.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_214_THE_SIX_FROGSa" id="X_214_THE_SIX_FROGSa"></a><a href="#X_214_THE_SIX_FROGS"><b>214.&mdash;THE SIX FROGS.&mdash;<i>solution</i></b></a></p>
+
+<p>Move the frogs in the following order: 2, 4, 6, 5, 3, 1 (repeat these
+moves in the same order twice more), 2, 4, 6. This is a solution in
+twenty-one moves&mdash;the fewest possible.</p>
+
+<p>If <i>n</i>, the number of frogs, be even, we require <sup>(<i>n</i>&sup2;+<i>n</i>)</sup>/<sub>2</sub> moves,
+of which <sup>(<i>n</i>&sup2;-<i>n</i>)</sup>/<sub>2</sub> will be leaps and <i>n</i> simple moves. If <i>n</i> be
+odd, we shall need (<sup>(<i>n</i>&sup2;+3<i>n</i>)</sup>/<sub>2</sub>)-4 moves, of which <sup>(<i>n</i>&sup2;-<i>n</i>)</sup>/<sub>2</sub> will
+be leaps and 2<i>n</i>-4 simple moves.</p>
+
+<p>In the even cases write, for the moves, all the even numbers in
+ascending order and the odd numbers in descending order. This series
+must be repeated &frac12;<i>n</i> times and followed by the even numbers in
+ascending order once only. Thus the solution for 14 frogs will be (2,
+4, 6, 8, 10, 12, 14, 13, 11, 9, 7, 5, 3, 1) repeated 7 times and
+followed by 2, 4, 6, 8, 10, 12, 14&nbsp;=&nbsp;105 moves.</p>
+
+<p>In the odd cases, write the even numbers in ascending order and the
+odd numbers in descending order, repeat this series &frac12;(<i>n</i>-1) times,
+follow with the even numbers in ascending order (omitting <i>n</i>-1), the
+odd numbers in descending order (omitting 1), and conclude with all
+the numbers (odd and even) in their natural order (omitting 1 and
+<i>n</i>). Thus for 11 frogs: (2, 4, 6, 8, 10, 11, 9, 7, 5, 3, 1) repeated
+5 times, 2, 4, 6, 8, 11, 9, 7, 5, 3, and 2, 3, 4, 5, 6, 7, 8, 9, 10 =
+73 moves.</p>
+
+<p>This complete general solution is published here for the first time.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_215_THE_GRASSHOPPER_PUZZLEa" id="X_215_THE_GRASSHOPPER_PUZZLEa"></a><a href="#X_215_THE_GRASSHOPPER_PUZZLE"><b>215.&mdash;THE GRASSHOPPER PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Move the counters in the following order. The moves in brackets are to
+be made four times in succession. 12, 1, 3, 2, 12, 11, 1, 3, 2 (5, 7,
+9, 10, 8, 6, 4), 3, 2, 12, 11, 2, 1, 2. The grasshoppers will then be
+reversed in forty-four moves.</p>
+
+<p>The general solution of this problem is very difficult. Of course it
+can always be solved by the method given in the solution of the last
+puzzle, if we have no desire to use the fewest possible moves. But to
+employ a full economy of moves we have two main points to consider.
+There are always what I call a lower movement (L) and an upper
+movement (U). L consists in exchanging certain of the highest numbers,
+such as 12, 11, 10 in our "Grasshopper Puzzle," with certain of the
+lower numbers, 1, 2, 3; the former moving in a clockwise direction,
+the latter in a non-clockwise direction. U consists in reversing the
+intermediate counters. In the above solution for 12, it will be seen
+that 12, 11, and 1, 2, 3 are engaged in the L movement, and 4, 5, 6,
+7, 8, 9, 10 in the U movement. The L movement needs 16 moves and U 28,
+making together 44. We might also involve 10 in the L movement, which
+would result in L 23, U 21, making also together 44 moves. These I
+call the first and second methods. But any other scheme will entail an
+increase of moves. You always get these two methods (of equal economy)
+for odd or even counters, but the point is to determine just how many
+to involve in L and how many in U. Here is the solution in table form.
+But first note, in giving values to <i>n</i>, that 2, 3, and 4 counters are
+special cases, requiring respectively 3, 3, and 6 moves, and that 5
+and 6 counters do not give a minimum solution by the second
+method&mdash;only by the first.</p>
+
+<h5>FIRST METHOD.</h5>
+
+<div class='center'>
+<table border="1" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' rowspan='2'>Total No.<br />of<br />Counters.</td><td align='center' colspan='2'>L MOVEMENT.</td><td align='center' colspan='2'>U MOVEMENT. </td><td align='center' rowspan='2'>Total No.<br />of Moves.</td></tr>
+<tr><td align='center'>No. of <br />Counters.</td><td align='center'>No. of <br />Moves.</td><td align='center'>No. of <br />Counters.</td><td align='center'>No. of <br />Moves.</td></tr>
+<tr><td align='center'>4n</td><td align='center'>n&nbsp;-&nbsp;1 and n</td><td align='center'>2(n&nbsp;-&nbsp;1)&sup2;&nbsp;+&nbsp;5n&nbsp;-&nbsp;7</td><td align='center'>2n&nbsp;+&nbsp;1</td><td align='center'>2n&sup2;&nbsp;+&nbsp;3n&nbsp;+&nbsp;1</td><td align='center'>4(n&sup2;&nbsp;+&nbsp;n&nbsp;-&nbsp;1)</td></tr>
+<tr><td align='center'>4n&nbsp;-&nbsp;2</td><td align='center'>n&nbsp;-&nbsp;1 " n</td><td align='center'>2(n&nbsp;-&nbsp;1)&sup2;&nbsp;+&nbsp;5n&nbsp;-&nbsp;7</td><td align='center'>2n&nbsp;-&nbsp;1</td><td align='center'>2(n&nbsp;-&nbsp;1)&sup2;&nbsp;+&nbsp;3n&nbsp;-&nbsp;2</td><td align='center'>4n&sup2;&nbsp;-&nbsp;5</td></tr>
+<tr><td align='center'>4n&nbsp;+&nbsp;1</td><td align='center'>n " n&nbsp;+&nbsp;1</td><td align='center'>2n&sup2;&nbsp;+&nbsp;5n&nbsp;-&nbsp;2</td><td align='center'>2n</td><td align='center'>2n&sup2;&nbsp;+&nbsp;3n&nbsp;-&nbsp;4</td><td align='center'>2(2n&sup2;&nbsp;+&nbsp;4n&nbsp;-&nbsp;3)</td></tr>
+<tr><td align='center'>4n&nbsp;-&nbsp;1</td><td align='center'>n&nbsp;-&nbsp;1 " n</td><td align='center'>2(n&nbsp;-&nbsp;1)&sup2;&nbsp;+&nbsp;5n&nbsp;-&nbsp;7</td><td align='center'>2n</td><td align='center'>2n&sup2;&nbsp;+&nbsp;3n&nbsp;-&nbsp;4</td><td align='center'>4n&sup2;&nbsp;+&nbsp;4n&nbsp;-&nbsp;9</td></tr>
+</table></div>
+
+<h5>SECOND METHOD.</h5>
+
+<div class='center'>
+<table border="1" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' rowspan='2'>Total No.<br />of<br />Counters.</td><td align='center' colspan='2'>L MOVEMENT.</td><td align='center' colspan='2'>U MOVEMENT. </td><td align='center' rowspan='2'>Total No.<br />of Moves.</td></tr>
+<tr><td align='center'>No. of <br />Counters.</td><td align='center'>No. of <br />Moves.</td><td align='center'>No. of <br />Counters.</td><td align='center'>No. of <br />Moves.</td></tr>
+<tr><td align='center'>4n</td><td align='center'>n and n</td><td align='center'>2n&sup2;&nbsp;+&nbsp;3n&nbsp;-&nbsp;4</td><td align='center'>2n</td><td align='center'>2(n&nbsp;-&nbsp;1)&sup2;&nbsp;+&nbsp;5n&nbsp;-&nbsp;2</td><td align='center'>4(n&sup2;&nbsp;+&nbsp;n&nbsp;-&nbsp;1)</td></tr>
+<tr><td align='center'>4n&nbsp;-&nbsp;2</td><td align='center'>n&nbsp;-&nbsp;1 " n&nbsp;-&nbsp;1</td><td align='center'>2(n&nbsp;-&nbsp;1)&sup2;&nbsp;+&nbsp;3n&nbsp;-&nbsp;7</td><td align='center'>2n</td><td align='center'>2(n&nbsp;-&nbsp;1)&sup2;&nbsp;+&nbsp;5n&nbsp;-&nbsp;2</td><td align='center'>4n&sup2;&nbsp;-&nbsp;5</td></tr>
+<tr><td align='center'>4n&nbsp;+&nbsp;1</td><td align='center'>n " n</td><td align='center'>2n&sup2;&nbsp;+&nbsp;3n&nbsp;-&nbsp;4</td><td align='center'>2n&nbsp;+&nbsp;1</td><td align='center'>2n&sup2;&nbsp;+&nbsp;5n&nbsp;-&nbsp;2</td><td align='center'>2(2n&sup2;&nbsp;+&nbsp;4n&nbsp;-&nbsp;3)</td></tr>
+<tr><td align='center'>4n&nbsp;-&nbsp;1</td><td align='center'>n " n</td><td align='center'>2n&sup2;&nbsp;+&nbsp;3n&nbsp;-&nbsp;4</td><td align='center'>2n&nbsp;-&nbsp;1</td><td align='center'>2(n&nbsp;-&nbsp;1)&sup2;&nbsp;+&nbsp;5n&nbsp;-&nbsp;7</td><td align='center'>4n&sup2;&nbsp;+&nbsp;4n-9</td></tr>
+</table></div>
+
+<p><span class='pagenum'>Pg 194<a name="Page_194" id="Page_194"></a></span>More generally we may say that with <i>m</i> counters, where <i>m</i> is even
+and greater than 4, we require <sup>(m&sup2;&nbsp;+&nbsp;4m&nbsp;-&nbsp;16)</sup>/<sub>4</sub> moves; and where <i>m</i> is
+odd and greater than 3, <sup>(m&sup2;&nbsp;+&nbsp;6m&nbsp;-&nbsp;31)</sup>/<sub>4</sub> moves. I have thus shown the
+reader how to find the minimum number of moves for any case, and the
+character and direction of the moves. I will leave him to discover for
+himself how the actual order of moves is to be determined. This is a
+hard nut, and requires careful adjustment of the L and the U
+movements, so that they may be mutually accommodating.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_216_THE_EDUCATED_FROGSa" id="X_216_THE_EDUCATED_FROGSa"></a><a href="#X_216_THE_EDUCATED_FROGS"><b>216.&mdash;THE EDUCATED FROGS.&mdash;<i>solution</i></b></a></p>
+
+<p>The following leaps solve the puzzle in ten moves: 2 to 1, 5 to 2, 3
+to 5, 6 to 3, 7 to 6, 4 to 7, 1 to 4, 3 to 1, 6 to 3, 7 to 6.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_217_THE_TWICKENHAM_PUZZLEa" id="X_217_THE_TWICKENHAM_PUZZLEa"></a><a href="#X_217_THE_TWICKENHAM_PUZZLE"><b>217.&mdash;THE TWICKENHAM PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Play the counters in the following order: K C E K W T C E H M K W T A
+N C E H M I K C E H M T, and there you are, at Twickenham. The
+position itself will always determine whether you are to make a leap
+or a simple move.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_218_THE_VICTORIA_CROSS_PUZZLEa" id="X_218_THE_VICTORIA_CROSS_PUZZLEa"></a><a href="#X_218_THE_VICTORIA_CROSS_PUZZLE"><b>218.&mdash;THE VICTORIA CROSS PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>In solving this puzzle there were two things to be achieved: first, so
+to manipulate the counters that the word VICTORIA should read round
+the cross in the same direction, only with the V on one of the dark
+arms; and secondly, to perform the feat in the fewest possible moves.
+Now, as a matter of fact, it would be impossible to perform the first
+part in any way whatever if all the letters of the word were
+different; but as there are two I's, it can be done by making these
+letters change places&mdash;that is, the first I changes from the 2nd place
+to the 7th, and the second I from the 7th place to the 2nd. But the
+point I referred to, when introducing the puzzle, as a little
+remarkable is this: that a solution in twenty-two moves is obtainable
+by moving the letters in the order of the following words: "A VICTOR!
+A VICTOR! A VICTOR I!"</p>
+
+<p>There are, however, just six solutions in eighteen moves, and the
+following is one of them: I (1), V, A, I (2), R, O, T, I (1), I (2),
+A, V, I (2), I (1), C, I (2), V, A, I (1). The first and second I in
+the word are distinguished by the numbers 1 and 2.</p>
+
+<p>It will be noticed that in the first solution given above one of the
+I's never moves, though the movements of the other letters cause it to
+change its relative position. There is another peculiarity I may point
+out&mdash;that there is a solution in twenty-eight moves requiring no
+letter to move to the central division except the I's. I may also
+mention that, in each of the solutions in eighteen moves, the letters
+C, T, O, R move once only, while the second I always moves four times,
+the V always being transferred to the right arm of the cross.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_219_THE_LETTER_BLOCK_PUZZLEa" id="X_219_THE_LETTER_BLOCK_PUZZLEa"></a><a href="#X_219_THE_LETTER_BLOCK_PUZZLE"><b>219.&mdash;THE LETTER BLOCK PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>This puzzle can be solved in 23 moves&mdash;the fewest possible. Move the
+blocks in the following order: A, B, F, E, C, A, B, F, E, C, A, B, D,
+H, G, A, B, D, H, G, D, E, F.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_220_A_LODGING-HOUSE_DIFFICULTYa" id="X_220_A_LODGING-HOUSE_DIFFICULTYa"></a><a href="#X_220_A_LODGING-HOUSE_DIFFICULTY"><b>220.&mdash;A LODGING-HOUSE DIFFICULTY.&mdash;<i>solution</i></b></a></p>
+
+<p>The shortest possible way is to move the articles in the following
+order: Piano, bookcase, wardrobe, piano, cabinet, chest of drawers,
+piano, wardrobe, bookcase, cabinet, wardrobe, piano, chest of drawers,
+wardrobe, cabinet, bookcase, piano. Thus seventeen removals are
+necessary. The landlady could then move chest of drawers, wardrobe,
+and cabinet. Mr. Dobson did not mind the wardrobe and chest of drawers
+changing rooms so long as he secured the piano.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_221_THE_EIGHT_ENGINESa" id="X_221_THE_EIGHT_ENGINESa"></a><a href="#X_221_THE_EIGHT_ENGINES"><b>221.&mdash;THE EIGHT ENGINES.&mdash;<i>solution</i></b></a></p>
+
+<p>The solution to the Eight Engines Puzzle is as follows: The engine
+that has had its fire drawn and therefore cannot move is No. 5. Move
+the other engines in the following order: 7, 6, 3, 7, 6, 1, 2, 4, 1,
+3, 8, 1, 3, 2, 4, 3, 2, seventeen moves in all, leaving the eight
+engines in the required order.</p>
+
+<p>There are two other slightly different solutions.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_222_A_RAILWAY_PUZZLEa" id="X_222_A_RAILWAY_PUZZLEa"></a><a href="#X_222_A_RAILWAY_PUZZLE"><b>222.&mdash;A RAILWAY PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>This little puzzle may be solved in as few as nine moves. Play the
+engines as follows: From 9 to 10, from 6 to 9, from 5 to 6, from 2 to
+5, from 1 to 2, from 7 to 1, from 8 to 7, from 9 to 8, and from 10 to
+9. You will then have engines A, B, and C on each of the three circles
+and on each of the three straight lines. This is the shortest solution
+that is possible.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_223_A_RAILWAY_MUDDLEa" id="X_223_A_RAILWAY_MUDDLEa"></a><a href="#X_223_A_RAILWAY_MUDDLE"><b>223.&mdash;A RAILWAY MUDDLE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a223.png" width="400" height="325" alt="" title="" />
+</div>
+
+<p>Only six reversals are necessary. The white train (from A to D) is
+divided into three sections, engine and 7 wagons, 8 wagons, and 1
+wagon. The black train (D to A) never uncouples anything throughout.
+Fig. 1 is original position <span class='pagenum'>Pg 195<a name="Page_195" id="Page_195"></a></span>with 8 and 1 uncoupled. The black train
+proceeds to position in Fig. 2 (no reversal). The engine and 7 proceed
+towards D, and black train backs, leaves 8 on loop, and takes up
+position in Fig. 3 (first reversal). Black train goes to position in
+Fig. 4 to fetch single wagon (second reversal). Black train pushes 8
+off loop and leaves single wagon there, proceeding on its journey, as
+in Fig. 5 (third and fourth reversals). White train now backs on to
+loop to pick up single car and goes right away to D (fifth and sixth
+reversals).</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_224_THE_MOTOR-GARAGE_PUZZLEa" id="X_224_THE_MOTOR-GARAGE_PUZZLEa"></a><a href="#X_224_THE_MOTOR-GARAGE_PUZZLE"><b>224.&mdash;THE MOTOR-GARAGE PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The exchange of cars can be made in forty-three moves, as follows:
+6-G, 2-B, 1-E, 3-H, 4-I, 3-L, 6-K, 4-G, 1-I, 2-J, 5-H, 4-A, 7-F, 8-E,
+4-D, 8-C, 7-A, 8-G, 5-C, 2-B, 1-E, 8-I, 1-G, 2-J, 7-H, 1-A, 7-G, 2-B,
+6-E, 3-H, 8-L, 3-I, 7-K, 3-G, 6-I, 2-J, 5-H, 3-C, 5-G, 2-B, 6-E, 5-I,
+6-J. Of course, "6-G" means that the car numbered "6" moves to the
+point "G." There are other ways in forty-three moves.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_225_THE_TEN_PRISONERSa" id="X_225_THE_TEN_PRISONERSa"></a><a href="#X_225_THE_TEN_PRISONERS"><b>225.&mdash;THE TEN PRISONERS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a225.png" width="400" height="394" alt="" title="" />
+</div>
+
+<p>It will be seen in the illustration how the prisoners may be arranged
+so as to produce as many as sixteen even rows. There are 4 such
+vertical rows, 4 horizontal rows, 5 diagonal rows in one direction,
+and 3 diagonal rows in the other direction. The arrows here show the
+movements of the four prisoners, and it will be seen that the infirm
+man in the bottom corner has not been moved.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_226_ROUND_THE_COASTa" id="X_226_ROUND_THE_COASTa"></a><a href="#X_226_ROUND_THE_COAST"><b>226.&mdash;ROUND THE COAST.&mdash;<i>solution</i></b></a></p>
+
+<p>In order to place words round the circle under the conditions, it is
+necessary to select words in which letters are repeated in certain
+relative positions. Thus, the word that solves our puzzle is
+"Swansea," in which the first and fifth letters are the same, and the
+third and seventh the same. We make out jumps as follows, taking the
+letters of the word in their proper order: 2-5, 7-2, 4-7, 1-4, 6-1,
+3-6, 8-3. Or we could place a word like "Tarapur" (in which the second
+and fourth letters, and the third and seventh, are alike) with these
+moves: 6-1, 7-4, 2-7, 5-2, 8-5, 3-6, 8-3. But "Swansea" is the only
+word, apparently, that will fulfil the conditions of the puzzle.</p>
+
+<p>This puzzle should be compared with Sharp's Puzzle, referred to in my
+solution to No. 341, "The Four Frogs." The condition "touch and jump
+over two" is identical with "touch and move along a line."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_227_CENTRAL_SOLITAIREa" id="X_227_CENTRAL_SOLITAIREa"></a><a href="#X_227_CENTRAL_SOLITAIRE"><b>227.&mdash;CENTRAL SOLITAIRE.&mdash;<i>solution</i></b></a></p>
+
+<p>Here is a solution in nineteen moves; the moves enclosed in brackets
+count as one move only: 19-17, 16-18, (29-17, 17-19), 30-18, 27-25,
+(22-24, 24-26), 31-23, (4-16, 16-28), 7-9, 10-8, 12-10, 3-11, 18-6,
+(1-3, 3-11), (13-27, 27-25), (21-7, 7-9), (33-31, 31-23), (10-8, 8-22,
+22-24, 24-26, 26-12, 12-10), 5-17. All the counters are now removed
+except one, which is left in the central hole. The solution needs
+judgment, as one is tempted to make several jumps in one move, where
+it would be the reverse of good play. For example, after playing the
+first 3-11 above, one is inclined to increase the length of the move
+by continuing with 11-25, 25-27, or with 11-9, 9-7.</p>
+
+<p>I do not think the number of moves can be reduced.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_228_THE_TEN_APPLESa" id="X_228_THE_TEN_APPLESa"></a><a href="#X_228_THE_TEN_APPLES"><b>228.&mdash;THE TEN APPLES.&mdash;<i>solution</i></b></a></p>
+
+<p>Number the plates (1, 2, 3, 4), (5, 6, 7, 8), (9, 10, 11, 12), (13,
+14, 15, 16) in successive rows from the top to the bottom. Then
+transfer the apple from 8 to 10 and play as follows, always removing
+the apple jumped over: 9-11, 1-9, 13-5, 16-8, 4-12, 12-10, 3-1, 1-9,
+9-11.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_229_THE_NINE_ALMONDSa" id="X_229_THE_NINE_ALMONDSa"></a><a href="#X_229_THE_NINE_ALMONDS"><b>229.&mdash;THE NINE ALMONDS.&mdash;<i>solution</i></b></a></p>
+
+<p>This puzzle may be solved in as few as four moves, in the following
+manner: Move 5 over 8, 9, 3, 1. Move 7 over 4. Move 6 over 2 and 7.
+Move 5 over 6, and all the counters are removed except 5, which is
+left in the central square that it originally occupied.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_230_THE_TWELVE_PENNIESa" id="X_230_THE_TWELVE_PENNIESa"></a><a href="#X_230_THE_TWELVE_PENNIES"><b>230.&mdash;THE TWELVE PENNIES.&mdash;<i>solution</i></b></a></p>
+
+<p>Here is one of several solutions. Move 12 to 3, 7 to 4, 10 to 6, 8 to
+1, 9 to 5, 11 to 2.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_231_PLATES_AND_COINSa" id="X_231_PLATES_AND_COINSa"></a><a href="#X_231_PLATES_AND_COINS"><b>231.&mdash;PLATES AND COINS.&mdash;<i>solution</i></b></a></p>
+
+<p>Number the plates from 1 to 12 in the order that the boy is seen to be
+going in the illustration. Starting from 1, proceed as follows, where
+"1 to 4" means that you take the coin from plate No. 1 and transfer it
+to plate No. 4: 1 to 4, 5 to 8, 9 to 12, 3 to 6, 7 to 10, 11 to 2, and
+complete the last revolution to 1, making three revolutions in all. Or
+you can proceed this way: 4 to 7, 8 to 11, 12 to 3, 2 to 5, 6 to 9, 10
+to 1. It is easy to solve in four revolutions, but the solutions in
+three are more difficult to discover.</p>
+
+<p>This is "The Riddle of the Fishpond" (No. 41, <i>Canterbury Puzzles</i>) in
+a different dress.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 196<a name="Page_196" id="Page_196"></a></span><a name="X_232_CATCHING_THE_MICEa" id="X_232_CATCHING_THE_MICEa"></a><a href="#X_232_CATCHING_THE_MICE"><b>232.&mdash;CATCHING THE MICE.&mdash;<i>solution</i></b></a></p>
+
+<p>In order that the cat should eat every thirteenth mouse, and the white
+mouse last of all, it is necessary that the count should begin at the
+seventh mouse (calling the white one the first)&mdash;that is, at the one
+nearest the tip of the cat's tail. In this case it is not at all
+necessary to try starting at all the mice in turn until you come to
+the right one, for you can just start anywhere and note how far
+distant the last one eaten is from the starting point. You will find
+it to be the eighth, and therefore must start at the eighth, counting
+backwards from the white mouse. This is the one I have indicated.</p>
+
+<p>In the case of the second puzzle, where you have to find the smallest
+number with which the cat may start at the white mouse and eat this
+one last of all, unless you have mastered the general solution of the
+problem, which is very difficult, there is no better course open to
+you than to try every number in succession until you come to one that
+works correctly. The smallest number is twenty-one. If you have to
+proceed by trial, you will shorten your labour a great deal by only
+counting out the remainders when the number is divided successively by
+13, 12, 11, 10, etc. Thus, in the case of 21, we have the remainders
+8, 9, 10, 1, 3, 5, 7, 3, 1, 1, 3, 1, 1. Note that I do not give the
+remainders of 7, 3, and 1 as nought, but as 7, 3, and 1. Now, count
+round each of these numbers in turn, and you will find that the white
+mouse is killed last of all. Of course, if we wanted simply any
+number, not the smallest, the solution is very easy, for we merely
+take the least common multiple of 13, 12, 11, 10, etc. down to 2. This
+is 360360, and you will find that the first count kills the thirteenth
+mouse, the next the twelfth, the next the eleventh, and so on down to
+the first. But the most arithmetically inclined cat could not be
+expected to take such a big number when a small one like twenty-one
+would equally serve its purpose.</p>
+
+<p>In the third case, the smallest number is 100. The number 1,000 would
+also do, and there are just seventy-two other numbers between these
+that the cat might employ with equal success.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_233_THE_ECCENTRIC_CHEESEMONGERa" id="X_233_THE_ECCENTRIC_CHEESEMONGERa"></a><a href="#X_233_THE_ECCENTRIC_CHEESEMONGER"><b>233.&mdash;THE ECCENTRIC CHEESEMONGER.&mdash;<i>solution</i></b></a></p>
+
+<p>To leave the three piles at the extreme ends of the rows, the cheeses
+may be moved as follows&mdash;the numbers refer to the cheeses and not to
+their positions in the row: 7-2, 8-7, 9-8, 10-15, 6-10, 5-6, 14-16,
+13-14, 12-13, 3-1, 4-3, 11-4. This is probably the easiest solution of
+all to find. To get three of the piles on cheeses 13, 14, and 15, play
+thus: 9-4, 10-9, 11-10, 6-14, 5-6, 12-15, 8-12, 7-8, 16-5, 3-13, 2-3,
+1-2. To leave the piles on cheeses 3, 5, 12, and 14, play thus: 8-3,
+9-14, 16-12, 1-5, 10-9, 7-10, 11-8, 2-1, 4-16, 13-2, 6-11, 15-4.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_234_THE_EXCHANGE_PUZZLEa" id="X_234_THE_EXCHANGE_PUZZLEa"></a><a href="#X_234_THE_EXCHANGE_PUZZLE"><b>234.&mdash;THE EXCHANGE PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Make the following exchanges of pairs: H-K, H-E, H-C, H-A, I-L, I-F,
+I-D, K-L, G-J, J-A, F-K, L-E, D-K, E-F, E-D, E-B, B-K. It will be
+found that, although the white counters can be moved to their proper
+places in 11 moves, if we omit all consideration of exchanges, yet the
+black cannot be so moved in fewer than 17 moves. So we have to
+introduce waste moves with the white counters to equal the minimum
+required by the black. Thus fewer than 17 moves must be impossible.
+Some of the moves are, of course, interchangeable.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_235_TORPEDO_PRACTICEa" id="X_235_TORPEDO_PRACTICEa"></a><a href="#X_235_TORPEDO_PRACTICE"><b>235.&mdash;TORPEDO PRACTICE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a235.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>If the enemy's fleet be anchored in the formation shown in the
+illustration, it will be seen that as many as ten out of the sixteen
+ships may be blown up by discharging the torpedoes in the order
+indicated by the numbers and in the directions indicated by the
+arrows. As each torpedo in succession passes under three ships and
+sinks the fourth, strike out each vessel with the pencil as it is
+sunk.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_236_THE_HAT_PUZZLEa" id="X_236_THE_HAT_PUZZLEa"></a><a href="#X_236_THE_HAT_PUZZLE"><b>236.&mdash;THE HAT PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a236.png" width="400" height="216" alt="" title="" />
+</div>
+
+<p>I suggested that the reader should try this puzzle with counters, so I
+give my solution in that form. The silk hats are represented by black
+counters and the felt hats by white counters. The first row shows the
+hats in their original positions, and then each succes<span class='pagenum'>Pg 197<a name="Page_197" id="Page_197"></a></span>sive row shows
+how they appear after one of the five manipulations. It will thus be
+seen that we first move hats 2 and 3, then 7 and 8, then 4 and 5, then
+10 and 11, and, finally, 1 and 2, leaving the four silk hats together,
+the four felt hats together, and the two vacant pegs at one end of the
+row. The first three pairs moved are dissimilar hats, the last two
+pairs being similar. There are other ways of solving the puzzle.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_237_BOYS_AND_GIRLSa" id="X_237_BOYS_AND_GIRLSa"></a><a href="#X_237_BOYS_AND_GIRLS"><b>237.&mdash;BOYS AND GIRLS.&mdash;<i>solution</i></b></a></p>
+
+<p>There are a good many different solutions to this puzzle. Any
+contiguous pair, except 7-8, may be moved first, and after the first
+move there are variations. The following solution shows the position
+from the start right through each successive move to the end:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>.</td><td align='center'>.</td><td align='center'>1</td><td align='center'>2</td><td align='center'>3</td><td align='center'>4</td><td align='center'>5</td><td align='center'>6</td><td align='center'>7</td><td align='center'>8</td></tr>
+<tr><td align='center'>4</td><td align='center'>3</td><td align='center'>1</td><td align='center'>2</td><td align='center'>.</td><td align='center'>.</td><td align='center'>5</td><td align='center'>6</td><td align='center'>7</td><td align='center'>8</td></tr>
+<tr><td align='center'>4</td><td align='center'>3</td><td align='center'>1</td><td align='center'>2</td><td align='center'>7</td><td align='center'>6</td><td align='center'>5</td><td align='center'>.</td><td align='center'>.</td><td align='center'>8</td></tr>
+<tr><td align='center'>4</td><td align='center'>3</td><td align='center'>1</td><td align='center'>2</td><td align='center'>7</td><td align='center'>.</td><td align='center'>.</td><td align='center'>5</td><td align='center'>6</td><td align='center'>8</td></tr>
+<tr><td align='center'>4</td><td align='center'>.</td><td align='center'>.</td><td align='center'>2</td><td align='center'>7</td><td align='center'>1</td><td align='center'>3</td><td align='center'>5</td><td align='center'>6</td><td align='center'>8</td></tr>
+<tr><td align='center'>4</td><td align='center'>8</td><td align='center'>6</td><td align='center'>2</td><td align='center'>7</td><td align='center'>1</td><td align='center'>3</td><td align='center'>5</td><td align='center'>.</td><td align='center'>.</td></tr>
+</table></div>
+
+<hr style="width: 30%;" />
+<p><a name="X_238_ARRANGING_THE_JAMPOTSa" id="X_238_ARRANGING_THE_JAMPOTSa"></a><a href="#X_238_ARRANGING_THE_JAMPOTS"><b>238.&mdash;ARRANGING THE JAMPOTS.&mdash;<i>solution</i></b></a></p>
+
+<p>Two of the pots, 13 and 19, were in their proper places. As every
+interchange may result in a pot being put in its place, it is clear
+that twenty-two interchanges will get them all in order. But this
+number of moves is not the fewest possible, the correct answer being
+seventeen. Exchange the following pairs: (3-1, 2-3), (15-4, 16-15),
+(17-7, 20-17), (24-10, 11-24, 12-11), (8-5, 6-8, 21-6, 23-21, 22-23,
+14-22, 9-14, 18-9). When you have made the interchanges within any
+pair of brackets, all numbers within those brackets are in their
+places. There are five pairs of brackets, and 5 from 22 gives the
+number of changes required&mdash;17.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_239_A_JUVENILE_PUZZLEa" id="X_239_A_JUVENILE_PUZZLEa"></a><a href="#X_239_A_JUVENILE_PUZZLE"><b>239.&mdash;A JUVENILE PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a239.png" width="400" height="230" alt="" title="" />
+</div>
+
+<p>As the conditions are generally understood, this puzzle is incapable
+of solution. This can be demonstrated quite easily. So we have to look
+for some catch or quibble in the statement of what we are asked to do.
+Now if you fold the paper and then push the point of your pencil down
+between the fold, you can with one stroke make the two lines CD and EF
+in our diagram. Then start at A, and describe the line ending at B.
+Finally put in the last line GH, and the thing is done strictly within
+the conditions, since folding the paper is not actually forbidden. Of
+course the lines are here left unjoined for the purpose of clearness.</p>
+
+<p>In the rubbing out form of the puzzle, first rub out A to B with a
+single finger in one stroke. Then rub out the line GH with one finger.
+Finally, rub out the remaining two vertical lines with two fingers at
+once! That is the old trick.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_240_THE_UNION_JACKa" id="X_240_THE_UNION_JACKa"></a><a href="#X_240_THE_UNION_JACK"><b>240.&mdash;THE UNION JACK.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a240.png" width="400" height="288" alt="" title="" />
+</div>
+
+<p>There are just sixteen points (all on the outside) where three roads
+may be said to join. These are called by mathematicians "odd nodes."
+There is a rule that tells us that in the case of a drawing like the
+present one, where there are sixteen odd nodes, it requires eight
+separate strokes or routes (that is, half as many as there are odd
+nodes) to complete it. As we have to produce as much as possible with
+only one of these eight strokes, it is clearly necessary to contrive
+that the seven strokes from odd node to odd node shall be as short as
+possible. Start at A and end at B, or go the reverse way.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_241_THE_DISSECTED_CIRCLEa" id="X_241_THE_DISSECTED_CIRCLEa"></a><a href="#X_241_THE_DISSECTED_CIRCLE"><b>241.&mdash;THE DISSECTED CIRCLE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a241.png" width="400" height="388" alt="" title="" />
+</div>
+
+<p>It can be done in twelve continuous strokes, thus: Start at A in the
+illustration, and eight <span class='pagenum'>Pg 198<a name="Page_198" id="Page_198"></a></span>strokes, forming the star, will bring you
+back to A; then one stroke round the circle to B, one stroke to C, one
+round the circle to D, and one final stroke to E&mdash;twelve in all. Of
+course, in practice the second circular stroke will be over the first
+one; it is separated in the diagram, and the points of the star not
+joined to the circle, to make the solution clear to the eye.</p>
+<hr style="width: 30%;" />
+<p><a name="X_242_THE_TUBE_INSPECTORS_PUZZLEa" id="X_242_THE_TUBE_INSPECTORS_PUZZLEa"></a><a href="#X_242_THE_TUBE_INSPECTORS_PUZZLE"><b>242.&mdash;THE TUBE INSPECTOR'S PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The inspector need only travel nineteen miles if he starts at B and
+takes the following route: B A D G D E F I F C B E H K L I H G J K. Thus the only
+portions of line travelled over twice are the two sections D to G and
+F to I. Of course, the route may be varied, but it cannot be
+shortened.</p>
+<hr style="width: 30%;" />
+<p><a name="X_243_VISITING_THE_TOWNSa" id="X_243_VISITING_THE_TOWNSa"></a><a href="#X_243_VISITING_THE_TOWNS"><b>243.&mdash;VISITING THE TOWNS.&mdash;<i>solution</i></b></a></p>
+
+<p>Note that there are six towns, from which only two roads issue. Thus 1
+must lie between 9 and 12 in the circular route. Mark these two roads
+as settled. Similarly mark 9, 5, 14, and 4, 8, 14, and 10, 6, 15, and
+10, 2, 13, and 3, 7, 13. All these roads must be taken. Then you will
+find that he must go from 4 to 15, as 13 is closed, and that he is
+compelled to take 3, 11, 16, and also 16, 12. Thus, there is only one
+route, as follows: 1, 9, 5, 14, 8, 4, 15, 6, 10, 2, 13, 7, 3, 11, 16,
+12, 1, or its reverse&mdash;reading the line the other way. Seven roads are
+not used.</p>
+<hr style="width: 30%;" />
+<p><a name="X_244_THE_FIFTEEN_TURNINGSa" id="X_244_THE_FIFTEEN_TURNINGSa"></a><a href="#X_244_THE_FIFTEEN_TURNINGS"><b>244.&mdash;THE FIFTEEN TURNINGS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a244.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>It will be seen from the illustration (where the roads not used are
+omitted) that the traveller can go as far as seventy miles in fifteen
+turnings. The turnings are all numbered in the order in which they are
+taken. It will be seen that he never visits nineteen of the towns. He
+might visit them all in fifteen turnings, never entering any town
+twice, and end at the black town from which he starts (see "The Rook's
+Tour," No. 320), but such a tour would only take him sixty-four miles.</p>
+<hr style="width: 30%;" />
+<p><a name="X_245_THE_FLY_ON_THE_OCTAHEDRONa" id="X_245_THE_FLY_ON_THE_OCTAHEDRONa"></a><a href="#X_245_THE_FLY_ON_THE_OCTAHEDRON"><b>245.&mdash;THE FLY ON THE OCTAHEDRON.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a245.png" width="400" height="350" alt="" title="" />
+</div>
+
+<p>Though we cannot really see all the sides of the octahedron at once,
+we can make a projection of it that suits our purpose just as well. In
+the diagram the six points represent the six angles of the octahedron,
+and four lines proceed from every point under exactly the same
+conditions as the twelve edges of the solid. Therefore if we start at
+the point A and go over all the lines once, we must always end our
+route at A. And the number of different routes is just 1,488, counting
+the reverse way of any route as different. It would take too much
+space to show how I make the count. It can be done in about five
+minutes, but an explanation of the method is difficult. The reader is
+therefore asked to accept my answer as correct.</p>
+<hr style="width: 30%;" />
+<p><a name="X_246_THE_ICOSAHEDRON_PUZZLEa" id="X_246_THE_ICOSAHEDRON_PUZZLEa"></a><a href="#X_246_THE_ICOSAHEDRON_PUZZLE"><b>246.&mdash;THE ICOSAHEDRON PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a246.png" width="400" height="340" alt="" title="" />
+</div>
+
+<p>There are thirty edges, of which eighteen were visible in the original
+illustration, represented <span class='pagenum'>Pg 199<a name="Page_199" id="Page_199"></a></span>in the following diagram by the hexagon
+NAESGD. By this projection of the solid we get an imaginary view of
+the remaining twelve edges, and are able to see at once their
+direction and the twelve points at which all the edges meet. The
+difference in the length of the lines is of no importance; all we want
+is to present their direction in a graphic manner. But in case the
+novice should be puzzled at only finding nineteen triangles instead of
+the required twenty, I will point out that the apparently missing
+triangle is the outline HIK.</p>
+
+<p>In this case there are twelve odd nodes; therefore six distinct and
+disconnected routes will be needful if we are not to go over any lines
+twice. Let us therefore find the greatest distance that we may so
+travel in one route.</p>
+
+<p>It will be noticed that I have struck out with little cross strokes
+five lines or edges in the diagram. These five lines may be struck out
+anywhere so long as they do not join one another, and so long as one
+of them does not connect with N, the North Pole, from which we are to
+start. It will be seen that the result of striking out these five
+lines is that all the nodes are now even except N and S. Consequently
+if we begin at N and stop at S we may go over all the lines, except
+the five crossed out, without traversing any line twice. There are
+many ways of doing this. Here is one route: N to H, I, K, S, I, E, S,
+G, K, D, H, A, N, B, A, E, F, B, C, G, D, N, C, F, S. By thus making
+five of the routes as short as is possible&mdash;simply from one node to
+the next&mdash;we are able to get the greatest possible length for our
+sixth line. A greater distance in one route, without going over the
+same ground twice, it is not possible to get.</p>
+
+<p>It is now readily seen that those five erased lines must be gone over
+twice, and they may be "picked up," so to speak, at any points of our
+route. Thus, whenever the traveller happens to be at I he can run up
+to A and back before proceeding on his route, or he may wait until he
+is at A and then run down to I and back to A. And so with the other
+lines that have to be traced twice. It is, therefore, clear that he
+can go over 25 of the lines once only (25&nbsp;&times;&nbsp;10,000 miles&nbsp;=&nbsp;250,000
+miles) and 5 of the lines twice (5&nbsp;&times;&nbsp;20,000 miles&nbsp;=&nbsp;100,000 miles), the
+total, 350,000 miles, being the length of his travels and the shortest
+distance that is possible in visiting the whole body.</p>
+
+<p>It will be noticed that I have made him end his travels at S, the
+South Pole, but this is not imperative. I might have made him finish
+at any of the other nodes, except the one from which he started.
+Suppose it had been required to bring him home again to N at the end
+of his travels. Then instead of suppressing the line AI we might leave
+that open and close IS. This would enable him to complete his 350,000
+miles tour at A, and another 10,000 miles would take him to his own
+fireside. There are a great many different routes, but as the lengths
+of the edges are all alike, one course is as good as another. To make
+the complete 350,000 miles tour from N to S absolutely clear to
+everybody, I will give it entire: N to H, I, A, I, K, H, K, S, I, E,
+S, G, F, G, K, D, C, D, H, A, N, B, E, B, A, E, F, B, C, G, D, N, C,
+F, S&mdash;that is, thirty-five lines of 10,000 miles each.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_247_INSPECTING_A_MINEa" id="X_247_INSPECTING_A_MINEa"></a><a href="#X_247_INSPECTING_A_MINE"><b>247.&mdash;INSPECTING A MINE.&mdash;<i>solution</i></b></a></p>
+
+<p>Starting from A, the inspector need only travel 36 furlongs if he
+takes the following route: A to B, G, H, C, D, I, H, M, N, I, J, O, N,
+S, R, M, L, G, F, K, L, Q, R, S, T, O, J, E, D, C, B, A, F, K, P, Q.
+He thus passes between A and B twice, between C and D twice, between F
+and K twice, between J and O twice, and between R and S twice&mdash;five
+repetitions. Therefore 31 passages plus 5 repeated equal 36 furlongs.
+The little pitfall in this puzzle lies in the fact that we start from
+an even node. Otherwise we need only travel 35 furlongs.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_248_THE_CYCLISTS_TOURa" id="X_248_THE_CYCLISTS_TOURa"></a><a href="#X_248_THE_CYCLISTS_TOUR"><b>248.&mdash;THE CYCLIST'S TOUR.&mdash;<i>solution</i></b></a></p>
+
+<p>When Mr. Maggs replied, "No way, I'm sure," he was not saying that the
+thing was impossible, but was really giving the actual route by which
+the problem can be solved. Starting from the star, if you visit the
+towns in the order, NO WAY, I'M SURE, you will visit every town once,
+and only once, and end at E. So both men were correct. This was the
+little joke of the puzzle, which is not by any means difficult.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_249_THE_SAILORS_PUZZLEa" id="X_249_THE_SAILORS_PUZZLEa"></a><a href="#X_249_THE_SAILORS_PUZZLE"><b>249.&mdash;THE SAILOR'S PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a249.png" width="400" height="505" alt="" title="" />
+</div>
+
+<p>There are only four different routes (or eight, if we count the
+reverse ways) by which the sailor can start at the island marked A,
+visit all the islands once, and once only, and return again to A. Here
+they are:&mdash;</p>
+
+<p><span class='pagenum'>Pg 200<a name="Page_200" id="Page_200"></a></span>A I P T L O E H R Q D C F U G N S K M B A A I P T S N G L O E U F C D
+K M B Q R H A A B M K S N G L T P I O E U F C D Q R H A A I P T L O E
+U G N S K M B Q D C F R H A</p>
+
+<p>Now, if the sailor takes the first route he will make C his 12th
+island (counting A as 1); by the second route he will make C his 13th
+island; by the third route, his 16th island; and by the fourth route,
+his 17th island. If he goes the reverse way, C will be respectively
+his 10th, 9th, 6th, and 5th island. As these are the only possible
+routes, it is evident that if the sailor puts off his visit to C as
+long as possible, he must take the last route reading from left to
+right. This route I show by the dark lines in the diagram, and it is
+the correct answer to the puzzle.</p>
+
+<p>The map may be greatly simplified by the "buttons and string" method,
+explained in the solution to <a href="#X_341_THE_FOUR_FROGSa">No. 341</a>, "The Four Frogs."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_250_THE_GRAND_TOURa" id="X_250_THE_GRAND_TOURa"></a><a href="#X_250_THE_GRAND_TOUR"><b>250.&mdash;THE GRAND TOUR.&mdash;<i>solution</i></b></a></p>
+
+<p>The first thing to do in trying to solve a puzzle like this is to
+attempt to simplify it. If you look at Fig. 1, you will see that it is
+a simplified version of the map. Imagine the circular towns to be
+buttons and the railways to be connecting strings. (See solution to
+<a href="#X_341_THE_FOUR_FROGSa">No. 341.</a>) Then, it will be seen, we have simply "straightened out" the
+previous diagram without affecting the conditions. Now we can further
+simplify by converting Fig. 1 into Fig. 2, which is a portion of a
+chessboard. Here the directions of the railways will resemble the
+moves of a rook in chess&mdash;that is, we may move in any direction
+parallel to the sides of the diagram, but not diagonally. Therefore
+the first town (or square) visited must be a black one; the second
+must be a white; the third must be a black; and so on. Every odd
+square visited will thus be black and every even one white. Now, we
+have 23 squares to visit (an odd number), so the last square visited
+must be black. But Z happens to be white, so the puzzle would seem to
+be impossible of solution.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a250.png" width="600" height="414" alt="" title="" />
+</div>
+
+<p>As we were told that the man "succeeded" in carrying put his plan, we
+must try to find some loophole in the conditions. He was to "enter
+every town once and only once," and we find no prohibition against his
+entering once the town A after leaving it, especially as he has never
+left it since he was born, and would thus be "entering" it for the
+first time in his life. But he must return at once from the first town
+he visits, and then he will have only 22 towns to visit, and as 22 is
+an even number, there is no reason why he should not end on the white
+square Z. A possible route for him is indicated by the dotted line
+from A to Z. This route is repeated by the dark lines in Fig. 1, and
+the reader will now have no difficulty in applying; it to the original
+map. We have thus proved that the puzzle can only be solved by a
+return to A immediately after leaving it.</p>
+<hr style="width: 30%;" />
+<p><a name="X_251_WATER_GAS_AND_ELECTRICITYa" id="X_251_WATER_GAS_AND_ELECTRICITYa"></a><a href="#X_251_WATER_GAS_AND_ELECTRICITY"><b>251.&mdash;WATER, GAS, AND ELECTRICITY.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a251.png" width="400" height="303" alt="" title="" />
+</div>
+
+<p>According to the conditions, in the strict sense in which one at first
+understands them, there<span class='pagenum'>Pg 201<a name="Page_201" id="Page_201"></a></span>is no possible solution to this puzzle. In such a dilemma one always
+has to look for some verbal quibble or trick. If the owner of house A
+will allow the water company to run their pipe for house C through his
+property (and we are not bound to assume that he would object), then
+the difficulty is got over, as shown in our illustration. It will be
+seen that the dotted line from W to C passes through house A, but no
+pipe ever crosses another pipe.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_252_A_PUZZLE_FOR_MOTORISTSa" id="X_252_A_PUZZLE_FOR_MOTORISTSa"></a><a href="#X_252_A_PUZZLE_FOR_MOTORISTS"><b>252.&mdash;A PUZZLE FOR MOTORISTS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/a252.png" width="500" height="307" alt="" title="" />
+</div>
+
+<p>The routes taken by the eight drivers are shown in the illustration,
+where the dotted line roads are omitted to make the paths clearer to
+the eye.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_253_A_BANK_HOLIDAY_PUZZLEa" id="X_253_A_BANK_HOLIDAY_PUZZLEa"></a><a href="#X_253_A_BANK_HOLIDAY_PUZZLE"><b>253.&mdash;A BANK HOLIDAY PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The simplest way is to write in the number of routes to all the towns
+in this manner. Put a 1 on all the towns in the top row and in the
+first column. Then the number of routes to any town will be the sum of
+the routes to the town immediately above and to the town immediately
+to the left. Thus the routes in the second row will be 1, 2, 3, 4, 5,
+6, etc., in the third row, 1, 3, 6, 10, 15, 21, etc.; and so on with
+the other rows. It will then be seen that the only town to which there
+are exactly 1,365 different routes is the twelfth town in the fifth
+row&mdash;the one immediately over the letter E. This town was therefore
+the cyclist's destination.</p>
+
+<p>The general formula for the number of routes from one corner to the
+corner diagonally opposite on any such rectangular reticulated
+arrangement, under the conditions as to direction, is <sup>(<i>m</i>&nbsp;+&nbsp;<i>n</i>)!</sup>/<sub><i>m</i>!<i>n</i>!</sub>,
+where m is the number of towns on one side, less one, and n the number
+on the other side, less one. Our solution involves the case where
+there are 12 towns by 5. Therefore <i>m</i>&nbsp;=&nbsp;11 and <i>n</i>&nbsp;=&nbsp;4. Then the formula
+gives us the answer 1,365 as above.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_254_THE_MOTOR-CAR_TOURa" id="X_254_THE_MOTOR-CAR_TOURa"></a><a href="#X_254_THE_MOTOR-CAR_TOUR"><b>254.&mdash;THE MOTOR-CAR TOUR.&mdash;<i>solution</i></b></a></p>
+
+<p>First of all I will ask the reader to compare the original square
+diagram with the circular one shown in Figs. 1, 2, and 3 below. If for
+the moment we ignore the shading (the purpose of which I shall proceed
+to explain), we find that the circular diagram in each case is merely
+a simplification of the original square one&mdash;that is, the roads from A
+lead to B, E, and M in both cases, the roads from L (London) lead to
+I, K, and S, and so on. The form below, being circular and
+symmetrical, answers my purpose better in applying a mechanical
+solution, and I therefore adopt it without altering in any way the
+conditions of the puzzle. If such a question as distances from town to
+town came into the problem, the new diagrams might require the
+addition of numbers to indicate these distances, or they might
+conceivably not be at all practicable.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a254.png" width="600" height="221" alt="" title="" />
+</div>
+
+<p>Now, I draw the three circular diagrams, as shown, on a sheet of paper
+and then cut out three pieces of cardboard of the forms indicated by
+the shaded parts of these diagrams. It can be shown that every route,
+if marked out with a red pencil, will form one or other of the designs
+indicated by the edges of the cards, or a reflection thereof. Let us
+direct our attention to Fig. 1. Here the card is so placed that the
+star is at the town T; it therefore gives us (by following the edge of
+the card) one of the circular routes from London: L, S, R, T, M, A, E,
+P, O, J, D, C, B, G, N, Q, K, H, F, I, L. If we went the other way, we
+should get L, I, F, H, K, Q, etc., but these reverse routes were not
+to be counted. When we have written out this first route we revolve
+the card until the <span class='pagenum'>Pg 202<a name="Page_202" id="Page_202"></a></span>star is at M, when we get another different route,
+at A a third route, at E a fourth route, and at P a fifth route. We
+have thus obtained five different routes by revolving the card as it
+lies. But it is evident that if we now take up the card and replace it
+with the other side uppermost, we shall in the same manner get five
+other routes by revolution.</p>
+
+<p>We therefore see how, by using the revolving card in Fig. 1, we may,
+without any difficulty, at once write out ten routes. And if we employ
+the cards in Figs. 2 and 3, we similarly obtain in each case ten other
+routes. These thirty routes are all that are possible. I do not give
+the actual proof that the three cards exhaust all the possible cases,
+but leave the reader to reason that out for himself. If he works out
+any route at haphazard, he will certainly find that it falls into one
+or other of the three categories.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_255_THE_LEVEL_PUZZLEa" id="X_255_THE_LEVEL_PUZZLEa"></a><a href="#X_255_THE_LEVEL_PUZZLE"><b>255.&mdash;THE LEVEL PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Let us confine our attention to the L in the top left-hand corner.
+Suppose we go by way of the E on the right: we must then go straight
+on to the V, from which letter the word may be completed in four ways,
+for there are four E's available through which we may reach an L.
+There are therefore four ways of reading through the right-hand E. It
+is also clear that there must be the same number of ways through the E
+that is immediately below our starting point. That makes eight. If,
+however, we take the third route through the E on the diagonal, we
+then have the option of any one of the three V's, by means of each of
+which we may complete the word in four ways. We can therefore spell
+LEVEL in twelve ways through the diagonal E. Twelve added to eight
+gives twenty readings, all emanating from the L in the top left-hand
+corner; and as the four corners are equal, the answer must be four
+times twenty, or eighty different ways.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_256_THE_DIAMOND_PUZZLEa" id="X_256_THE_DIAMOND_PUZZLEa"></a><a href="#X_256_THE_DIAMOND_PUZZLE"><b>256.&mdash;THE DIAMOND PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>There are 252 different ways. The general formula is that, for words
+of <i>n</i> letters (not palindromes, as in the case of the next puzzle),
+when grouped in this manner, there are always 2<sup>(n+1)</sup>&nbsp;-&nbsp;4 different
+readings. This does not allow diagonal readings, such as you would get
+if you used instead such a word as DIGGING, where it would be possible
+to pass from one G to another G by a diagonal step.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_257_THE_DEIFIED_PUZZLEa" id="X_257_THE_DEIFIED_PUZZLEa"></a><a href="#X_257_THE_DEIFIED_PUZZLE"><b>257.&mdash;THE DEIFIED PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The correct answer is 1,992 different ways. Every F is either a corner
+F or a side F&mdash;standing next to a corner in its own square of F's.
+Now, FIED may be read <i>from</i> a corner F in 16 ways; therefore DEIF may
+be read <i>into</i> a corner F also in 16 ways; hence DEIFIED may be read
+<i>through</i> a corner F in 16&nbsp;&times;&nbsp;16&nbsp;=&nbsp;256 ways. Consequently, the four
+corner F's give 4&nbsp;&times;&nbsp;256&nbsp;=&nbsp;1,024 ways. Then FIED may be read from a
+side F in 11 ways, and DEIFIED therefore in 121 ways. But there are
+eight side F's; consequently these give together 8&nbsp;&times;&nbsp;121&nbsp;=&nbsp;968 ways.
+Add 968 to 1,024 and we get the answer, 1,992.</p>
+
+<p>In this form the solution will depend on whether the number of letters
+in the palindrome be odd or even. For example, if you apply the word
+NUN in precisely the same manner, you will get 64 different readings;
+but if you use the word NOON, you will only get 56, because you cannot
+use the same letter twice in immediate succession (since you must
+"always pass from one letter to another") or diagonal readings, and
+every reading must involve the use of the central N.</p>
+
+<p>The reader may like to find for himself the general formula in this
+case, which is complex and difficult. I will merely add that for such
+a case as MADAM, dealt with in the same way as DEIFIED, the number of
+readings is 400.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_258_THE_VOTERS_PUZZLEa" id="X_258_THE_VOTERS_PUZZLEa"></a><a href="#X_258_THE_VOTERS_PUZZLE"><b>258.&mdash;THE VOTERS' PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>THE number of readings here is 63,504, as in the case of "WAS IT A RAT
+I SAW" (No. 30, <i>Canterbury Puzzles</i>). The general formula is that for
+palindromic sentences containing 2<i>n</i>&nbsp;+&nbsp;1 letters there are [4(2<sup><i>n</i></sup> -
+1)]&sup2; readings.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_259_HANNAHS_PUZZLEa" id="X_259_HANNAHS_PUZZLEa"></a><a href="#X_259_HANNAHS_PUZZLE"><b>259.&mdash;HANNAH'S PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Starting from any one of the N's, there are 17 different readings of
+NAH, or 68 (4 times 17) for the 4 N's. Therefore there are also 68
+ways of spelling HAN. If we were allowed to use the same N twice in a
+spelling, the answer would be 68 times 68, or 4,624 ways. But the
+conditions were, "always passing from one letter to another."
+Therefore, for every one of the 17 ways of spelling HAN with a
+particular N, there would be 51 ways (3 times 17) of completing the
+NAH, or 867 (17 times 51) ways for the complete word. Hence, as there
+are four N's to use in HAN, the correct solution of the puzzle is
+3,468 (4 times 867) different ways.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_260_THE_HONEYCOMB_PUZZLEa" id="X_260_THE_HONEYCOMB_PUZZLEa"></a><a href="#X_260_THE_HONEYCOMB_PUZZLE"><b>260.&mdash;THE HONEYCOMB PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The required proverb is, "There is many a slip 'twixt the cup and the
+lip." Start at the T on the outside at the bottom right-hand corner,
+pass to the H above it, and the rest is easy.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_261_THE_MONK_AND_THE_BRIDGESa" id="X_261_THE_MONK_AND_THE_BRIDGESa"></a><a href="#X_261_THE_MONK_AND_THE_BRIDGES"><b>261.&mdash;THE MONK AND THE BRIDGES.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 300px;">
+<img src="images/a261.png" width="300" height="425" alt="" title="" />
+</div>
+
+<p>The problem of the Bridges may be reduced to the simple diagram shown
+in illustration. The <span class='pagenum'>Pg 203<a name="Page_203" id="Page_203"></a></span>point M represents the Monk, the point I the
+Island, and the point Y the Monastery. Now the only direct ways from M
+to I are by the bridges <i>a</i> and <i>b</i>; the only direct ways from I to Y
+are by the bridges <i>c</i> and <i>d</i>; and there is a direct way from M to Y
+by the bridge <i>e</i>. Now, what we have to do is to count all the routes
+that will lead from M to Y, passing over all the bridges, <i>a</i>, <i>b</i>,
+<i>c</i>, <i>d</i>, and <i>e</i> once and once only. With the simple diagram under
+the eye it is quite easy, without any elaborate rule, to count these
+routes methodically. Thus, starting from <i>a</i>, <i>b</i>, we find there are
+only two ways of completing the route; with <i>a, c</i>, there are only two
+routes; with <i>a</i>, <i>d</i>, only two routes; and so on. It will be found
+that there are sixteen such routes in all, as in the following list:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="2" cellspacing="0" summary="">
+<tr><td align='center'><i>a b e c d</i></td></tr>
+<tr><td align='center'><i>a b e d c</i></td></tr>
+<tr><td align='center'><i>a c d b e</i></td></tr>
+<tr><td align='center'><i>a c e b d</i></td></tr>
+<tr><td align='center'><i>a d e b c</i></td></tr>
+<tr><td align='center'><i>a d c b e</i></td></tr>
+<tr><td align='center'><i>b a e c d</i></td></tr>
+<tr><td align='center'><i>b a e d c</i></td></tr>
+<tr><td align='center'><i>b c d a e</i></td></tr>
+<tr><td align='center'><i>b c e a d</i></td></tr>
+<tr><td align='center'><i>b d c a e</i></td></tr>
+<tr><td align='center'><i>b d e a c</i></td></tr>
+<tr><td align='center'><i>e c a b d</i></td></tr>
+<tr><td align='center'><i>e c b a d</i></td></tr>
+<tr><td align='center'><i>e d a b c</i></td></tr>
+<tr><td align='center'><i>e d b a c</i></td></tr>
+</table></div>
+
+<p>If the reader will transfer the letters indicating the bridges from
+the diagram to the corresponding bridges in the original illustration,
+everything will be quite obvious.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_262_THOSE_FIFTEEN_SHEEPa" id="X_262_THOSE_FIFTEEN_SHEEPa"></a><a href="#X_262_THOSE_FIFTEEN_SHEEP"><b>262.&mdash;THOSE FIFTEEN SHEEP.&mdash;<i>solution</i></b></a></p>
+
+<p>If we read the exact words of the writer in the cyclop&aelig;dia, we find
+that we are not told that the pens were all necessarily empty! In
+fact, if the reader will refer back to the illustration, he will see
+that one sheep is already in one of the pens. It was just at this
+point that the wily farmer said to me, "<i>Now</i> I'm going to start
+placing the fifteen sheep." He thereupon proceeded to drive three from
+his flock into the already occupied pen, and then placed four sheep in
+each of the other three pens. "There," says he, "you have seen me
+place fifteen sheep in four pens so that there shall be the same
+number of sheep in every pen." I was, of course, forced to admit that
+he was perfectly correct, according to the exact wording of the
+question.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_263_KING_ARTHURS_KNIGHTSa" id="X_263_KING_ARTHURS_KNIGHTSa"></a><a href="#X_263_KING_ARTHURS_KNIGHTS"><b>263.&mdash;KING ARTHUR'S KNIGHTS.&mdash;<i>solution</i></b></a></p>
+
+<p>On the second evening King Arthur arranged the knights and himself in
+the following order round the table: A, F, B, D, G, E, C. On the third
+evening they sat thus, A, E, B, G, C, F, D. He thus had B next but one
+to him on both occasions (the nearest possible), and G was the third
+from him at both sittings (the furthest position possible). No other
+way of sitting the knights would have been so satisfactory.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_264_THE_CITY_LUNCHEONSa" id="X_264_THE_CITY_LUNCHEONSa"></a><a href="#X_264_THE_CITY_LUNCHEONS"><b>264.&mdash;THE CITY LUNCHEONS.&mdash;<i>solution</i></b></a></p>
+
+<p>The men may be grouped as follows, where each line represents a day
+and each column a table:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>AB</td><td align='center'>CD</td><td align='center'>EF</td><td align='center'>GH</td><td align='center'>IJ</td><td align='center'>KL</td></tr>
+<tr><td align='center'>AE</td><td align='center'>DL</td><td align='center'>GK</td><td align='center'>FI</td><td align='center'>CB</td><td align='center'>HJ</td></tr>
+<tr><td align='center'>AG</td><td align='center'>LJ</td><td align='center'>FH</td><td align='center'>KC</td><td align='center'>DE</td><td align='center'>IB</td></tr>
+<tr><td align='center'>AF</td><td align='center'>JB</td><td align='center'>KI</td><td align='center'>HD</td><td align='center'>LG</td><td align='center'>CE</td></tr>
+<tr><td align='center'>AK</td><td align='center'>BE</td><td align='center'>HC</td><td align='center'>IL</td><td align='center'>JF</td><td align='center'>DG</td></tr>
+<tr><td align='center'>AH</td><td align='center'>EG</td><td align='center'>ID</td><td align='center'>CJ</td><td align='center'>BK</td><td align='center'>LF</td></tr>
+<tr><td align='center'>AI</td><td align='center'>GF</td><td align='center'>CL</td><td align='center'>DB</td><td align='center'>EH</td><td align='center'>JK</td></tr>
+<tr><td align='center'>AC</td><td align='center'>FK</td><td align='center'>DJ</td><td align='center'>LE</td><td align='center'>GI</td><td align='center'>BH</td></tr>
+<tr><td align='center'>AD</td><td align='center'>KH</td><td align='center'>LB</td><td align='center'>JG</td><td align='center'>FC</td><td align='center'>EI</td></tr>
+<tr><td align='center'>AL</td><td align='center'>HI</td><td align='center'>JE</td><td align='center'>BF</td><td align='center'>KD</td><td align='center'>GC</td></tr>
+<tr><td align='center'>AJ</td><td align='center'>IC</td><td align='center'>BG</td><td align='center'>EK</td><td align='center'>HL</td><td align='center'>FD</td></tr>
+</table></div>
+
+<p>Note that in every column (except in the case of the A's) all the
+letters descend cyclically in the same order, B, E, G, F, up to J,
+which is followed by B.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_265_A_PUZZLE_FOR_CARD-PLAYERSa" id="X_265_A_PUZZLE_FOR_CARD-PLAYERSa"></a><a href="#X_265_A_PUZZLE_FOR_CARD-PLAYERS"><b>265.&mdash;A PUZZLE FOR CARD-PLAYERS.&mdash;<i>solution</i></b></a></p>
+
+<p>In the following solution each of the eleven lines represents a
+sitting, each column a table, and each pair of letters a pair of
+partners.</p>
+
+<div class='center'>
+<table border="0" cellpadding="6" cellspacing="0" summary="">
+<tr><td align='center'>A B &mdash; I L</td><td align='center'>E J &mdash; G K</td><td align='center'>F H &mdash; C D</td></tr>
+<tr><td align='center'>A C &mdash; J B</td><td align='center'>F K &mdash; H L</td><td align='center'>G I &mdash; D E</td></tr>
+<tr><td align='center'>A D &mdash; K C</td><td align='center'>G L &mdash; I B</td><td align='center'>H J &mdash; E F</td></tr>
+<tr><td align='center'>A E &mdash; L D</td><td align='center'>H B &mdash; J C</td><td align='center'>I K &mdash; F G</td></tr>
+<tr><td align='center'>A F &mdash; B E</td><td align='center'>I C &mdash; K D</td><td align='center'>J L &mdash; G H</td></tr>
+<tr><td align='center'>A G &mdash; C F</td><td align='center'>J D &mdash; L E</td><td align='center'>K B &mdash; H I</td></tr>
+<tr><td align='center'>A H &mdash; D G</td><td align='center'>K E &mdash; B F</td><td align='center'>L C &mdash; I J</td></tr>
+<tr><td align='center'>A I &mdash; E H</td><td align='center'>L F &mdash; C G</td><td align='center'>B D &mdash; J K</td></tr>
+<tr><td align='center'>A J &mdash; F I</td><td align='center'>B G &mdash; D H</td><td align='center'>C E &mdash; K L</td></tr>
+<tr><td align='center'>A K &mdash; G J</td><td align='center'>C H &mdash; E I</td><td align='center'>D F &mdash; L B</td></tr>
+<tr><td align='center'>A L &mdash; H K</td><td align='center'>D I &mdash; F J</td><td align='center'>E G &mdash; B C</td></tr>
+</table></div>
+
+<p>It will be seen that the letters B, C, D ...L descend cyclically. The
+solution given above is absolutely perfect in all respects. It will be
+found that every player has every other player once as his partner and
+twice as his opponent.</p>
+<hr style="width: 30%;" />
+<p><a name="X_266_A_TENNIS_TOURNAMENTa" id="X_266_A_TENNIS_TOURNAMENTa"></a><a href="#X_266_A_TENNIS_TOURNAMENT"><b>266.&mdash;A TENNIS TOURNAMENT.&mdash;<i>solution</i></b></a></p>
+
+<p>Call the men A, B, D, E, and their wives a, b, d, e. Then they may
+play as follows without any person ever playing twice with or against
+any other person:&mdash;</p>
+
+<div class='center'>
+<table border="1" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'></td><td align='center'>First Court.</td><td align='center'>Second Court.</td></tr>
+<tr><td align='center'>1st Day</td><td align='center'>A d against B e</td><td align='center'>D a against E b</td></tr>
+<tr><td align='center'>2nd Day</td><td align='center'>A e against D b</td><td align='center'>E a against B d</td></tr>
+<tr><td align='center'>3rd Day</td><td align='center'>A b against E d</td><td align='center'>B a against D e</td></tr>
+</table></div>
+
+<p>It will be seen that no man ever plays with or against his own
+wife&mdash;an ideal arrangement. If the reader wants a hard puzzle, let him
+try to arrange eight married couples (in four courts on seven days)
+under exactly similar conditions. It can be done, but I leave the
+reader in this case the pleasure of seeking the answer and the general
+solution.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_267_THE_WRONG_HATSa" id="X_267_THE_WRONG_HATSa"></a><a href="#X_267_THE_WRONG_HATS"><b>267.&mdash;THE WRONG HATS.&mdash;<i>solution</i></b></a></p>
+
+<p>The number of different ways in which eight persons, with eight hats,
+can each take the wrong hat, is 14,833.</p>
+
+<p>Here are the successive solutions for any number of persons from one
+to eight:&mdash;</p>
+
+<p><span class='pagenum'>Pg 204<a name="Page_204" id="Page_204"></a></span></p><p>
+<span style="margin-left: 2em;">1&nbsp;=&nbsp;0</span><br />
+<span style="margin-left: 2em;">2&nbsp;=&nbsp;1</span><br />
+<span style="margin-left: 2em;">3&nbsp;=&nbsp;2</span><br />
+<span style="margin-left: 2em;">4&nbsp;=&nbsp;9</span><br />
+<span style="margin-left: 2em;">5&nbsp;=&nbsp;44</span><br />
+<span style="margin-left: 2em;">6&nbsp;=&nbsp;265</span><br />
+<span style="margin-left: 2em;">7&nbsp;=&nbsp;1,854</span><br />
+<span style="margin-left: 2em;">8&nbsp;=&nbsp;14,833</span><br />
+</p>
+
+<p>To get these numbers, multiply successively by 2, 3, 4, 5, etc. When
+the multiplier is even, add 1; when odd, deduct 1. Thus, 3&nbsp;&times;&nbsp;1&nbsp;-&nbsp;1&nbsp;=&nbsp;2,
+4&nbsp;&times;&nbsp;2&nbsp;+&nbsp;1&nbsp;=&nbsp;9; 5&nbsp;&times;&nbsp;9&nbsp;-&nbsp;1&nbsp;=&nbsp;44; and so on. Or you can multiply the sum of the
+number of ways for n-1 and n-2 persons by n-1, and so get the solution
+for n persons. Thus, 4(2&nbsp;+&nbsp;9)&nbsp;=&nbsp;44; 5(9&nbsp;+&nbsp;44)&nbsp;=&nbsp;265; and so on.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_268_THE_PEAL_OF_BELLSa" id="X_268_THE_PEAL_OF_BELLSa"></a><a href="#X_268_THE_PEAL_OF_BELLS"><b>268.&mdash;THE PEAL OF BELLS.&mdash;<i>solution</i></b></a></p>
+
+<p>The bells should be rung as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>1 2 3 4</td></tr>
+<tr><td align='center'>2 1 4 3</td></tr>
+<tr><td align='center'>2 4 1 3</td></tr>
+<tr><td align='center'>4 2 3 1</td></tr>
+<tr><td align='center'>4 3 2 1</td></tr>
+<tr><td align='center'>3 4 1 2</td></tr>
+<tr><td align='center'>3 1 4 2</td></tr>
+<tr><td align='center'>1 3 2 4</td></tr>
+<tr><td align='center'>3 1 2 4</td></tr>
+<tr><td align='center'>1 3 4 2</td></tr>
+<tr><td align='center'>1 4 3 2</td></tr>
+<tr><td align='center'>4 1 2 3</td></tr>
+<tr><td align='center'>4 2 1 3</td></tr>
+<tr><td align='center'>2 4 3 1</td></tr>
+<tr><td align='center'>2 3 4 1</td></tr>
+<tr><td align='center'>3 2 1 4</td></tr>
+<tr><td align='center'>2 3 1 4</td></tr>
+<tr><td align='center'>3 2 4 1</td></tr>
+<tr><td align='center'>3 4 2 1</td></tr>
+<tr><td align='center'>4 3 1 2</td></tr>
+<tr><td align='center'>4 1 3 2</td></tr>
+<tr><td align='center'>1 4 2 3</td></tr>
+<tr><td align='center'>1 2 4 3</td></tr>
+<tr><td align='center'>2 1 3 4</td></tr>
+</table></div>
+
+<p>I have constructed peals for five and six bells respectively, and a
+solution is possible for any number of bells under the conditions
+previously stated.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_269_THREE_MEN_IN_A_BOATa" id="X_269_THREE_MEN_IN_A_BOATa"></a><a href="#X_269_THREE_MEN_IN_A_BOAT"><b>269.&mdash;THREE MEN IN A BOAT.&mdash;<i>solution</i></b></a></p>
+
+<p>If there were no conditions whatever, except that the men were all to
+go out together, in threes, they could row in an immense number of
+different ways. If the reader wishes to know how many, the number is
+455<sup>7</sup>. And with the condition that no two may ever be together more
+than once, there are no fewer than 15,567,552,000 different
+solutions&mdash;that is, different ways of arranging the men. With one
+solution before him, the reader will realize why this must be, for
+although, as an example, A must go out once with B and once with C, it
+does not necessarily follow that he must go out with C on the same
+occasion that he goes with B. He might take any other letter with him
+on that occasion, though the fact of his taking other than B would
+have its effect on the arrangement of the other triplets.</p>
+
+<p>Of course only a certain number of all these arrangements are
+available when we have that other condition of using the smallest
+possible number of boats. As a matter of fact we need employ only ten
+different boats. Here is one the arrangements:&mdash;</p>
+
+<div class='center'>
+<table border="1" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'></td><td align='center'>1</td><td align='center'>2</td><td align='center'>3</td><td align='center'>4</td><td align='center'>5</td></tr>
+<tr><td align='center'>1st Day</td><td align='center'>(ABC)</td><td align='center'>(DBF)</td><td align='center'>(GHI)</td><td align='center'>(JKL)</td><td align='center'>(MNO)</td></tr>
+<tr><td align='center'></td><td align='center'>8</td><td align='center'>6</td><td align='center'>7</td><td align='center'>9</td><td align='center'>10</td></tr>
+<tr><td align='center'>2nd Day</td><td align='center'>(ADG)</td><td align='center'>(BKN)</td><td align='center'>(COL)</td><td align='center'>(JEI)</td><td align='center'>(MHF)</td></tr>
+<tr><td align='center'></td><td align='center'>3</td><td align='center'>5</td><td align='center'>4</td><td align='center'>1</td><td align='center'>2</td></tr>
+<tr><td align='center'>3rd Day</td><td align='center'>(AJM)</td><td align='center'>(BEH)</td><td align='center'>(CFI)</td><td align='center'>(DKO)</td><td align='center'>(GNL)</td></tr>
+<tr><td align='center'></td><td align='center'>7</td><td align='center'>6</td><td align='center'>8</td><td align='center'>9</td><td align='center'>1</td></tr>
+<tr><td align='center'>4th Day</td><td align='center'>(AEK)</td><td align='center'>(CGM)</td><td align='center'>(BOI)</td><td align='center'>(DHL)</td><td align='center'>(JNF)</td></tr>
+<tr><td align='center'></td><td align='center'>4</td><td align='center'>5</td><td align='center'>3</td><td align='center'>10</td><td align='center'>2</td></tr>
+<tr><td align='center'>5th Day</td><td align='center'>(AHN)</td><td align='center'>(CDJ)</td><td align='center'>(BFL)</td><td align='center'>(GEO)</td><td align='center'>(MKI)</td></tr>
+<tr><td align='center'></td><td align='center'>6</td><td align='center'>7</td><td align='center'>8</td><td align='center'>10</td><td align='center'>1</td></tr>
+<tr><td align='center'>6th Day</td><td align='center'>(AFO)</td><td align='center'>(BGJ)</td><td align='center'>(CKH)</td><td align='center'>(DNI)</td><td align='center'>(MEL)</td></tr>
+<tr><td align='center'></td><td align='center'>5</td><td align='center'>4</td><td align='center'>3</td><td align='center'>9</td><td align='center'>2</td></tr>
+<tr><td align='center'>7th Day</td><td align='center'>(AIL)</td><td align='center'>(BDM)</td><td align='center'>(CEN)</td><td align='center'>(GKF)</td><td align='center'>(JHO)</td></tr>
+</table></div>
+
+<p>It will be found that no two men ever go out twice together, and that
+no man ever goes out twice in the same boat.</p>
+
+<p>This is an extension of the well-known problem of the "Fifteen
+Schoolgirls," by Kirkman. The original conditions were simply that
+fifteen girls walked out on seven days in triplets without any girl
+ever walking twice in a triplet with another girl. Attempts at a
+general solution of this puzzle had exercised the ingenuity of
+mathematicians since 1850, when the question was first propounded,
+until recently. In 1908 and the two following years I indicated (see
+<i>Educational Times Reprints</i>, Vols. XIV., XV., and XVII.) that all our
+trouble had arisen from a failure to discover that 15 is a special
+case (too small to enter into the general law for all higher numbers
+of girls of the form 6<i>n</i>&nbsp;+&nbsp;3), and showed what that general law is and
+how the groups should be posed for any number of girls. I gave actual
+arrangements for numbers that had previously baffled all attempts to
+manipulate, and the problem may now be considered generally solved.
+Readers will find an excellent full account of the puzzle in W.W.
+Rouse Ball's <i>Mathematical Recreations</i>, 5th edition.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_270_THE_GLASS_BALLSa" id="X_270_THE_GLASS_BALLSa"></a><a href="#X_270_THE_GLASS_BALLS"><b>270.&mdash;THE GLASS BALLS.&mdash;<i>solution</i></b></a></p>
+
+<p>There are, in all, sixteen balls to be broken, or sixteen places in
+the order of breaking. Call the four strings A, B, C, and D&mdash;order is
+here of no importance. The breaking of the balls on A may occupy any 4
+out of these 16 places&mdash;that is, the combinations of 16 things, taken
+4 together, will be</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' class='bb'>13&nbsp;&times;&nbsp;14&nbsp;&times;&nbsp;15&nbsp;&times;&nbsp;16</td><td align='center' rowspan='2'>&nbsp;=&nbsp;1,820</td></tr>
+<tr><td align='center'>1&nbsp;&times;&nbsp;2&nbsp;&times;&nbsp;3&nbsp;&times;&nbsp;4</td></tr>
+</table></div>
+
+<p>ways for A. In every one of these cases B may occupy any 4 out of the
+remaining 12 places, making</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' class='bb'>9&nbsp;&times;&nbsp;10&nbsp;&times;&nbsp;11&nbsp;&times;&nbsp;12</td><td align='center' rowspan='2'>&nbsp;=&nbsp;495</td></tr>
+<tr><td align='center'>1&nbsp;&times;&nbsp;2&nbsp;&times;&nbsp;3&nbsp;&times;&nbsp;4</td></tr>
+</table></div>
+
+<p>ways. Thus 1,820&nbsp;&times;&nbsp;495&nbsp;=&nbsp;900,900 different placings are open to A and B.
+But for every one of these cases C may occupy</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' class='bb'>5&nbsp;&times;&nbsp;6&nbsp;&times;&nbsp;7&nbsp;&times;&nbsp;8</td><td align='center' rowspan='2'>&nbsp;=&nbsp;70</td></tr>
+<tr><td align='center'>1&nbsp;&times;&nbsp;2&nbsp;&times;&nbsp;3&nbsp;&times;&nbsp;4</td></tr>
+</table></div>
+
+
+<p>different places; so that 900,900&nbsp;&times;&nbsp;70&nbsp;=&nbsp;63,063,000 different placings are
+open to A, B, and C. In every one of these cases, D has no choice but
+to take the four places that remain. Therefore the correct answer is
+that the balls may be broken in 63,063,000 different ways under the
+conditions. Readers should compare this problem with No. 345, "The Two
+Pawns," which they will then know how to solve for cases where there
+are three, four, or more pawns on the board.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 205<a name="Page_205" id="Page_205"></a></span><a name="X_271_FIFTEEN_LETTER_PUZZLEa" id="X_271_FIFTEEN_LETTER_PUZZLEa"></a><a href="#X_271_FIFTEEN_LETTER_PUZZLE"><b>271.&mdash;FIFTEEN LETTER PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The following will be found to comply with the conditions of
+grouping:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>ALE</td><td align='center'>MET</td><td align='center'>MOP</td><td align='center'>BLM</td></tr>
+<tr><td align='center'>BAG</td><td align='center'>CAP</td><td align='center'>YOU</td><td align='center'>CLT</td></tr>
+<tr><td align='center'>IRE</td><td align='center'>OIL</td><td align='center'>LUG</td><td align='center'>LNR</td></tr>
+<tr><td align='center'>NAY</td><td align='center'>BIT</td><td align='center'>BUN</td><td align='center'>BPR</td></tr>
+<tr><td align='center'>AIM</td><td align='center'>BEY</td><td align='center'>RUM</td><td align='center'>GMY</td></tr>
+<tr><td align='center'>OAR</td><td align='center'>GIN</td><td align='center'>PLY</td><td align='center'>CGR</td></tr>
+<tr><td align='center'>PEG</td><td align='center'>ICY</td><td align='center'>TRY</td><td align='center'>CMN</td></tr>
+<tr><td align='center'>CUE</td><td align='center'>COB</td><td align='center'>TAU</td><td align='center'>PNT</td></tr>
+<tr><td align='center'>ONE</td><td align='center'>GOT</td><td align='center'>PIU</td></tr>
+</table></div>
+
+<p>The fifteen letters used are A, E, I, O, U, Y, and B, C, G, L, M, N,
+P, R, T. The number of words is 27, and these are all shown in the
+first three columns. The last word, PIU, is a musical term in common
+use; but although it has crept into some of our dictionaries, it is
+Italian, meaning "a little; slightly." The remaining twenty-six are
+good words. Of course a TAU-cross is a T-shaped cross, also called the
+cross of St. Anthony, and borne on a badge in the Bishop's Palace at
+Exeter. It is also a name for the toad-fish.</p>
+
+<p>We thus have twenty-six good words and one doubtful, obtained under
+the required conditions, and I do not think it will be easy to improve
+on this answer. Of course we are not bound by dictionaries but by
+common usage. If we went by the dictionary only in a case of this
+kind, we should find ourselves involved in prefixes, contractions, and
+such absurdities as I.O.U., which Nuttall actually gives as a word.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_272_THE_NINE_SCHOOLBOYSa" id="X_272_THE_NINE_SCHOOLBOYSa"></a><a href="#X_272_THE_NINE_SCHOOLBOYS"><b>272.&mdash;THE NINE SCHOOLBOYS.&mdash;<i>solution</i></b></a></p>
+
+<p>The boys can walk out as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>1st Day.</td><td align='center'>2nd Day.</td><td align='center'>3rd Day.</td><td align='center'>4th Day.</td><td align='center'>5th Day.</td><td align='center'>6th Day.</td></tr>
+<tr><td align='center'>A B C</td><td align='center'>B F H</td><td align='center'>F A G</td><td align='center'>A D H</td><td align='center'>G B I</td><td align='center'>D C A</td></tr>
+<tr><td align='center'>D E F</td><td align='center'>E I A</td><td align='center'>I D B</td><td align='center'>B E G</td><td align='center'>C F D</td><td align='center'>E H B</td></tr>
+<tr><td align='center'>G H I</td><td align='center'>C G D</td><td align='center'>H C E</td><td align='center'>F I C</td><td align='center'>H A E</td><td align='center'>I G F</td></tr>
+</table></div>
+
+<p>Every boy will then have walked by the side of every other boy once
+and once only.</p>
+
+<p>Dealing with the problem generally, 12<i>n</i>&nbsp;+&nbsp;9 boys may walk out in
+triplets under the conditions on 9<i>n</i>&nbsp;+&nbsp;6 days, where n may be nought or
+any integer. Every possible pair will occur once. Call the number of
+boys <i>m</i>. Then every boy will pair <i>m</i>&nbsp;-&nbsp;1 times, of which <sup>(<i>m</i>&nbsp;-&nbsp;1)</sup>/<sub>4</sub> times he
+will be in the middle of a triplet and <sup>(<i>m</i>&nbsp;-&nbsp;1)</sup>/<sub>2</sub> times on the outside.
+Thus, if we refer to the solution above, we find that every boy is in
+the middle twice (making 4 pairs) and four times on the outside
+(making the remaining 4 pairs of his 8). The reader may now like to
+try his hand at solving the two next cases of 21 boys on 15 days, and
+33 boys on 24 days. It is, perhaps, interesting to note that a school
+of 489 boys could thus walk out daily in one leap year, but it would
+take 731 girls (referred to in the solution to No. 269) to perform
+their particular feat by a daily walk in a year of 365 days.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_273_THE_ROUND_TABLEa" id="X_273_THE_ROUND_TABLEa"></a><a href="#X_273_THE_ROUND_TABLE"><b>273.&mdash;THE ROUND TABLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The history of this problem will be found in <i>The Canterbury Puzzles</i>
+(No. 90). Since the publication of that book in 1907, so far as I
+know, nobody has succeeded in solving the case for that unlucky number
+of persons, 13, seated at a table on 66 occasions. A solution is
+possible for any number of persons, and I have recorded schedules for
+every number up to 25 persons inclusive and for 33. But as I know a
+good many mathematicians are still considering the case of 13, I will
+not at this stage rob them of the pleasure of solving it by showing
+the answer. But I will now display the solutions for all the cases up
+to 12 persons inclusive. Some of these solutions are now published for
+the first time, and they may afford useful clues to investigators.</p>
+
+<p>The solution for the case of 3 persons seated on 1 occasion needs no
+remark.</p>
+
+<p>A solution for the case of 4 persons on 3 occasions is as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>1 2 3 4</td></tr>
+<tr><td align='center'>1 3 4 2</td></tr>
+<tr><td align='center'>1 4 2 3</td></tr>
+</table></div>
+
+<p>Each line represents the order for a sitting, and the person
+represented by the last number in a line must, of course, be regarded
+as sitting next to the first person in the same line, when placed at
+the round table.</p>
+
+<p>The case of 5 persons on 6 occasions may be solved as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>1 2 3 4 5</td></tr>
+<tr><td align='center'>1 2 4 5 3</td></tr>
+<tr><td align='center'>1 2 5 3 4</td></tr>
+<tr><td align='center'>1 3 2 5 4</td></tr>
+<tr><td align='center'>1 4 2 3 5</td></tr>
+<tr><td align='center'>1 5 2 4 3</td></tr>
+</table></div>
+
+<p>The case for 6 persons on 10 occasions is solved thus:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>1 2 3 6 4 5</td></tr>
+<tr><td align='center'>1 3 4 2 5 6</td></tr>
+<tr><td align='center'>1 4 5 3 6 2</td></tr>
+<tr><td align='center'>1 5 6 4 2 3</td></tr>
+<tr><td align='center'>1 6 2 5 3 4</td></tr>
+<tr><td align='center'>1 2 4 5 6 3</td></tr>
+<tr><td align='center'>1 3 5 6 2 4</td></tr>
+<tr><td align='center'>1 4 6 2 3 5</td></tr>
+<tr><td align='center'>1 5 2 3 4 6</td></tr>
+<tr><td align='center'>1 6 3 4 5 2</td></tr>
+</table></div>
+
+<p>It will now no longer be necessary to give the solutions in full, for
+reasons that I will explain. It will be seen in the examples above
+that the 1 (and, in the case of 5 persons, also the 2) <span class='pagenum'>Pg 206<a name="Page_206" id="Page_206"></a></span>is repeated
+down the column. Such a number I call a "repeater." The other numbers
+descend in cyclical order. Thus, for 6 persons we get the cycle, 2, 3,
+4, 5, 6, 2, and so on, in every column. So it is only necessary to
+give the two lines 1 2 3 6 4 5 and 1 2 4 5 6 3, and denote the cycle
+and repeaters, to enable any one to write out the full solution
+straight away. The reader may wonder why I do not start the last
+solution with the numbers in their natural order, 1 2 3 4 5 6. If I
+did so the numbers in the descending cycle would not be in their
+natural order, and it is more convenient to have a regular cycle than
+to consider the order in the first line.</p>
+
+<p>The difficult case of 7 persons on 15 occasions is solved as follows,
+and was given by me in <i>The Canterbury Puzzles</i>:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>1 2 3 4 5 7 6</td></tr>
+<tr><td align='center'>1 6 2 7 5 3 4</td></tr>
+<tr><td align='center'>1 3 5 2 6 7 4</td></tr>
+<tr><td align='center'>1 5 7 4 3 6 2</td></tr>
+<tr><td align='center'>1 5 2 7 3 4 6</td></tr>
+</table></div>
+
+<p>In this case the 1 is a repeater, and there are <i>two</i> separate cycles,
+2, 3, 4, 2, and 5, 6, 7, 5. We thus get five groups of three lines
+each, for a fourth line in any group will merely repeat the first
+line.</p>
+
+<p>A solution for 8 persons on 21 occasions is as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>1 8 6 3 4 5 2 7</td></tr>
+<tr><td align='center'>1 8 4 5 7 2 3 6</td></tr>
+<tr><td align='center'>1 8 2 7 3 6 4 5</td></tr>
+</table></div>
+
+<p>The 1 is here a repeater, and the cycle 2, 3, 4, 5, 6, 7, 8. Every one
+of the 3 groups will give 7 lines.</p>
+
+<p>Here is my solution for 9 persons on 28 occasions:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>2 1 9 7 4 5 6 3 8</td></tr>
+<tr><td align='center'>2 9 5 1 6 8 3 4 7</td></tr>
+<tr><td align='center'>2 9 3 1 8 4 7 5 6</td></tr>
+<tr><td align='center'>2 9 1 5 6 4 7 8 3</td></tr>
+</table></div>
+
+<p>There are here two repeaters, 1 and 2, and the cycle is 3, 4, 5, 6, 7,
+8, 9. We thus get 4 groups of 7 lines each.</p>
+
+<p>The case of 10 persons on 36 occasions is solved as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>1 10 8 3 6 5 4 7 2 9</td></tr>
+<tr><td align='center'>1 10 6 5 2 9 7 4 3 8</td></tr>
+<tr><td align='center'>1 10 2 9 3 8 6 5 7 4</td></tr>
+<tr><td align='center'>1 10 7 4 8 3 2 9 5 6</td></tr>
+</table></div>
+
+<p>The repeater is 1, and the cycle, 2, 3, 4, 5, 6, 7, 8, 9, 10. We here
+have 4 groups of 9 lines each.</p>
+
+<p>My solution for 11 persons on 45 occasions is as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>2</td><td align='right'>11</td><td align='right'>9</td><td align='right'>4</td><td align='right'>7</td><td align='right'>6</td><td align='right'>5</td><td align='right'>1</td><td align='right'>8</td><td align='right'>3</td><td align='right'>10</td></tr>
+<tr><td align='right'>2</td><td align='right'>1</td><td align='right'>11</td><td align='right'>7</td><td align='right'>6</td><td align='right'>3</td><td align='right'>10</td><td align='right'>8</td><td align='right'>5</td><td align='right'>4</td><td align='right'>9</td></tr>
+<tr><td align='right'>2</td><td align='right'>11</td><td align='right'>10</td><td align='right'>3</td><td align='right'>9</td><td align='right'>4</td><td align='right'>8</td><td align='right'>5</td><td align='right'>1</td><td align='right'>7</td><td align='right'>6</td></tr>
+<tr><td align='right'>2</td><td align='right'>11</td><td align='right'>5</td><td align='right'>8</td><td align='right'>1</td><td align='right'>3</td><td align='right'>10</td><td align='right'>6</td><td align='right'>7</td><td align='right'>9</td><td align='right'>4</td></tr>
+<tr><td align='right'>2</td><td align='right'>11</td><td align='right'>1</td><td align='right'>10</td><td align='right'>3</td><td align='right'>4</td><td align='right'>9</td><td align='right'>6</td><td align='right'>7</td><td align='right'>5</td><td align='right'>8</td></tr>
+</table></div>
+
+<p>There are two repeaters, 1 and 2, and the cycle is, 3, 4, 5,... 11. We
+thus get 5 groups of 9 lines each.</p>
+
+<p>The case of 12 persons on 55 occasions is solved thus:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>1</td><td align='right'>2</td><td align='right'>3</td><td align='right'>12</td><td align='right'>4</td><td align='right'>11</td><td align='right'>5</td><td align='right'>10</td><td align='right'>6</td><td align='right'>9</td><td align='right'>7</td><td align='right'>8</td></tr>
+<tr><td align='right'>1</td><td align='right'>2</td><td align='right'>4</td><td align='right'>11</td><td align='right'>6</td><td align='right'>9</td><td align='right'>8</td><td align='right'>7</td><td align='right'>10</td><td align='right'>5</td><td align='right'>12</td><td align='right'>3</td></tr>
+<tr><td align='right'>1</td><td align='right'>2</td><td align='right'>5</td><td align='right'>10</td><td align='right'>8</td><td align='right'>7</td><td align='right'>11</td><td align='right'>4</td><td align='right'>3</td><td align='right'>12</td><td align='right'>6</td><td align='right'>9</td></tr>
+<tr><td align='right'>1</td><td align='right'>2</td><td align='right'>6</td><td align='right'>9</td><td align='right'>10</td><td align='right'>5</td><td align='right'>3</td><td align='right'>12</td><td align='right'>7</td><td align='right'>8</td><td align='right'>11</td><td align='right'>4</td></tr>
+<tr><td align='right'>1</td><td align='right'>2</td><td align='right'>7</td><td align='right'>8</td><td align='right'>12</td><td align='right'>3</td><td align='right'>6</td><td align='right'>9</td><td align='right'>11</td><td align='right'>4</td><td align='right'>5</td><td align='right'>10</td></tr>
+</table></div>
+
+<p>Here 1 is a repeater, and the cycle is 2, 3, 4, 5,... 12. We thus get
+5 groups of 11 lines each.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_274_THE_MOUSE-TRAP_PUZZLEa" id="X_274_THE_MOUSE-TRAP_PUZZLEa"></a><a href="#X_274_THE_MOUSE-TRAP_PUZZLE"><b>274.&mdash;THE MOUSE-TRAP PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>If we interchange cards 6 and 13 and begin our count at 14, we may
+take up all the twenty-one cards&mdash;that is, make twenty-one
+"catches"&mdash;in the following order: 6, 8, 13, 2, 10, 1, 11, 4, 14, 3,
+5, 7, 21, 12, 15, 20, 9, 16, 18, 17, 19. We may also exchange 10 and
+14 and start at 16, or exchange 6 and 8 and start at 19.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_275_THE_SIXTEEN_SHEEPa" id="X_275_THE_SIXTEEN_SHEEPa"></a><a href="#X_275_THE_SIXTEEN_SHEEP"><b>275.&mdash;THE SIXTEEN SHEEP.&mdash;<i>solution</i></b></a></p>
+
+<p>The six diagrams on next page show solutions for the cases where we
+replace 2, 3, 4, 5, 6, and 7 hurdles. The dark lines indicate the
+hurdles that have been replaced. There are, of course, other ways of
+making the removals.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a275.png" width="600" height="413" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_276_THE_EIGHT_VILLASa" id="X_276_THE_EIGHT_VILLASa"></a><a href="#X_276_THE_EIGHT_VILLAS"><b>276.&mdash;THE EIGHT VILLAS.&mdash;<i>solution</i></b></a></p>
+
+<p>There are several ways of solving the puzzle, but there is very little
+difference between them. The solver should, however, first of all bear
+in mind that in making his calculations he need only consider the four
+villas that stand at the corners, because the intermediate villas can
+never vary when the corners are known. One way is to place the numbers
+nought to 9 one at a time in the top left-hand corner, and then
+consider each case in turn.</p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/a276.png" width="500" height="343" alt="" title="" />
+</div>
+
+<p>Now, if we place 9 in the corner as shown in the Diagram A, two of the
+corners cannot be occupied, while the corner that is diagonally
+opposite may be filled by 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 persons. We
+thus see that there are 10<span class='pagenum'>Pg 207<a name="Page_207" id="Page_207"></a></span>
+solutions with a 9 in the corner. If, however, we substitute 8, the
+two corners in the same row and column may contain 0, 0, or 1, 1, or
+0, 1, or 1, 0. In the case of B, ten different selections may be made
+for the fourth corner; but in each of the cases C, D, and E, only nine
+selections are possible, because we cannot use the 9. Therefore with 8
+in the top left-hand corner there are 10&nbsp;+&nbsp;(3&nbsp;&times;&nbsp;9)&nbsp;=&nbsp;37 different
+solutions. If we then try 7 in the corner, the result will be 10&nbsp;+&nbsp;27
++ 40, or 77 solutions. With 6 we get 10&nbsp;+&nbsp;27&nbsp;+&nbsp;40&nbsp;+&nbsp;49&nbsp;=&nbsp;126; with 5,
+10&nbsp;+&nbsp;27&nbsp;+&nbsp;40&nbsp;+&nbsp;49&nbsp;+&nbsp;54&nbsp;=&nbsp;180; with 4, the same as with 5,&nbsp;+&nbsp;55&nbsp;=&nbsp;235
+; with 3, the same as with 4,&nbsp;+&nbsp;52&nbsp;=&nbsp;287; with 2, the same as with 3,
++ 45&nbsp;=&nbsp;332; with 1, the same as with 2,&nbsp;+&nbsp;34&nbsp;=&nbsp;366, and with nought in
+the top left-hand corner the number of solutions will be found to be
+10&nbsp;+&nbsp;27&nbsp;+&nbsp;40&nbsp;+&nbsp;49&nbsp;+&nbsp;54&nbsp;+&nbsp;55&nbsp;+&nbsp;52&nbsp;+&nbsp;45&nbsp;+&nbsp;34&nbsp;+&nbsp;19&nbsp;=&nbsp;385. As there is no
+other number to be placed in the top left-hand corner, we have now
+only to add these totals together thus, 10&nbsp;+&nbsp;37&nbsp;+&nbsp;77&nbsp;+&nbsp;126&nbsp;+&nbsp;180&nbsp;+&nbsp;235
++ 287&nbsp;+&nbsp;332&nbsp;+&nbsp;366&nbsp;+&nbsp;385&nbsp;=&nbsp;2,035. We therefore find that the total
+number of ways in which tenants may occupy some or all of the eight
+villas so that there shall be always nine persons living along each
+side of the square is 2,035. Of course, this method must obviously
+cover all the reversals and reflections, since each corner in turn is
+occupied by every number in all possible combinations with the other
+two corners that are in line with it.</p>
+
+<p>Here is a general formula for solving the puzzle: <sup>(<i>n</i>&sup2;&nbsp;+&nbsp;3<i>n</i> +
+2)(<i>n</i>&sup2;&nbsp;+&nbsp;3<i>n</i>&nbsp;+&nbsp;3)</sup>/<sub>6</sub>. Whatever may be the stipulated number of
+residents along each of the sides (which number is represented by
+<i>n</i>), the total number of different arrangements may be thus
+ascertained. In our particular case the number of residents was nine.
+Therefore (81&nbsp;+&nbsp;27&nbsp;+&nbsp;2)&nbsp;&times;&nbsp;(81&nbsp;+&nbsp;27&nbsp;+&nbsp;3) and the product, divided by 6,
+gives 2,035. If the number of residents had been 0, 1, 2, 3, 4, 5, 6,
+7, or 8, the total arrangements would be 1, 7, 26, 70, 155, 301, 532,
+876, or 1,365 respectively.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_277_COUNTER_CROSSESa" id="X_277_COUNTER_CROSSESa"></a><a href="#X_277_COUNTER_CROSSES"><b>277.&mdash;COUNTER CROSSES.&mdash;<i>solution</i></b></a></p>
+
+<p>Let us first deal with the Greek Cross. There are
+just eighteen forms in which the numbers may be paired for the two
+arms. Here they are:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>12978</td><td align='center'>13968</td><td align='center'>14958</td></tr>
+<tr><td align='center'>34956</td><td align='center'>24957</td><td align='center'>23967</td></tr>
+<tr><td align='center'>23958</td><td align='center'>13769</td><td align='center'>14759</td></tr>
+<tr><td align='center'>14967</td><td align='center'>24758</td><td align='center'>23768</td></tr>
+<tr><td align='center'>12589</td><td align='center'>23759</td><td align='center'>13579</td></tr>
+<tr><td align='center'>34567</td><td align='center'>14768</td><td align='center'>24568</td></tr>
+<tr><td align='center'>14569</td><td align='center'>23569</td><td align='center'>14379</td></tr>
+<tr><td align='center'>23578</td><td align='center'>14578</td><td align='center'>25368</td></tr>
+<tr><td align='center'>15369</td><td align='center'>24369</td><td align='center'>23189</td></tr>
+<tr><td align='center'>24378</td><td align='center'>15378</td><td align='center'>45167</td></tr>
+<tr><td align='center'>24179</td><td align='center'>25169</td><td align='center'>34169</td></tr>
+<tr><td align='center'>35168</td><td align='center'>34178</td><td align='center'>25178</td></tr>
+</table></div>
+
+<p><span class='pagenum'>Pg 208<a name="Page_208" id="Page_208"></a></span>Of course, the number in the middle is common to both arms. The first
+pair is the one I gave as an example. I will suppose that we have
+written out all these crosses, always placing the first row of a pair
+in the upright and the second row in the horizontal arm. Now, if we
+leave the central figure fixed, there are 24 ways in which the numbers
+in the upright may be varied, for the four counters may be changed in
+1&nbsp;&times;&nbsp;2&nbsp;&times;&nbsp;3&nbsp;&times;&nbsp;4&nbsp;=&nbsp;24 ways. And as the four in the horizontal may also be
+changed in 24 ways for every arrangement on the other arm, we find
+that there are 24&nbsp;&times;&nbsp;24&nbsp;=&nbsp;576 variations for every form; therefore, as
+there are 18 forms, we get 18&nbsp;&times;&nbsp;576&nbsp;=&nbsp;10,368 ways. But this will
+include half the four reversals and half the four reflections that we
+barred, so we must divide this by 4 to obtain the correct answer to
+the Greek Cross, which is thus 2,592 different ways. The division is
+by 4 and not by 8, because we provided against half the reversals and
+reflections by always reserving one number for the upright and the
+other for the horizontal.</p>
+
+<p>In the case of the Latin Cross, it is obvious that we have to deal
+with the same 18 forms of pairing. The total number of different ways
+in this case is the full number, 18&nbsp;&times;&nbsp;576. Owing to the fact that the
+upper and lower arms are unequal in length, permutations will repeat
+by reflection, but not by reversal, for we cannot reverse. Therefore
+this fact only entails division by 2. But in every pair we may
+exchange the figures in the upright with those in the horizontal
+(which we could not do in the case of the Greek Cross, as the arms are
+there all alike); consequently we must multiply by 2. This
+multiplication by 2 and division by 2 cancel one another. Hence 10,368
+is here the correct answer.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_278_A_DORMITORY_PUZZLEa" id="X_278_A_DORMITORY_PUZZLEa"></a><a href="#X_278_A_DORMITORY_PUZZLE"><b>278.&mdash;A DORMITORY PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a278.png" width="400" height="276" alt="" title="" />
+</div>
+
+<p>Arrange the nuns from day to day as shown in the six diagrams. The
+smallest possible number of nuns would be thirty-two, and the
+arrangements on the last three days admit of variation.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_279_THE_BARRELS_OF_BALSAMa" id="X_279_THE_BARRELS_OF_BALSAMa"></a><a href="#X_279_THE_BARRELS_OF_BALSAM"><b>279.&mdash;THE BARRELS OF BALSAM.&mdash;<i>solution</i></b></a></p>
+
+<p>This is quite easy to solve for any number of barrels&mdash;if you know
+how. This is the way to do it. There are five barrels in each row
+Multiply the numbers 1, 2, 3, 4, 5 together; and also multiply 6, 7,
+8, 9, 10 together. Divide one result by the other, and we get the
+number of different combinations or selections of ten things taken
+five at a time. This is here 252. Now, if we divide this by 6 (1 more
+than the number in the row) we get 42, which is the correct answer to
+the puzzle, for there are 42 different ways of arranging the barrels.
+Try this method of solution in the case of six barrels, three in each
+row, and you will find the answer is 5 ways. If you check this by
+trial, you will discover the five arrangements with 123, 124, 125,
+134, 135 respectively in the top row, and you will find no others.</p>
+
+<p>The general solution to the problem is, in fact, this:</p>
+
+<div class='center'>
+<table border="0" cellpadding="2" cellspacing="0" summary="">
+<tr><td align='center' rowspan='2' class='bb'><div style="font-size: larger"><b>C</b></div></td><td align='center'><sub>n</sub></td></tr>
+<tr><td align='center' class='bb'><sup>2<i>n</i></sup></td></tr>
+<tr><td align='center' colspan='2'><i>n</i>&nbsp;+&nbsp;1</td></tr>
+</table></div>
+
+<p>where 2<i>n</i> equals the number of barrels. The symbol <b>C</b>, of course,
+implies that we have to find how many combinations, or selections, we
+can make of 2<i>n</i> things, taken <i>n</i> at a time.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_280_BUILDING_THE_TETRAHEDRONa" id="X_280_BUILDING_THE_TETRAHEDRONa"></a><a href="#X_280_BUILDING_THE_TETRAHEDRON"><b>280.&mdash;BUILDING THE TETRAHEDRON.&mdash;<i>solution</i></b></a></p>
+
+<p>Take your constructed pyramid and hold it so that one stick only lies
+on the table. Now, four sticks must branch off from it in different
+directions&mdash;two at each end. Any one of the five sticks may be left
+out of this connection; therefore the four may be selected in 5
+different ways. But these four matches may be placed in 24 different
+orders. And as any match may be joined at either of its ends, they may
+further be varied (after their situations are settled for any
+particular arrangement) in 16 different ways. In every arrangement the
+sixth stick may be added in 2 different ways. Now multiply these
+results together, and we get 5&nbsp;&times;&nbsp;24&nbsp;&times;&nbsp;16&nbsp;&times;&nbsp;2&nbsp;=&nbsp;3,840 as the exact
+number of ways in which the pyramid may be constructed. This method
+excludes all possibility of error.</p>
+
+<p>A common cause of error is this. If you calculate your combinations by
+working upwards from a basic triangle lying on the table, you will get
+half the correct number of ways, because you overlook the fact that an
+equal number of pyramids may be built on that triangle downwards, so
+to speak, through the table. They are, in fact, reflections of the
+others, and examples from the two sets of pyramids cannot be set up to
+resemble one another&mdash;except under fourth dimensional conditions!</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_281_PAINTING_A_PYRAMIDa" id="X_281_PAINTING_A_PYRAMIDa"></a><a href="#X_281_PAINTING_A_PYRAMID"><b>281.&mdash;PAINTING A PYRAMID.&mdash;<i>solution</i></b></a></p>
+
+<p>It will be convenient to imagine that we are painting our pyramids on
+the flat cardboard, as in the diagrams, before folding up. Now, if we
+take any <i>four</i> colours (say red, blue, green, and yellow), they may
+be applied in only 2 distinctive ways, as shown in Figs, 1 and 2. Any
+other way will only result in one of these when the pyramids are
+folded up. If we take any <i>three</i> colours, they may be applied in the
+3 ways shown in Figs. 3, 4, and 5. If we take any <i>two</i> colours, they
+may be applied in the 3 <span class='pagenum'>Pg 209<a name="Page_209" id="Page_209"></a></span>ways shown in Figs. 6, 7, and 8. If we take
+any <i>single</i> colour, it may obviously be applied in only 1 way. But
+four colours may be selected in 35 ways out of seven; three in 35
+ways; two in 21 ways; and one colour in 7 ways. Therefore 35 applied
+in 2 ways&nbsp;=&nbsp;70; 35 in 3 ways&nbsp;=&nbsp;105; 21 in 3 ways&nbsp;=&nbsp;63; and 7 in 1 way
+= 7. Consequently the pyramid may be painted in 245 different ways (70
++ 105&nbsp;+&nbsp;63&nbsp;+&nbsp;7), using the seven colours of the solar spectrum in
+accordance with the conditions of the puzzle.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a281.png" width="400" height="345" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_282_THE_ANTIQUARYS_CHAINa" id="X_282_THE_ANTIQUARYS_CHAINa"></a><a href="#X_282_THE_ANTIQUARYS_CHAIN"><b>282.&mdash;THE ANTIQUARY'S CHAIN.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a282.png" width="400" height="239" alt="" title="" />
+</div>
+
+<p>The number of ways in which nine things may be arranged in a row
+without any restrictions is 1&nbsp;&times;&nbsp;2&nbsp;&times;&nbsp;3&nbsp;&times;&nbsp;4&nbsp;&times;&nbsp;5&nbsp;&times;&nbsp;6&nbsp;&times;&nbsp;7&nbsp;&times;&nbsp;8&nbsp;&times;&nbsp;9 =
+362,880. But we are told that the two circular rings must never be
+together; therefore we must deduct the number of times that this would
+occur. The number is 1&nbsp;&times;&nbsp;2&nbsp;&times;&nbsp;3&nbsp;&times;&nbsp;4&nbsp;&times;&nbsp;5&nbsp;&times;&nbsp;6&nbsp;&times;&nbsp;7&nbsp;&times;&nbsp;8&nbsp;=&nbsp;40,320&nbsp;&times;&nbsp;2 =
+80,640, because if we consider the two circular links to be
+inseparably joined together they become as one link, and eight links
+are capable of 40,320 arrangements; but as these two links may always
+be put on in the orders AB or BA, we have to double this number, it
+being a question of arrangement and not of design. The deduction
+required reduces our total to 282,240. Then one of our links is of a
+peculiar form, like an 8. We have therefore the option of joining on
+either one end or the other on every occasion, so we must double the
+last result. This brings up our total to 564,480.</p>
+
+<p>We now come to the point to which I directed the reader's
+attention&mdash;that every link may be put on in one of two ways. If we
+join the first finger and thumb of our left hand horizontally, and
+then link the first finger and thumb of the right hand, we see that
+the right thumb may be either above or below. But in the case of our
+chain we must remember that although that 8-shaped link has two
+independent <i>ends</i> it is like every other link in having only two
+<i>sides</i>&mdash;that is, you cannot turn over one end without turning the
+other at the same time.</p>
+
+<p>We will, for convenience, assume that each link has a black side and a
+side painted white. Now, if it were stipulated that (with the chain
+lying on the table, and every successive link falling over its
+predecessor in the same way, as in the diagram) only the white sides
+should be uppermost as in A, then the answer would be 564,480, as
+above&mdash;ignoring for the present all reversals of the completed chain.
+If, however, the first link were allowed to be placed either side up,
+then we could have either A or B, and the answer would be 2&nbsp;&times;&nbsp;564,480
+= 1,128,960; if two links might be placed either way up, the answer
+would be 4&nbsp;&times;&nbsp;564,480; if three links, then 8&nbsp;&times;&nbsp;564,480, and so on.
+Since, therefore, every link may be placed either side up, the number
+will be 564,480 multiplied by 2<sup>9</sup>, or by 512. This raises our total to
+289,013,760.</p>
+
+<p>But there is still one more point to be considered. We have not yet
+allowed for the fact that with any given arrangement three of the
+other arrangements may be obtained by simply turning the chain over
+through its entire length and by reversing the ends. Thus C is really
+the same as A, and if we turn this page upside down, then A and C give
+two other arrangements that are still really identical. Thus to get
+the correct answer to the puzzle we must divide our last total by 4,
+when we find that there are just 72,253,440 different ways in which
+the smith might have put those links together. In other words, if the
+nine links had originally formed a piece of chain, and it was known
+that the two circular links were separated, then it would be
+72,253,439 chances to 1 that the smith would not have put the links
+together again precisely as they were arranged before!</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_283_THE_FIFTEEN_DOMINOESa" id="X_283_THE_FIFTEEN_DOMINOESa"></a><a href="#X_283_THE_FIFTEEN_DOMINOES"><b>283.&mdash;THE FIFTEEN DOMINOES.&mdash;<i>solution</i></b></a></p>
+
+<p>The reader may have noticed that at each end of the line I give is a
+four, so that, if we like, we can form a ring instead of a line. It
+can easily be proved that this must always be so. Every line
+arrangement will make a circular arrangement if we like to join the
+ends. Now, curious as it may at first appear, the following diagram
+exactly represents the conditions when we leave the doubles out of the
+question and devote our attention to forming circular arrangements.
+Each number, or half domino, is in line with every other number, so
+that if we start at any one of the five numbers and go over all the
+lines of the pentagon once and once only we shall come back to the
+starting place, and the order of our route will give us one of the
+circular arrangements for the ten <span class='pagenum'>Pg 210<a name="Page_210" id="Page_210"></a></span>dominoes. Take your pencil and
+follow out the following route, starting at the 4: 41304210234. You
+have been over all the lines once only, and by repeating all these
+figures in this way, 41&mdash;13&mdash;30&mdash;04&mdash;42&mdash;21&mdash;10&mdash;02&mdash;23&mdash;34, you get
+an arrangement of the dominoes (without the doubles) which will be
+perfectly clear. Take other routes and you will get other
+arrangements. If, therefore, we can ascertain just how many of these
+circular routes are obtainable from the pentagon, then the rest is
+very easy.</p>
+
+<p>Well, the number of different circular routes over the pentagon is
+264. How I arrive at these figures I will not at present explain,
+because it would take a lot of space. The dominoes may, therefore, be
+arranged in a circle in just 264 different ways, leaving out the
+doubles. Now, in any one of these circles the five doubles may be
+inserted in 2<sup>5</sup>&nbsp;=&nbsp;32 different ways. Therefore when we include the
+doubles there are 264&nbsp;&times;&nbsp;32&nbsp;=&nbsp;8,448 different circular arrangements.
+But each of those circles may be broken (so as to form our straight
+line) in any one of 15 different places. Consequently, 8,448&nbsp;&times;&nbsp;15
+gives 126,720 different ways as the correct answer to the puzzle.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a283.png" width="400" height="382" alt="" title="" />
+</div>
+
+<p>I purposely refrained from asking the reader to discover in just how
+many different ways the full set of twenty-eight dominoes may be
+arranged in a straight line in accordance with the ordinary rules of
+the game, left to right and right to left of any arrangement counting
+as different ways. It is an exceedingly difficult problem, but the
+correct answer is 7,959,229,931,520 ways. The method of solving is
+very complex.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_284_THE_CROSS_TARGETa" id="X_284_THE_CROSS_TARGETa"></a><a href="#X_284_THE_CROSS_TARGET"><b>284.&mdash;THE CROSS TARGET.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a284.png" width="400" height="394" alt="" title="" />
+</div>
+
+<p>Twenty-one different squares may be selected. Of these nine will be of
+the size shown by the four A's in the diagram, four of the size shown
+by the B's, four of the size shown by the C's, two of the size shown
+by the D's, and two of the size indicated by the upper single A, the
+upper single E, the lower single C, and the EB. It is an interesting
+fact that you cannot form any one of these twenty-one squares without
+using at least one of the six circles marked E.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_285_THE_FOUR_POSTAGE_STAMPSa" id="X_285_THE_FOUR_POSTAGE_STAMPSa"></a><a href="#X_285_THE_FOUR_POSTAGE_STAMPS"><b>285.&mdash;THE FOUR POSTAGE STAMPS.&mdash;<i>solution</i></b></a></p>
+
+<p>Referring to the original diagram, the four stamps may be given in the
+shape 1, 2, 3, 4, in three ways; in the shape 1, 2, 5, 6, in six ways;
+in the shape 1, 2, 3, 5, or 1, 2, 3, 7, or 1, 5, 6, 7, or 3, 5, 6, 7,
+in twenty-eight ways; in shape 1, 2, 3, 6, or 2, 5, 6, 7, in fourteen
+ways; in shape 1, 2, 6, 7, or 2, 3, 5, 6, or 1, 5, 6, 10, or 2, 5, 6,
+9, in fourteen ways. Thus there are sixty-five ways in all.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_286_PAINTING_THE_DIEa" id="X_286_PAINTING_THE_DIEa"></a><a href="#X_286_PAINTING_THE_DIE"><b>286.&mdash;PAINTING THE DIE.&mdash;<i>solution</i></b></a></p>
+
+<p>The 1 can be marked on any one of six different sides. For every side
+occupied by 1 we have a selection of four sides for the 2. For every
+situation of the 2 we have two places for the 3. (The 6, 5, and 4 need
+not be considered, as their positions are determined by the 1, 2, and
+3.) Therefore 6, 4, and 2 multiplied together make 48 different
+ways&mdash;the correct answer.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_287_AN_ACROSTIC_PUZZLEa" id="X_287_AN_ACROSTIC_PUZZLEa"></a><a href="#X_287_AN_ACROSTIC_PUZZLE"><b>287.&mdash;AN ACROSTIC PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>There are twenty-six letters in the alphabet, giving 325 different
+pairs. Every one of these pairs may be reversed, making 650 ways. But
+every initial letter may be repeated as the final, producing 26 other
+ways. The total is therefore 676 different pairs. In other words, the
+answer is the square of the number of letters in the alphabet.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_288_CHEQUERED_BOARD_DIVISIONSa" id="X_288_CHEQUERED_BOARD_DIVISIONSa"></a><a href="#X_288_CHEQUERED_BOARD_DIVISIONS"><b>288.&mdash;CHEQUERED BOARD DIVISIONS.&mdash;<i>solution</i></b></a></p>
+
+<p>There are 255 different ways of cutting the board into two pieces of
+exactly the same size <span class='pagenum'>Pg 211<a name="Page_211" id="Page_211"></a></span>and shape. Every way must involve one of the
+five cuts shown in Diagrams A, B, C, D, and E. To avoid repetitions by
+reversal and reflection, we need only consider cuts that enter at the
+points a, b, and c. But the exit must always be at a point in a
+straight line from the entry through the centre. This is the most
+important condition to remember. In case B you cannot enter at a, or
+you will get the cut provided for in E. Similarly in C or D, you must
+not enter the key-line in the same direction as itself, or you will
+get A or B. If you are working on A or C and entering at a, you must
+consider joins at one end only of the key-line, or you will get
+repetitions. In other cases you must consider joins at both ends of
+the key; but after leaving a in case D, turn always either to right or
+left&mdash;use one direction only. Figs. 1 and 2 are examples under A; 3
+and 4 are examples under B; 5 and 6 come under C; <span class='pagenum'>Pg 212<a name="Page_212" id="Page_212"></a></span>and 7 is a pretty example of D. Of course, E is a peculiar type, and
+obviously admits of only one way of cutting, for you clearly cannot
+enter at b or c.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a288.png" width="600" height="791" alt="" title="" />
+</div>
+
+<p>Here is a table of the results:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'></td><td></td><td align='center'>a</td><td></td><td align='center'>b</td><td></td><td align='center'>c</td><td></td><td align='center'>Ways.</td></tr>
+<tr><td align='right'>A</td><td align='right'>=</td><td align='right'>8</td><td align='right'>+</td><td align='right'>17</td><td align='right'>+</td><td align='right'>21</td><td align='right'>=</td><td align='right'>46</td></tr>
+<tr><td align='right'>B</td><td align='right'>=</td><td align='right'>0</td><td align='right'>+</td><td align='right'>17</td><td align='right'>+</td><td align='right'>21</td><td align='right'>=</td><td align='right'>38</td></tr>
+<tr><td align='right'>C</td><td align='right'>=</td><td align='right'>15</td><td align='right'>+</td><td align='right'>31</td><td align='right'>+</td><td align='right'>39</td><td align='right'>=</td><td align='right'>85</td></tr>
+<tr><td align='right'>D</td><td align='right'>=</td><td align='right'>17</td><td align='right'>+</td><td align='right'>29</td><td align='right'>+</td><td align='right'>39</td><td align='right'>=</td><td align='right'>85</td></tr>
+<tr><td align='right'>E</td><td align='right'>=</td><td align='right'>1</td><td align='right'>+</td><td align='right'>0</td><td align='right'>+</td><td align='right'>0</td><td align='right'>=</td><td align='right'>1</td></tr>
+<tr><td align='right'></td><td></td><td align='right' class='bt'>41</td><td></td><td align='right' class='bt'>94</td><td></td><td align='right' class='bt'>120</td><td></td><td align='right' class='bt'>255</td></tr>
+</table></div>
+
+<p>I have not attempted the task of enumerating the ways of dividing a
+board 8x8&mdash;that is, an ordinary chessboard. Whatever the method
+adopted, the solution would entail considerable labour.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_289_LIONS_AND_CROWNSa" id="X_289_LIONS_AND_CROWNSa"></a><a href="#X_289_LIONS_AND_CROWNS"><b>289.&mdash;LIONS AND CROWNS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a289.png" width="400" height="396" alt="" title="" />
+</div>
+
+<p>Here is the solution. It will be seen that each of the four pieces
+(after making the cuts along the thick lines) is of exactly the same
+size and shape, and that each piece contains a lion and a crown. Two
+of the pieces are shaded so as to make the solution quite clear to the
+eye.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_290_BOARDS_WITH_AN_ODD_NUMBER_OF_SQUARESa" id="X_290_BOARDS_WITH_AN_ODD_NUMBER_OF_SQUARESa"></a><a href="#X_290_BOARDS_WITH_AN_ODD_NUMBER_OF_SQUARES"><b>290.&mdash;BOARDS WITH AN ODD NUMBER OF SQUARES.&mdash;<i>solution</i></b></a></p>
+
+<p>There are fifteen different ways of cutting the 5x5 board (with the
+central square removed) into two pieces of the same size and shape.
+Limitations of space will not allow me to give diagrams of all these,
+but I will enable the reader to draw them all out for himself without
+the slightest difficulty. At whatever point on the edge your cut
+enters, it must always end at a point on the edge, exactly opposite in
+a line through the centre of the square. Thus, if you enter at point 1
+(see Fig. 1) at the top, you must leave at point 1 at the bottom. Now,
+1 and 2 are the only two really different points of entry; if we use
+any others they will simply produce similar solutions. The directions
+of the cuts in the following fifteen</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a290.png" width="400" height="234" alt="" title="" />
+</div>
+
+<p>solutions are indicated by the numbers on the diagram. The duplication
+of the numbers can lead to no confusion, since every successive number
+is contiguous to the previous one. But whichever direction you take
+from the top downwards you must repeat from the bottom upwards, one
+direction being an exact reflection of the other.</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='left'>1, 4, 8.</td></tr>
+<tr><td align='left'>1, 4, 3, 7, 8.</td></tr>
+<tr><td align='left'>1, 4, 3, 7, 10, 9.</td></tr>
+<tr><td align='left'>1, 4, 3, 7, 10, 6, 5, 9.</td></tr>
+<tr><td align='left'>1, 4, 5, 9.</td></tr>
+<tr><td align='left'>1, 4, 5, 6, 10, 9.</td></tr>
+<tr><td align='left'>1, 4, 5, 6, 10, 7, 8.</td></tr>
+<tr><td align='left'>2, 3, 4, 8.</td></tr>
+<tr><td align='left'>2, 3, 4, 5, 9.</td></tr>
+<tr><td align='left'>2, 3, 4, 5, 6, 10, 9.</td></tr>
+<tr><td align='left'>2, 3, 4, 5, 6, 10, 7, 8.</td></tr>
+<tr><td align='left'>2, 3, 7, 8.</td></tr>
+<tr><td align='left'>2, 3, 7, 10, 9.</td></tr>
+<tr><td align='left'>2, 3, 7, 10, 6, 5, 9.</td></tr>
+<tr><td align='left'>2, 3, 7, 10, 6, 5, 4, 8.</td></tr>
+</table></div>
+
+<p>It will be seen that the fourth direction (1, 4, 3, 7, 10, 6, 5, 9)
+produces the solution shown in Fig. 2. The thirteenth produces the
+solution given in propounding the puzzle, where the cut entered at the
+side instead of at the top. The pieces, however, will be of the same
+shape if turned over, which, as it was stated in the conditions, would
+not constitute a different solution.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_291_THE_GRAND_LAMAS_PROBLEMa" id="X_291_THE_GRAND_LAMAS_PROBLEMa"></a><a href="#X_291_THE_GRAND_LAMAS_PROBLEM"><b>291.&mdash;THE GRAND LAMA'S PROBLEM.&mdash;<i>solution</i></b></a></p>
+
+<p>The method of dividing the chessboard so that each of the four parts
+shall be of exactly the same size and shape, and contain one of the
+gems, is shown in the diagram. The method of shading the squares is
+adopted to make the shape of the pieces clear to the eye. Two of the
+pieces are shaded and two left white.</p>
+
+<p>The reader may find it interesting to compare this puzzle with that of
+the "Weaver" (No. 14, <i>Canterbury Puzzles</i>).</p>
+
+<p><span class='pagenum'>Pg 213<a name="Page_213" id="Page_213"></a></span></p><div class="figcenter" style="width: 400px;">
+<img src="images/a291.png" width="400" height="398" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_292_THE_ABBOTS_WINDOWa" id="X_292_THE_ABBOTS_WINDOWa"></a><a href="#X_292_THE_ABBOTS_WINDOW"><b>292.&mdash;THE ABBOT'S WINDOW.&mdash;<i>solution</i></b></a></p>
+
+<p>The man who was "learned in strange mysteries" pointed out to Father
+John that the orders of the Lord Abbot of St. Edmondsbury might be
+easily carried out by blocking up twelve of the lights in the window
+as shown by the dark squares in the following sketch:&mdash;</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a292.png" width="400" height="388" alt="" title="" />
+</div>
+
+<p>Father John held that the four corners should also be darkened, but
+the sage explained that it was desired to obstruct no more light than
+was absolutely necessary, and he said, anticipating Lord Dundreary, "A
+single pane can no more be in a <i>line</i> with itself than one bird can
+go into a corner and flock in solitude. The Abbot's condition was that
+no diagonal <i>lines</i> should contain an odd number of lights."</p>
+
+<p>Now, when the holy man saw what had been done he was well pleased, and
+said, "Truly, Father John, thou art a man of deep wisdom, in that thou
+hast done that which seemed impossible, and yet withal adorned our
+window with a device of the cross of St. Andrew, whose name I received
+from my godfathers and godmothers." Thereafter he slept well and arose
+refreshed. The window might be seen intact to-day in the monastery of
+St. Edmondsbury, if it existed, which, alas! the window does not.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_293_THE_CHINESE_CHESSBOARDa" id="X_293_THE_CHINESE_CHESSBOARDa"></a><a href="#X_293_THE_CHINESE_CHESSBOARD"><b>293.&mdash;THE CHINESE CHESSBOARD.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a293.png" width="400" height="799" alt="" title="" />
+</div>
+
+<p>Eighteen is the maximum number of pieces. I give two solutions. The
+numbered diagram is so cut that the eighteenth piece has the largest
+area&mdash;eight squares&mdash;that is possible under the conditions. The second
+diagram was prepared under the added condition that no piece should
+contain more than five squares.</p>
+
+<p>No. 74 in <i>The Canterbury Puzzles</i> shows how to cut the board into
+twelve pieces, all differ<span class='pagenum'>Pg 214<a name="Page_214" id="Page_214"></a></span>ent, each containing five squares, with one
+square piece of four squares.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_294_THE_CHESSBOARD_SENTENCEa" id="X_294_THE_CHESSBOARD_SENTENCEa"></a><a href="#X_294_THE_CHESSBOARD_SENTENCE"><b>294.&mdash;THE CHESSBOARD SENTENCE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a294.png" width="400" height="397" alt="" title="" />
+</div>
+
+<p>The pieces may be fitted together, as shown in the illustration, to
+form a perfect chessboard.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_295_THE_EIGHT_ROOKSa" id="X_295_THE_EIGHT_ROOKSa"></a><a href="#X_295_THE_EIGHT_ROOKS"><b>295.&mdash;THE EIGHT ROOKS.&mdash;<i>solution</i></b></a></p>
+
+<p>Obviously there must be a rook in every row and every column. Starting
+with the top row, it is clear that we may put our first rook on any
+one of eight different squares. Wherever it is placed, we have the
+option of seven squares for the second rook in the second row. Then we
+have six squares from which to select the third row, five in the
+fourth, and so on. Therefore the number of our different ways must be
+8 � 7 � 6 � 5 � 4 � 3 � 2 � 1&nbsp;=&nbsp;40,320 (that is 8!), which is the
+correct answer.</p>
+
+<p>How many ways there are if mere reversals and reflections are not
+counted as different has not yet been determined; it is a difficult
+problem. But this point, on a smaller square, is considered in the
+next puzzle.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_296_THE_FOUR_LIONSa" id="X_296_THE_FOUR_LIONSa"></a><a href="#X_296_THE_FOUR_LIONS"><b>296.&mdash;THE FOUR LIONS.&mdash;<i>solution</i></b></a></p>
+
+<p>There are only seven different ways under the conditions. They are as
+follows: 1 2 3 4, 1 2 4 3, 1 3 2 4, 1 3 4 2, 1 4 3 2, 2 1 4 3, 2 4 1
+3. Taking the last example, this notation means that we place a lion
+in the second square of first row, fourth square of second row, first
+square of third row, and third square of fourth row. The first example
+is, of course, the one we gave when setting the puzzle.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_297_BISHOPSmdashUNGUARDEDa" id="X_297_BISHOPSmdashUNGUARDEDa"></a><a href="#X_297_BISHOPSmdashUNGUARDED"><b>297.&mdash;BISHOPS&mdash;UNGUARDED.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a297.png" width="400" height="394" alt="" title="" />
+</div>
+
+<p>This cannot be done with fewer bishops than eight, and the simplest
+solution is to place the bishops in line along the fourth or fifth row
+of the board (see diagram). But it will be noticed that no bishop is
+here guarded by another, so we consider that point in the next puzzle.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_298_BISHOPSmdashGUARDEDa" id="X_298_BISHOPSmdashGUARDEDa"></a><a href="#X_298_BISHOPSmdashGUARDED"><b>298.&mdash;BISHOPS&mdash;GUARDED.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a298.png" width="400" height="405" alt="" title="" />
+</div>
+
+<p>This puzzle is quite easy if you first of all give it a little
+thought. You need only consider squares of one colour, for whatever
+can be done in the case of the white squares can always be repeated on
+the black, and they are here quite independent of one another. This
+equality, of course, is in consequence of the fact that the number of
+squares on an ordinary chessboard, sixty-four, is an even number. If a
+square chequered board has an odd number of squares, then there will
+always be one more square of one colour than of the other.</p>
+
+<p>Ten bishops are necessary in order that every square shall be attacked
+and every bishop guarded by another bishop. I give one way of
+arranging them in the diagram. It will be noticed that the two central
+bishops in the group <span class='pagenum'>Pg 215<a name="Page_215" id="Page_215"></a></span>of six on the left-hand side of the board serve
+no purpose, except to protect those bishops that are on adjoining
+squares. Another solution would therefore be obtained by simply
+raising the upper one of these one square and placing the other a
+square lower down.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_299_BISHOPS_IN_CONVOCATIONa" id="X_299_BISHOPS_IN_CONVOCATIONa"></a><a href="#X_299_BISHOPS_IN_CONVOCATION"><b>299.&mdash;BISHOPS IN CONVOCATION.&mdash;<i>solution</i></b></a></p>
+
+<p>The fourteen bishops may be placed in 256 different ways. But every
+bishop must always be placed on one of the sides of the board&mdash;that
+is, somewhere on a row or file on the extreme edge. The puzzle,
+therefore, consists in counting the number of different ways that we can
+arrange the fourteen round the edge of the board without attack. This is
+not a difficult matter. On a chessboard of <i>n</i><sup>2</sup> squares
+<i>2n&nbsp;-&nbsp;2</i> bishops (the maximum number) may always be placed in 2<sup>(<i>n</i>)</sup> ways
+without attacking. On an ordinary chessboard <i>n</i> would be 8; therefore
+14 bishops may be placed in 256 different ways. It is rather curious
+that the general result should come out in so simple a form.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_300_THE_EIGHT_QUEENSa" id="X_300_THE_EIGHT_QUEENSa"></a><a href="#X_300_THE_EIGHT_QUEENS"><b>300.&mdash;THE EIGHT QUEENS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a300.png" width="400" height="387" alt="" title="" />
+</div>
+
+<p>The solution to this puzzle is shown in the diagram. It will be found
+that no queen attacks another, and also that no three queens are in a
+straight line in any oblique direction. This is the only arrangement
+out of the twelve fundamentally different ways of placing eight queens
+without attack that fulfils the last condition.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_301_THE_EIGHT_STARSa" id="X_301_THE_EIGHT_STARSa"></a><a href="#X_301_THE_EIGHT_STARS"><b>301.&mdash;THE EIGHT STARS.&mdash;<i>solution</i></b></a></p>
+
+<p>The solution of this puzzle is shown in the first diagram. It is the
+only possible solution within the conditions stated. But if one of the
+eight stars had not already been placed as shown, there would then
+have been eight ways of arranging the stars according to this scheme,
+if we count reversals and reflections as different. If you turn this
+page round so that each side is in turn at the bottom, you will get
+the four reversals; and if you reflect each of these in a mirror, you
+will get the four reflections. These are, therefore, merely eight
+aspects of one "fundamental solution." But without that first star
+being so placed, there is another fundamental solution, as shown in
+the second diagram. But this arrangement being in a way symmetrical,
+only produces four different aspects by reversal and reflection.</p>
+
+<div class="figcenter" style="width: 500px;">
+<img src="images/a301.png" width="500" height="242" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_302_A_PROBLEM_IN_MOSAICSa" id="X_302_A_PROBLEM_IN_MOSAICSa"></a><a href="#X_302_A_PROBLEM_IN_MOSAICS"><b>302.&mdash;A PROBLEM IN MOSAICS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a302.png" width="400" height="392" alt="" title="" />
+</div>
+
+<p>The diagram shows how the tiles may be rearranged. As before, one
+yellow and one <span class='pagenum'>Pg 216<a name="Page_216" id="Page_216"></a></span>purple tile are dispensed with. I will here point out
+that in the previous arrangement the yellow and purple tiles in the
+seventh row might have changed places, but no other arrangement was
+possible.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_303_UNDER_THE_VEILa" id="X_303_UNDER_THE_VEILa"></a><a href="#X_303_UNDER_THE_VEIL"><b>303.&mdash;UNDER THE VEIL.&mdash;<i>solution</i></b></a></p>
+
+<p>Some schemes give more diagonal readings of four letters than others,
+and we are at first tempted to favour these; but this is a false
+scent, because what you appear to gain in this direction you lose in
+others. Of course it immediately occurs to the solver that every LIVE
+or EVIL is worth twice as much as any other word, since it reads both
+ways and always counts as 2. This is an important consideration,
+though sometimes those arrangements that contain most readings of
+these two words are fruitless in other words, and we lose in the
+general count.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a303.png" width="400" height="363" alt="" title="" />
+</div>
+
+<p>The above diagram is in accordance with the conditions requiring no
+letter to be in line with another similar letter, and it gives twenty
+readings of the five words&mdash;six horizontally, six vertically, four in
+the diagonals indicated by the arrows on the left, and four in the
+diagonals indicated by the arrows on the right. This is the maximum.</p>
+
+<p>Four sets of eight letters may be placed on the board of sixty-four
+squares in as many as 604 different ways, without any letter ever
+being in line with a similar one. This does not count reversals and
+reflections as different, and it does not take into consideration the
+actual permutations of the letters among themselves; that is, for
+example, making the L's change places with the E's. Now it is a
+singular fact that not only do the twenty word-readings that I have
+given prove to be the real maximum, but there is actually only that
+one arrangement from which this maximum may be obtained. But if you
+make the V's change places with the I's, and the L's with the E's, in
+the solution given, you still get twenty readings&mdash;the same number as
+before in every direction. Therefore there are two ways of getting the
+maximum from the same arrangement. The minimum number of readings is
+zero&mdash;that is, the letters can be so arranged that no word can be read
+in any of the directions.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_304_BACHETS_SQUAREa" id="X_304_BACHETS_SQUAREa"></a><a href="#X_304_BACHETS_SQUARE"><b>304.&mdash;BACHET'S SQUARE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a304.png" width="400" height="426" alt="" title="" />
+</div>
+
+<p>Let us use the letters A, K, Q, J, to denote ace, king, queen, jack;
+and D, S, H, C, to denote diamonds, spades, hearts, clubs. In Diagrams
+1 and 2 we have the two available ways of arranging either group of
+letters so that no two similar letters shall be in line&mdash;though a
+quarter-turn of 1 will give us the arrangement in 2. If we superimpose
+or combine these two squares, we get the arrangement of Diagram 3,
+which is one solution. But in each square we may put the letters in
+the top line in twenty-four different ways without altering the scheme
+of arrangement. Thus, in Diagram 4 the S's are similarly placed to the
+D's in 2, the H's to the S's, the C's to the H's, and the D's to the
+C's. It clearly follows that there must be 24&nbsp;&times;&nbsp;24&nbsp;=&nbsp;576 ways of
+combining the two primitive arrangements. But the error that Labosne
+fell into was that of assuming that the A, K, Q, J must be arranged in
+the form 1, and the D, S, H, C in the form 2. He thus included
+reflections and half-turns, but not quarter-turns. They may obviously
+be interchanged. So that the correct answer is 2&nbsp;&times;&nbsp;576&nbsp;=&nbsp;1,152, counting
+reflections and reversals as different. Put in another manner, the
+pairs in the top row may be written in 16&nbsp;&times;&nbsp;9 &times;4&nbsp;&times;&nbsp;1&nbsp;=&nbsp;576 different ways,
+and the square then completed in 2 ways, making 1,152 ways in all.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_305_THE_THIRTY_SIX_LETTER_BLOCKSa" id="X_305_THE_THIRTY_SIX_LETTER_BLOCKSa"></a><a href="#X_305_THE_THIRTY_SIX_LETTER_BLOCKS"><b>305.&mdash;THE THIRTY-SIX LETTER BLOCKS.&mdash;<i>solution</i></b></a></p>
+
+<p>I pointed out that it was impossible to get all the letters into the
+box under the conditions, but the puzzle was to place as many as
+possible.</p>
+
+<p><span class='pagenum'>Pg 217<a name="Page_217" id="Page_217"></a></span>This requires a little judgment and careful investigation, or we are
+liable to jump to the hasty conclusion that the proper way to solve
+the puzzle must be first to place all six of one letter, then all six
+of another letter, and so on. As there is only one scheme (with its
+reversals) for placing six similar letters so that no two shall be in
+a line in any direction, the reader will find that after he has placed
+four different kinds of letters, six times each, every place is
+occupied except those twelve that form the two long diagonals. He is,
+therefore, unable to place more than two each of his last two letters,
+and there are eight blanks left. I give such an arrangement in Diagram
+1.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a305.png" width="400" height="214" alt="" title="" />
+</div>
+
+<p>The secret, however, consists in not trying thus to place all six of
+each letter. It will be found that if we content ourselves with
+placing only five of each letter, this number (thirty in all) may be
+got into the box, and there will be only six blanks. But the correct
+solution is to place six of each of two letters and five of each of
+the remaining four. An examination of Diagram 2 will show that there
+are six each of C and D, and five each of A, B, E, and F. There are,
+therefore, only four blanks left, and no letter is in line with a
+similar letter in any direction.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_306_THE_CROWDED_CHESSBOARDa" id="X_306_THE_CROWDED_CHESSBOARDa"></a><a href="#X_306_THE_CROWDED_CHESSBOARD"><b>306.&mdash;THE CROWDED CHESSBOARD.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a306.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>Here is the solution. Only 8 queens or 8 rooks can be placed on the
+board without attack, while the greatest number of bishops is 14, and
+of knights 32. But as all these knights must be placed on squares of
+the same colour, while the queens occupy four of each colour and the
+bishops 7 of each colour, it follows that only 21 knights can be
+placed on the same colour in this puzzle. More than 21 knights can be
+placed alone on the board if we use both colours, but I have not
+succeeded in placing more than 21 on the "crowded chessboard." I
+believe the above solution contains the maximum number of pieces, but
+possibly some ingenious reader may succeed in getting in another
+knight.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_307_THE_COLOURED_COUNTERSa" id="X_307_THE_COLOURED_COUNTERSa"></a><a href="#X_307_THE_COLOURED_COUNTERS"><b>307.&mdash;THE COLOURED COUNTERS.&mdash;<i>solution</i></b></a></p>
+
+<p>The counters may be arranged in this order:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>R1,</td><td align='center'>B2,</td><td align='center'>Y3,</td><td align='center'>O4,</td><td align='center'>GS.</td></tr>
+<tr><td align='center'>Y4,</td><td align='center'>O5,</td><td align='center'>G1,</td><td align='center'>R2,</td><td align='center'>B3.</td></tr>
+<tr><td align='center'>G2,</td><td align='center'>R3,</td><td align='center'>B4,</td><td align='center'>Y5,</td><td align='center'>O1.</td></tr>
+<tr><td align='center'>B5,</td><td align='center'>Y1,</td><td align='center'>O2,</td><td align='center'>G3,</td><td align='center'>R4.</td></tr>
+<tr><td align='center'>O3,</td><td align='center'>G4,</td><td align='center'>R5,</td><td align='center'>B1,</td><td align='center'>Y2.</td></tr>
+</table></div>
+
+<hr style="width: 30%;" />
+<p><a name="X_308_THE_GENTLE_ART_OF_STAMP-LICKINGa" id="X_308_THE_GENTLE_ART_OF_STAMP-LICKINGa"></a><a href="#X_308_THE_GENTLE_ART_OF_STAMP-LICKING"><b>308.&mdash;THE GENTLE ART OF STAMP-LICKING.&mdash;<i>solution</i></b></a></p>
+
+<p>The following arrangement shows how sixteen stamps may be stuck on the
+card, under the conditions, of a total value of fifty pence, or 4<i>s</i>.
+2<i>d</i>.:&mdash;</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a308.png" width="400" height="322" alt="" title="" />
+</div>
+
+<p>If, after placing the four 5<i>d</i>. stamps, the reader is tempted to place
+four 4<i>d</i>. stamps also, he can afterwards only place two of each of the
+three other denominations, thus losing two spaces and counting no more
+than forty-eight pence, or 4<i>s</i>. This is the pitfall that was hinted at.
+(Compare with No. 43, <i>Canterbury Puzzles</i>.)</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_309_THE_FORTY-NINE_COUNTERSa" id="X_309_THE_FORTY-NINE_COUNTERSa"></a><a href="#X_309_THE_FORTY-NINE_COUNTERS"><b>309.&mdash;THE FORTY-NINE COUNTERS.&mdash;<i>solution</i></b></a></p>
+
+<p>The counters may be arranged in this order:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>A1,</td><td align='center'>B2,</td><td align='center'>C3,</td><td align='center'>D4,</td><td align='center'>E5,</td><td align='center'>F6,</td><td align='center'>G7.</td></tr>
+<tr><td align='center'>F4,</td><td align='center'>G5,</td><td align='center'>A6,</td><td align='center'>B7,</td><td align='center'>C1,</td><td align='center'>D2,</td><td align='center'>E3.</td></tr>
+<tr><td align='center'>D7,</td><td align='center'>E1,</td><td align='center'>F2,</td><td align='center'>G3,</td><td align='center'>A4,</td><td align='center'>B5,</td><td align='center'>C6.</td></tr>
+<tr><td align='center'>B3,</td><td align='center'>C4,</td><td align='center'>D5,</td><td align='center'>E6,</td><td align='center'>F7,</td><td align='center'>G1,</td><td align='center'>A2.</td></tr>
+<tr><td align='center'>G6,</td><td align='center'>A7,</td><td align='center'>B1,</td><td align='center'>C2,</td><td align='center'>D3,</td><td align='center'>E4,</td><td align='center'>F5.</td></tr>
+<tr><td align='center'>E2,</td><td align='center'>F3,</td><td align='center'>G4,</td><td align='center'>A5,</td><td align='center'>B6,</td><td align='center'>C7,</td><td align='center'>D1.</td></tr>
+<tr><td align='center'>C5,</td><td align='center'>D6,</td><td align='center'>E7,</td><td align='center'>F1,</td><td align='center'>G2,</td><td align='center'>A3,</td><td align='center'>B4.</td></tr>
+</table></div>
+
+<hr style="width: 30%;" />
+<p><a name="X_310_THE_THREE_SHEEPa" id="X_310_THE_THREE_SHEEPa"></a><a href="#X_310_THE_THREE_SHEEP"><b>310.&mdash;THE THREE SHEEP.&mdash;<i>solution</i></b></a></p>
+
+<p>The number of different ways in which the three sheep may be placed so
+that every pen <span class='pagenum'>Pg 218<a name="Page_218" id="Page_218"></a></span>shall always be either occupied or in line with at
+least one sheep is forty-seven.</p>
+
+<p>The following table, if used with the key in Diagram 1, will enable
+the reader to place them in all these ways:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' class='bb br'>Two Sheep.</td><td align='center' class='bb br'>Third Sheep.</td><td align='center' class='bb'>No. of <br />Ways.</td></tr>
+<tr><td align='center' class='br'>A and B</td><td align='left' class='br'>C, E, G, K, L, N, or P</td><td align='center'>7</td></tr>
+<tr><td align='center' class='br'>A and C</td><td align='left' class='br'>I, J, K, or O</td><td align='center'>4</td></tr>
+<tr><td align='center' class='br'>A and D</td><td align='left' class='br'>M, N, or J</td><td align='center'>3</td></tr>
+<tr><td align='center' class='br'>A and F</td><td align='left' class='br'>J, K, L, or P</td><td align='center'>4</td></tr>
+<tr><td align='center' class='br'>A and G</td><td align='left' class='br'>H, J, K, N, O, or P</td><td align='center'>6</td></tr>
+<tr><td align='center' class='br'>A and H</td><td align='left' class='br'>K, L, N, or O</td><td align='center'>4</td></tr>
+<tr><td align='center' class='br'>A and O</td><td align='left' class='br'>K or L</td><td align='center'>2</td></tr>
+<tr><td align='center' class='br'>B and C</td><td align='left' class='br'>N</td><td align='center'>1</td></tr>
+<tr><td align='center' class='br'>B and E</td><td align='left' class='br'>F, H, K, or L</td><td align='center'>4</td></tr>
+<tr><td align='center' class='br'>B and F</td><td align='left' class='br'>G, J, N, or O</td><td align='center'>4</td></tr>
+<tr><td align='center' class='br'>B and G</td><td align='left' class='br'>K, L, or N</td><td align='center'>3</td></tr>
+<tr><td align='center' class='br'>B and H</td><td align='left' class='br'>J or N</td><td align='center'>2</td></tr>
+<tr><td align='center' class='br'>B and J</td><td align='left' class='br'>K or L</td><td align='center'>2</td></tr>
+<tr><td align='center' class='br'>F and G</td><td align='left' class='br'>J</td><td align='center'>1</td></tr>
+<tr><td align='center' class='br'></td><td align='left' class='br'></td><td align='center' class='bt'>47</td></tr>
+</table></div>
+
+<p>This, of course, means that if you place sheep in the pens marked A
+and B, then there are seven different pens in which you may place the
+third sheep, giving seven different solutions. It was understood that
+reversals and reflections do not count as different.</p>
+
+<p>If one pen at least is to be <i>not</i> in line with a sheep, there would
+be thirty solutions to that problem. If we counted all the reversals
+and reflections of these 47 and 30 cases respectively as different,
+their total would be 560, which is the number of different ways in
+which the sheep may be placed in three pens without any conditions. I
+will remark that there are three ways in which two sheep may be placed
+so that every pen is occupied or in line, as in Diagrams 2, 3, and 4,
+but in every case each sheep is in line with its companion. There are
+only two ways in which three sheep may be so placed that every pen
+shall be occupied or in line, but no sheep in line with another. These
+I show in Diagrams 5 and 6. Finally, there is only one way in which
+three sheep may be placed so that at least one pen shall not be in
+line with a sheep and yet no sheep in line with another. Place the
+sheep in C, E, L. This is practically all there is to be said on this
+pleasant pastoral subject.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a310.png" width="600" height="449" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_311_THE_FIVE_DOGS_PUZZLEa" id="X_311_THE_FIVE_DOGS_PUZZLEa"></a><a href="#X_311_THE_FIVE_DOGS_PUZZLE"><b>311.&mdash;THE FIVE DOGS PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The diagrams show four fundamentally different solutions. In the case
+of A we can reverse <span class='pagenum'>Pg 219<a name="Page_219" id="Page_219"></a></span>the order, so that the single dog is in the
+bottom row and the other four shifted up two squares. Also we may use
+the next column to the right and both of the two central horizontal
+rows. Thus A gives 8 solutions. Then B may be reversed and placed in
+either diagonal, giving 4 solutions. Similarly C will give 4
+solutions. The line in D being symmetrical, its reversal will not be
+different, but it may be disposed in 4 different directions. We thus
+have in all 20 different solutions.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a311.png" width="400" height="418" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_312_THE_FIVE_CRESCENTS_OF_BYZANTIUMa" id="X_312_THE_FIVE_CRESCENTS_OF_BYZANTIUMa"></a><a href="#X_312_THE_FIVE_CRESCENTS_OF_BYZANTIUM"><b>312.&mdash;THE FIVE CRESCENTS OF BYZANTIUM.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 300px;">
+<img src="images/a312.png" width="300" height="300" alt="" title="" />
+</div>
+
+<p>If that ancient architect had arranged his five crescent tiles in the
+manner shown in the following diagram, every tile would have been
+watched over by, or in a line with, at least one crescent, and space
+would have been reserved for a perfectly square carpet equal in area
+to exactly half of the pavement. It is a very curious fact that,
+although there are two or three solutions allowing a carpet to be laid
+down within the conditions so as to cover an area of nearly
+twenty-nine of the tiles, this is the only possible solution giving
+exactly half the area of the pavement, which is the largest space
+obtainable.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_313_QUEENS_AND_BISHOP_PUZZLEa" id="X_313_QUEENS_AND_BISHOP_PUZZLEa"></a><a href="#X_313_QUEENS_AND_BISHOP_PUZZLE"><b>313.&mdash;QUEENS AND BISHOP PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a313a.png" width="400" height="396" alt="" title="" />
+</div>
+
+<p>The bishop is on the square originally occupied by the rook, and the
+four queens are so placed that every square is either occupied or
+attacked by a piece. (Fig. 1.)</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a313b.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>I pointed out in 1899 that if four queens are placed as shown in the
+diagram (Fig. 2), then the fifth queen may be placed on any one of the
+twelve squares marked a, b, c, d, and e; or a rook on the two squares,
+c; or a bishop on the eight squares, a, b, and e; or a pawn on <span class='pagenum'>Pg 220<a name="Page_220" id="Page_220"></a></span>the
+square b; or a king on the four squares, b, c, and e. The only known
+arrangement for four queens and a knight is that given by Mr. J.
+Wallis in <i>The Strand Magazine</i> for August 1908, here reproduced.
+(Fig. 3.)</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a313c.png" width="400" height="394" alt="" title="" />
+</div>
+
+<p>I have recorded a large number of solutions with four queens and a
+rook, or bishop, but the only arrangement, I believe, with three
+queens and two rooks in which all the pieces are guarded is that of
+which I give an illustration (Fig. 4), first published by Dr. C.
+Planck. But I have since found the accompanying solution with three
+queens, a rook, and a bishop, though the pieces do not protect one
+another. (Fig. 5.)</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a313d.png" width="400" height="398" alt="" title="" />
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a313e.png" width="400" height="396" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_314_THE_SOUTHERN_CROSSa" id="X_314_THE_SOUTHERN_CROSSa"></a><a href="#X_314_THE_SOUTHERN_CROSS"><b>314.&mdash;THE SOUTHERN CROSS.&mdash;<i>solution</i></b></a></p>
+
+<p>My readers have been so familiarized with the fact that it requires at
+least five planets to attack every one of a square arrangement of
+sixty-four stars that many of them have, perhaps, got to believe that
+a larger square arrangement of stars must need an increase of planets.
+It was to correct this possible error of reasoning, and so warn
+readers against another of those numerous little pitfalls in the world
+of puzzledom, that I devised this new stellar problem. Let me then
+state at once that, in the case of a square arrangement of eighty one
+stars, there are several ways of placing five planets so that every
+star shall be in line with at least one planet vertically,
+horizontally, or diagonally. Here is the solution to the "Southern
+Cross": &mdash;</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a314.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p><span class='pagenum'>Pg 221<a name="Page_221" id="Page_221"></a></span>It will be remembered that I said that the five planets in their new
+positions "will, of course, obscure five other stars in place of those
+at present covered." This was to exclude an easier solution in which
+only four planets need be moved.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_315_THE_HAT-PEG_PUZZLEa" id="X_315_THE_HAT-PEG_PUZZLEa"></a><a href="#X_315_THE_HAT-PEG_PUZZLE"><b>315.&mdash;THE HAT-PEG PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The moves will be made quite clear by a reference to the diagrams,
+which show the position on the board after each of the four moves. The
+darts indicate the successive removals that have been made. It will be
+seen that at every stage all the squares are either attacked or
+occupied, and that after the fourth move no queen attacks any other.
+In the case of the last move the queen in the top row might also have
+been moved one square farther to the left. This is, I believe, the
+only solution to the puzzle.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a315.png" width="600" height="634" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_316_THE_AMAZONSa" id="X_316_THE_AMAZONSa"></a><a href="#X_316_THE_AMAZONS"><b>316.&mdash;THE AMAZONS.&mdash;<i>solution</i></b></a></p>
+
+<p>It will be seen that only three queens have been removed from their
+positions on the edge of the board, and that, as a consequence, eleven
+squares (indicated by the black dots) are left unattacked by any
+queen. I will hazard the statement that eight queens cannot be placed
+on the chessboard so as to leave more than eleven squares unattacked.
+It is true that we have no rigid proof of this yet, but I have
+<span class='pagenum'>Pg 222<a name="Page_222" id="Page_222"></a></span>entirely convinced myself of the truth of the statement. There are at
+least five different ways of arranging the queens so as to leave
+eleven squares unattacked.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a316.png" width="400" height="411" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_317_A_PUZZLE_WITH_PAWNSa" id="X_317_A_PUZZLE_WITH_PAWNSa"></a><a href="#X_317_A_PUZZLE_WITH_PAWNS"><b>317.&mdash;A PUZZLE WITH PAWNS.&mdash;<i>solution</i></b></a></p>
+
+<p>Sixteen pawns may be placed so that no three shall be in a straight
+line in any possible direction, as in the diagram. We regard, as the
+conditions required, the pawns as mere points on a plane.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a317.png" width="400" height="396" alt="" title="" />
+</div>
+<hr style="width: 30%;" />
+<p><a name="X_318_LION-HUNTINGa" id="X_318_LION-HUNTINGa"></a><a href="#X_318_LION-HUNTING"><b>318.&mdash;LION-HUNTING.&mdash;<i>solution</i></b></a></p>
+
+<p>There are 6,480 ways of placing the man and the lion, if there are no
+restrictions whatever except that they must be on different spots.
+This is obvious, because the man may be placed on any one of the 81
+spots, and in every case there are 80 spots remaining for the lion;
+therefore 81&nbsp;&times;&nbsp;80&nbsp;=&nbsp;6,480. Now, if we deduct the number of ways in
+which the lion and the man may be placed on the same path, the result
+must be the number of ways in which they will not be on the same path.
+The number of ways in which they may be in line is found without much
+difficulty to be 816. Consequently, 6,480&nbsp;-&nbsp;816&nbsp;=&nbsp;5,664, the required
+answer.</p>
+
+<p>The general solution is this: <sup>1</sup>/<sub>3</sub>n(n&nbsp;-&nbsp;1) (3n<sup>2</sup>&nbsp;-&nbsp;n&nbsp;+&nbsp;2). This is, of
+course, equivalent to saying that if we call the number of squares on
+the side of a "chessboard" <i>n</i>, then the formula shows the number of
+ways in which two bishops may be placed without attacking one another.
+Only in this case we must divide by two, because the two bishops have
+no distinct individuality, and cannot produce a different solution by
+mere exchange of places.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_319_THE_KNIGHT-GUARDSa" id="X_319_THE_KNIGHT-GUARDSa"></a><a href="#X_319_THE_KNIGHT-GUARDS"><b>319.&mdash;THE KNIGHT-GUARDS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a319a.png" width="400" height="406" alt="" title="" />
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a319b.png" width="400" height="420" alt="" title="" />
+</div>
+
+<p>The smallest possible number of knights with which this puzzle can be
+solved is fourteen.</p>
+
+<p><span class='pagenum'>Pg 223<a name="Page_223" id="Page_223"></a></span>It has sometimes been assumed that there are a great many different
+solutions. As a matter of fact, there are only three arrangements&mdash;not
+counting mere reversals and reflections as different. Curiously
+enough, nobody seems ever to have hit on the following simple proof,
+or to have thought of dealing with the black and the white squares
+separately.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a319c.png" width="400" height="429" alt="" title="" />
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a319d.png" width="400" height="427" alt="" title="" />
+</div>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a319e.png" width="400" height="430" alt="" title="" />
+</div>
+
+<p>Seven knights can be placed on the board on white squares so as to
+attack every black square in two ways only. These are shown in
+Diagrams 1 and 2. Note that three knights occupy the same position in
+both arrangements. It is therefore clear that if we turn the board so
+that a black square shall be in the top left-hand corner instead of a
+white, and place the knights in exactly the same positions, we shall
+have two similar ways of attacking all the white squares. I will
+assume the reader has made the two last described diagrams on
+transparent paper, and marked them <i>1a</i> and <i>2a</i>. Now, by placing the
+transparent Diagram <i>1a</i> over 1 you will be able to obtain the
+solution in Diagram 3, by placing <i>2a</i> over 2 you will get Diagram 4,
+and by placing <i>2a</i> over 1 you will get Diagram 5. You may now try all
+possible combinations of those two pairs of diagrams, but you will
+only get the three arrangements I have given, or their reversals and
+reflections. Therefore these three solutions are all that exist.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_320_THE_ROOKS_TOURa" id="X_320_THE_ROOKS_TOURa"></a><a href="#X_320_THE_ROOKS_TOUR"><b>320.&mdash;THE ROOK'S TOUR.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a320a.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>The only possible minimum solutions are shown in the two diagrams,
+where it will be seen that only sixteen moves are required to perform
+the feat. Most people find it difficult to reduce the number of moves
+below seventeen.</p>
+
+<p><span class='pagenum'>Pg 224<a name="Page_224" id="Page_224"></a></span></p>
+<div class="figcenter" style="width: 400px;">
+<img src="images/a320b.png" width="400" height="396" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_321_THE_ROOKS_JOURNEYa" id="X_321_THE_ROOKS_JOURNEYa"></a><a href="#X_321_THE_ROOKS_JOURNEY"><b>321.&mdash;THE ROOK'S JOURNEY.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a321.png" width="400" height="392" alt="" title="" />
+</div>
+
+<p>I show the route in the diagram. It will be seen that the tenth move
+lands us at the square marked "10," and that the last move, the
+twenty-first, brings us to a halt on square "21."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_322_THE_LANGUISHING_MAIDENa" id="X_322_THE_LANGUISHING_MAIDENa"></a><a href="#X_322_THE_LANGUISHING_MAIDEN"><b>322.&mdash;THE LANGUISHING MAIDEN.&mdash;<i>solution</i></b></a></p>
+
+<p>The dotted line shows the route in twenty-two straight paths by which
+the knight may rescue the maiden. It is necessary, after entering the
+first cell, immediately to return before entering another. Otherwise a
+solution would not be possible. (See "<a href="#X_250_THE_GRAND_TOURa">The Grand Tour</a>," <a href="#Page_200">p. 200</a>.)</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a322.png" width="400" height="397" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_323_A_DUNGEON_PUZZLEa" id="X_323_A_DUNGEON_PUZZLEa"></a><a href="#X_323_A_DUNGEON_PUZZLE"><b>323.&mdash;A DUNGEON PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>If the prisoner takes the route shown in the diagram&mdash;where for
+clearness the doorways are omitted&mdash;he will succeed in visiting every
+cell once, and only once, in as many as fifty-seven straight lines. No
+rook's path over the chessboard can exceed this number of moves.</p>
+
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a323.png" width="400" height="396" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_324_THE_LION_AND_THE_MANa" id="X_324_THE_LION_AND_THE_MANa"></a><a href="#X_324_THE_LION_AND_THE_MAN"><b>324.&mdash;THE LION AND THE MAN.&mdash;<i>solution</i></b></a></p>
+
+<p>First of all, the fewest possible straight lines in each case are
+twenty-two, and in order that no cell may be visited twice it is
+absolutely necessary that each should pass into one cell and then
+immediately "visit" the one from which he started, afterwards
+proceeding by way of the second available cell. In the following
+diagram the man's route is indicated by the unbroken lines, and the
+lion's by the dotted lines. It will be found, if the two routes are
+followed cell by cell with two pencil points, that the lion and the
+man never meet. But there was one little point that ought not to be
+overlooked&mdash;"they occasionally got glimpses of one another." Now, if
+we take one route for the <span class='pagenum'>Pg 225<a name="Page_225" id="Page_225"></a></span>man and merely reverse it for the lion, we
+invariably find that, going at the same speed, they never get a
+glimpse of one another. But in our diagram it will be found that the
+man and the lion are in the cells marked A at the same moment, and may
+see one another through the open doorways; while the same happens when
+they are in the two cells marked B, the upper letters indicating the
+man and the lower the lion. In the first case the lion goes straight
+for the man, while the man appears to attempt to get in the rear of
+the lion; in the second case it looks suspiciously like running away
+from one another!</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a324.png" width="400" height="397" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_325_AN_EPISCOPAL_VISITATIONa" id="X_325_AN_EPISCOPAL_VISITATIONa"></a><a href="#X_325_AN_EPISCOPAL_VISITATION"><b>325.&mdash;AN EPISCOPAL VISITATION.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a325.png" width="400" height="401" alt="" title="" />
+</div>
+
+<p>In the diagram I show how the bishop may be made to visit every one of
+his white parishes in seventeen moves. It is obvious that we must
+start from one corner square and end at the one that is diagonally
+opposite to it. The puzzle cannot be solved in fewer than seventeen
+moves.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_326_A_NEW_COUNTER_PUZZLEa" id="X_326_A_NEW_COUNTER_PUZZLEa"></a><a href="#X_326_A_NEW_COUNTER_PUZZLE"><b>326.&mdash;A NEW COUNTER PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Play as follows: 2&mdash;3, 9&mdash;4, 10&mdash;7, 3&mdash;8, 4&mdash;2, 7&mdash;5, 8&mdash;6, 5&mdash;10,
+6&mdash;9, 2&mdash;5, 1&mdash;6, 6&mdash;4, 5&mdash;3, 10&mdash;8, 4&mdash;7, 3&mdash;2, 8&mdash;1, 7&mdash;10. The
+white counters have now changed places with the red ones, in eighteen
+moves, without breaking the conditions.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_327_A_NEW_BISHOPS_PUZZLEa" id="X_327_A_NEW_BISHOPS_PUZZLEa"></a><a href="#X_327_A_NEW_BISHOPS_PUZZLE"><b>327.&mdash;A NEW BISHOP'S PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a327.png" width="400" height="255" alt="" title="" />
+</div>
+
+<p>Play as follows, using the notation indicated by the numbered squares
+in Diagram A:&mdash;</p>
+
+<div class='center'>
+<table border="1" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>White.</td><td align='center'>Black.</td></tr>
+<tr><td align='left'>1. 18&mdash;15</td><td align='left'>1. 3&mdash;6</td></tr>
+<tr><td align='left'>2. 17&mdash;8</td><td align='left'>2. 4&mdash;13</td></tr>
+<tr><td align='left'>3. 19&mdash;14</td><td align='left'>3. 2&mdash;7</td></tr>
+<tr><td align='left'>4. 15&mdash;5</td><td align='left'>4. 6&mdash;16</td></tr>
+<tr><td align='left'>5. 8&mdash;3</td><td align='left'>5. 13-18</td></tr>
+<tr><td align='left'>6. 14&mdash;9</td><td align='left'>6. 7&mdash;12</td></tr>
+<tr><td align='left'>7. 5&mdash;10</td><td align='left'>7. 16-11</td></tr>
+<tr><td align='left'>8. 9&mdash;19</td><td align='left'>8. 12&mdash;2</td></tr>
+<tr><td align='left'>9. 10&mdash;4</td><td align='left'>9. 11-17</td></tr>
+<tr><td align='left'>10. 20&mdash;10</td><td align='left'>10. 1&mdash;11</td></tr>
+<tr><td align='left'>11. 3&mdash;9</td><td align='left'>11. 18&mdash;12</td></tr>
+<tr><td align='left'>12. 10&mdash;13</td><td align='left'>12. 11&mdash;8</td></tr>
+<tr><td align='left'>13. 19&mdash;16</td><td align='left'>13. 2&mdash;5</td></tr>
+<tr><td align='left'>14. 16&mdash;1</td><td align='left'>14. 5&mdash;20</td></tr>
+<tr><td align='left'>15. 9&mdash;6</td><td align='left'>15. 12&mdash;15</td></tr>
+<tr><td align='left'>16. 13-7</td><td align='left'>16. 8&mdash;14</td></tr>
+<tr><td align='left'>17. 6&mdash;3</td><td align='left'>17. 15-18</td></tr>
+<tr><td align='left'>18. 7&mdash;2</td><td align='left'>18. 14&mdash;19</td></tr>
+</table></div>
+
+<p>Diagram B shows the position after the ninth move. Bishops at 1 and 20
+have not yet moved, but 2 and 19 have sallied forth and returned. In
+the end, 1 and 19, 2 and 20, 3 and 17, and 4 and 18 will have
+exchanged places. Note the position after the thirteenth move.</p>
+<hr style="width: 30%;" />
+<p><a name="X_328_THE_QUEENS_TOURa" id="X_328_THE_QUEENS_TOURa"></a><a href="#X_328_THE_QUEENS_TOUR"><b>328.&mdash;THE QUEEN'S TOUR.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a328.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p><span class='pagenum'>Pg 226<a name="Page_226" id="Page_226"></a></span>The annexed diagram shows a second way of performing the Queen's
+Tour. If you break the line at the point J and erase the shorter
+portion of that line, you will have the required path solution for any
+J square. If you break the line at I, you will have a non-re-entrant
+solution starting from any I square. And if you break the line at G,
+you will have a solution for any G square. The Queen's Tour previously
+given may be similarly broken at three different places, but I seized
+the opportunity of exhibiting a second tour.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_329_THE_STAR_PUZZLEa" id="X_329_THE_STAR_PUZZLEa"></a><a href="#X_329_THE_STAR_PUZZLE"><b>329.&mdash;THE STAR PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The illustration explains itself. The stars are all struck out in
+fourteen straight strokes, starting and ending at a white star.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a329.png" width="400" height="392" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_330_THE_YACHT_RACEa" id="X_330_THE_YACHT_RACEa"></a><a href="#X_330_THE_YACHT_RACE"><b>330.&mdash;THE YACHT RACE.&mdash;<i>solution</i></b></a></p>
+
+<p>The diagram explains itself. The numbers will show the direction of
+the lines in their proper order, and it will be seen that the seventh
+course ends at the flag-buoy, as stipulated.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a330.png" width="400" height="326" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_331_THE_SCIENTIFIC_SKATERa" id="X_331_THE_SCIENTIFIC_SKATERa"></a><a href="#X_331_THE_SCIENTIFIC_SKATER"><b>331.&mdash;THE SCIENTIFIC SKATER.&mdash;<i>solution</i></b></a></p>
+
+<p>In this case we go beyond the boundary of the square. Apart from that,
+the moves are all queen moves. There are three or four ways in which
+it can be done.</p>
+
+<p>Here is one way of performing the feat:&mdash;</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a331.png" width="400" height="394" alt="" title="" />
+</div>
+
+<p>It will be seen that the skater strikes out all the stars in one
+continuous journey of fourteen straight lines, returning to the point
+from which he started. To follow the skater's course in the diagram it
+is necessary always to go as far as we can in a straight line before
+turning.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_332_THE_FORTY-NINE_STARSa" id="X_332_THE_FORTY-NINE_STARSa"></a><a href="#X_332_THE_FORTY-NINE_STARS"><b>332.&mdash;THE FORTY-NINE STARS.&mdash;<i>solution</i></b></a></p>
+
+<p>The illustration shows how all the stars may be struck out in twelve
+straight strokes, beginning and ending at a black star.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a332.png" width="400" height="396" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 227<a name="Page_227" id="Page_227"></a></span><a name="X_333_THE_QUEENS_JOURNEYa" id="X_333_THE_QUEENS_JOURNEYa"></a><a href="#X_333_THE_QUEENS_JOURNEY"><b>333.&mdash;THE QUEEN'S JOURNEY.&mdash;<i>solution</i></b></a></p>
+
+<p>The correct solution to this puzzle is shown in the diagram by the
+dark line. The five moves indicated will take the queen the greatest
+distance that it is possible for her to go in five moves, within the
+conditions. The dotted line shows the route that most people suggest,
+but it is not quite so long as the other. Let us assume that the
+distance from the centre of any square to the centre of the next in
+the same horizontal or vertical line is 2 inches, and that the queen
+travels from the centre of her original square to the centre of the
+one at which she rests. Then the first route will be found to exceed
+67.9 inches, while the dotted route is less than 67.8 inches. The
+difference is small, but it is sufficient to settle the point as to
+the longer route. All other routes are shorter still than these two.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a333.png" width="400" height="400" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_334_ST_GEORGE_AND_THE_DRAGONa" id="X_334_ST_GEORGE_AND_THE_DRAGONa"></a><a href="#X_334_ST_GEORGE_AND_THE_DRAGON"><b>334.&mdash;ST. GEORGE AND THE DRAGON.&mdash;<i>solution</i></b></a></p>
+
+<p>We select for the solution of this puzzle one of the prettiest designs
+that can be formed by representing the moves of the knight by lines
+from square to square. The chequering of the squares is omitted to
+give greater clearness. St. George thus slays the Dragon in strict
+accordance with the conditions and in the elegant manner we should
+expect of him.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a334.png" width="400" height="403" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_335_FARMER_LAWRENCES_CORNFIELDSa" id="X_335_FARMER_LAWRENCES_CORNFIELDSa"></a><a href="#X_335_FARMER_LAWRENCES_CORNFIELDS"><b>335.&mdash;FARMER LAWRENCE'S CORNFIELDS.&mdash;<i>solution</i></b></a></p>
+
+<p>There are numerous solutions to this little agricultural problem. The
+version I give in the next column is rather curious on account of the
+long parallel straight lines formed by some of the moves.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a335.png" width="400" height="406" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_336_THE_GREYHOUND_PUZZLEa" id="X_336_THE_GREYHOUND_PUZZLEa"></a><a href="#X_336_THE_GREYHOUND_PUZZLE"><b>336.&mdash;THE GREYHOUND PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>There are several interesting points involved in this question. In the
+first place, if we had made no stipulation as to the positions of the
+two ends of the string, it is quite impossible to form any such string
+unless we begin and end in the top and bottom row of kennels. We may
+begin in the top row and end in the bottom (or, of course, the
+reverse), or we may begin in one of these rows and end in the same.
+But we can never begin or end in one of the two central rows. Our
+places of starting and ending, however, were fixed for us. Yet the
+first half of our route must be confined entirely to those squares
+that are distinguished in the following diagram by circles, and the
+second half will therefore be confined to the squares that are not
+circled. The squares reserved for the two half-strings will be seen to
+be symmetrical and similar.</p>
+
+<p>The next point is that the first half-string must end in one of the
+central rows, and the <span class='pagenum'>Pg 228<a name="Page_228" id="Page_228"></a></span>second half-string must begin in one of these
+rows. This is now obvious, because they have to link together to form
+the complete string, and every square on an outside row is connected
+by a knight's move with similar squares only&mdash;that is, circled or
+non-circled as the case may be. The half-strings can, therefore, only
+be linked in the two central rows.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a336.png" width="400" height="337" alt="" title="" />
+</div>
+
+<p>Now, there are just eight different first half-strings, and
+consequently also eight second half-strings. We shall see that these
+combine to form twelve complete strings, which is the total number
+that exist and the correct solution of our puzzle. I do not propose to
+give all the routes at length, but I will so far indicate them that if
+the reader has dropped any he will be able to discover which they are
+and work them out for himself without any difficulty. The following
+numbers apply to those in the above diagram.</p>
+
+<p>The eight first half-strings are: 1 to 6 (2 routes); 1 to 8 (1 route);
+1 to 10 (3 routes); 1 to 12 (1 route); and 1 to 14 (1 route). The
+eight second half-strings are: 7 to 20 (1 route); 9 to 20 (1 route);
+11 to 20 (3 routes); 13 to 20 (1 route); and 15 to 20 (2 routes).
+Every different way in which you can link one half-string to another
+gives a different solution. These linkings will be found to be as
+follows: 6 to 13 (2 cases); 10 to 13 (3 cases); 8 to 11 (3 cases); 8
+to 15 (2 cases); 12 to 9 (1 case); and 14 to 7 (1 case). There are,
+therefore, twelve different linkings and twelve different answers to
+the puzzle. The route given in the illustration with the greyhound
+will be found to consist of one of the three half-strings 1 to 10,
+linked to the half-string 13 to 20. It should be noted that ten of the
+solutions are produced by five distinctive routes and their
+reversals&mdash;that is, if you indicate these five routes by lines and
+then turn the diagrams upside down you will get the five other routes.
+The remaining two solutions are symmetrical (these are the cases where
+12 to 9 and 14 to 7 are the links), and consequently they do not
+produce new solutions by reversal.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_337_THE_FOUR_KANGAROOSa" id="X_337_THE_FOUR_KANGAROOSa"></a><a href="#X_337_THE_FOUR_KANGAROOS"><b>337.&mdash;THE FOUR KANGAROOS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a337.png" width="400" height="394" alt="" title="" />
+</div>
+
+<p>A pretty symmetrical solution to this puzzle is shown in the diagram.
+Each of the four kangaroos makes his little excursion and returns to
+his corner, without ever entering a square that has been visited by
+another kangaroo and without crossing the central line. It will at
+once occur to the reader, as a possible improvement of the puzzle, to
+divide the board by a central vertical line and make the condition
+that this also shall not be crossed. This would mean that each
+kangaroo had to confine himself to a square 4 by 4, but it would be
+quite impossible, as I shall explain in the next two puzzles.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_338_THE_BOARD_IN_COMPARTMENTSa" id="X_338_THE_BOARD_IN_COMPARTMENTSa"></a><a href="#X_338_THE_BOARD_IN_COMPARTMENTS"><b>338.&mdash;THE BOARD IN COMPARTMENTS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a338.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p>In attempting to solve this problem it is first necessary to take the
+two distinctive compartments of twenty and twelve squares respectively
+and analyse them with a view to deter<span class='pagenum'>Pg 229<a name="Page_229" id="Page_229"></a></span>mining where the necessary
+points of entry and exit lie. In the case of the larger compartment it
+will be found that to complete a tour of it we must begin and end on
+two of the outside squares on the long sides. But though you may start
+at any one of these ten squares, you are restricted as to those at
+which you can end, or (which is the same thing) you may end at
+whichever of these you like, provided you begin your tour at certain
+particular squares. In the case of the smaller compartment you are
+compelled to begin and end at one of the six squares lying at the two
+narrow ends of the compartments, but similar restrictions apply as in
+the other instance. A very little thought will show that in the case
+of the two small compartments you must begin and finish at the ends
+that lie together, and it then follows that the tours in the larger
+compartments must also start and end on the contiguous sides.</p>
+
+<p>In the diagram given of one of the possible solutions it will be seen
+that there are eight places at which we may start this particular
+tour; but there is only one route in each case, because we must
+complete the compartment in which we find ourself before passing into
+another. In any solution we shall find that the squares distinguished
+by stars must be entering or exit points, but the law of reversals
+leaves us the option of making the other connections either at the
+diamonds or at the circles. In the solution worked out the diamonds
+are used, but other variations occur in which the circle squares are
+employed instead. I think these remarks explain all the essential
+points in the puzzle, which is distinctly instructive and interesting.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_339_THE_FOUR_KNIGHTS_TOURSa" id="X_339_THE_FOUR_KNIGHTS_TOURSa"></a><a href="#X_339_THE_FOUR_KNIGHTS_TOURS"><b>339.&mdash;THE FOUR KNIGHTS' TOURS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a339.png" width="400" height="396" alt="" title="" />
+</div>
+
+<p>It will be seen in the illustration how a chessboard may be divided
+into four parts, each of the same size and shape, so that a complete
+re-entrant knight's tour may be made on each portion. There is only
+one possible route for each knight and its reversal.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_340_THE_CUBIC_KNIGHTS_TOURa" id="X_340_THE_CUBIC_KNIGHTS_TOURa"></a><a href="#X_340_THE_CUBIC_KNIGHTS_TOUR"><b>340.&mdash;THE CUBIC KNIGHT'S TOUR.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a340.png" width="400" height="515" alt="" title="" />
+</div>
+
+<p>If the reader should cut out the above diagram, fold it in the form of
+a cube, and stick it together by the strips left for that purpose at
+the edges, he would have an interesting little curiosity. Or he can
+make one on a larger scale for himself. It will be found that if we
+imagine the cube to have a complete chessboard on each of its sides,
+we may start with the knight on any one of the 384 squares, and make a
+complete tour of the cube, always returning to the starting-point. The
+method of passing from one side of the cube to another is easily
+understood, but, of course, the difficulty consisted in finding the
+proper points of entry and exit on each board, the order in which the
+different boards should be taken, and in getting arrangements that
+would comply with the required conditions.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_341_THE_FOUR_FROGSa" id="X_341_THE_FOUR_FROGSa"></a><a href="#X_341_THE_FOUR_FROGS"><b>341.&mdash;THE FOUR FROGS.&mdash;<i>solution</i></b></a></p>
+
+<p>The fewest possible moves, counting every move separately, are
+sixteen. But the puzzle may be solved in seven plays, as follows, if
+any number of successive moves by one frog count as a single play. All
+the moves contained within a bracket are a single play; the numbers
+refer to the toadstools: (1&mdash;5), (3&mdash;7, 7&mdash;1), (8&mdash;4, 4&mdash;3, 3&mdash;7),
+(6&mdash;2, 2&mdash;8, 8&mdash;4, 4&mdash;3), (5&mdash;6, 6&mdash;2, 2&mdash;8), (1&mdash;5, 5&mdash;6), (7&mdash;1).</p>
+
+<p>This is the familiar old puzzle by Guarini, propounded in 1512, and I
+give it here in order to explain my "buttons and string" method of
+solving this class of moving-counter problem.</p>
+
+<p><span class='pagenum'>Pg 230<a name="Page_230" id="Page_230"></a></span>Diagram A shows the old way of presenting Guarini's puzzle, the point
+being to make the white knights change places with the black ones. In
+"The Four Frogs" presentation of the idea the possible directions of
+the moves are indicated by lines, to obviate the necessity of the
+reader's understanding the nature of the knight's move in chess. But
+it will at once be seen that the two problems are identical. The
+central square can, of course, be ignored, since no knight can ever
+enter it. Now, regard the toadstools as buttons and the connecting
+lines as strings, as in Diagram B. Then by disentangling these strings
+we can clearly present the diagram in the form shown in Diagram C,
+where the relationship between the buttons is precisely the same as in
+B. Any solution on C will be applicable to B, and to A. Place your
+white knights on 1 and 3 and your black knights on 6 and 8 in the C
+diagram, and the simplicity of the solution will be very evident. You
+have simply to move the knights round the circle in one direction or
+the other. Play over the moves given above, and you will find that
+every little difficulty has disappeared.</p>
+
+<div class="figcenter" style="width: 600px;">
+<img src="images/a341.png" width="600" height="357" alt="" title="" />
+</div>
+
+<p>In Diagram D I give another familiar puzzle that first appeared in a
+book published in Brussels in 1789, <i>Les Petites Aventures de Jerome
+Sharp</i>. Place seven counters on seven of the eight points in the
+following manner. You must always touch a point that is vacant with a
+counter, and then move it along a straight line leading from that
+point to the next vacant point (in either direction), where you
+deposit the counter. You proceed in the same way until all the
+counters are placed. Remember you always touch a vacant place and
+slide the counter from it to the next place, which must be also
+vacant. Now, by the "buttons and string" method of simplification we
+can transform the diagram into E. Then the solution becomes obvious.
+"Always move <i>to</i> the point that you last moved <i>from</i>." This is not,
+of course, the only way of placing the counters, but it is the
+simplest solution to carry in the mind.</p>
+
+<p>There are several puzzles in this book that the reader will find lend
+themselves readily to this method.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_342_THE_MANDARINS_PUZZLEa" id="X_342_THE_MANDARINS_PUZZLEa"></a><a href="#X_342_THE_MANDARINS_PUZZLE"><b>342.&mdash;THE MANDARIN'S PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The rather perplexing point that the solver has to decide for himself
+in attacking this puzzle is whether the shaded numbers (those that are
+shown in their right places) are mere dummies or not. Ninety-nine
+persons out of a hundred might form the opinion that there can be no
+advantage in moving any of them, but if so they would be wrong.</p>
+
+<p>The shortest solution without moving any shaded number is in
+thirty-two moves. But the puzzle can be solved in thirty moves. The
+trick lies in moving the 6, or the 15, on the second move and
+replacing it on the nineteenth move. Here is the solution:
+2, 6, 13, 4, 1, 21, 4, 1, 10, 2, 21, 10, 2, 5, 22, 16, 1, 13, 6, 19, 11, 2, 5, 22, 16, 5, 13, 4, 10, 21.
+Thirty moves.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_343_EXERCISE_FOR_PRISONERSa" id="X_343_EXERCISE_FOR_PRISONERSa"></a><a href="#X_343_EXERCISE_FOR_PRISONERS"><b>343.&mdash;EXERCISE FOR PRISONERS.&mdash;<i>solution</i></b></a></p>
+
+<p>There are eighty different arrangements of the numbers in the form of
+a perfect knight's path, but only forty of these can be reached
+without two men ever being in a cell at the same time. Two is the
+greatest number of men that can be given a complete rest, and though
+the <span class='pagenum'>Pg 231<a name="Page_231" id="Page_231"></a></span>knight's path can be arranged so as to leave either 7 and 13, 8
+and 13, 5 and 7, or 5 and 13 in their original positions, the
+following four arrangements, in which 7 and 13 are unmoved, are the
+only ones that can be reached under the moving conditions. It
+therefore resolves itself into finding the fewest possible moves that
+will lead up to one of these positions. This is certainly no easy
+matter, and no rigid rules can be laid down for arriving at the
+correct answer. It is largely a matter for individual judgment,
+patient experiment, and a sharp eye for revolutions and position.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a343.png" width="400" height="445" alt="" title="" />
+</div>
+
+<p>As a matter of fact, the position C can be reached in as few as
+sixty-six moves in the following manner: 12, 11, 15, 12, 11, 8, 4, 3,
+2, 6, 5, 1, 6, 5, 10, 15, 8, 4, 3, 2, 5, 10, 15, 8, 4, 3, 2, 5, 10,
+15, 8, 4, 12, 11, 3, 2, 5, 10, 15, 6, 1, 8, 4, 9, 8, 1, 6, 4, 9, 12,
+2, 5, 10, 15, 4, 9, 12, 2, 5, 3, 11, 14, 2, 5, 14, 11&nbsp;=&nbsp;66 moves.
+Though this is the shortest that I know of, and I do not think it can
+be beaten, I cannot state positively that there is not a shorter way
+yet to be discovered. The most tempting arrangement is certainly A;
+but things are not what they seem, and C is really the easiest to
+reach.</p>
+
+<p>If the bottom left-hand corner cell might be left vacant, the
+following is a solution in forty-five moves by Mr. R. Elrick: 15, 11,
+10, 9, 13, 14, 11, 10, 7, 8, 4, 3, 8, 6, 9, 7, 12, 4, 6, 9, 5, 13, 7,
+5, 13, 1, 2, 13, 5, 7, 1, 2, 13, 8, 3, 6, 9, 12, 7, 11, 14, 1, 11, 14,
+1. But every man has moved.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_344_THE_KENNEL_PUZZLEa" id="X_344_THE_KENNEL_PUZZLEa"></a><a href="#X_344_THE_KENNEL_PUZZLE"><b>344.&mdash;THE KENNEL PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The first point is to make a choice of the most promising knight's
+string and then consider the question of reaching the arrangement in
+the fewest moves. I am strongly of opinion that the best string is the
+one represented in the following diagram, in which it will be seen
+that each successive number is a knight's move from the preceding one,
+and that five of the dogs (1, 5, 10, 15, and 20) never leave their
+original kennels.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a344.png" width="400" height="402" alt="" title="" />
+</div>
+
+<p>This position may be arrived at in as few as forty-six moves, as
+follows: 16&mdash;21, 16&mdash;22, 16&mdash;23, 17&mdash;16, 12&mdash;17, 12&mdash;22, 12&mdash;21,7&mdash;12,
+7&mdash;17, 7&mdash;22, 11&mdash;12, 11&mdash;17, 2&mdash;7, 2&mdash;12, 6&mdash;11, 8&mdash;7, 8&mdash;6, 13&mdash;8,
+18&mdash;13, 11&mdash;18, 2&mdash;17, 18&mdash;12, 18&mdash;7, 18&mdash;2, 13&mdash;7, 3&mdash;8, 3&mdash;13, 4&mdash;3,
+4&mdash;8, 9&mdash;4, 9&mdash;3, 14&mdash;9, 14&mdash;4, 19&mdash;14, 19&mdash;9, 3&mdash;14, 3&mdash;19, 6&mdash;12,
+6&mdash;13, 6&mdash;14, 17&mdash;11, 12&mdash;16, 2&mdash;12, 7&mdash;17, 11&mdash;13, 16&mdash;18&nbsp;=&nbsp;46 moves.
+I am, of course, not able to say positively that a solution cannot be
+discovered in fewer moves, but I believe it will be found a very hard
+task to reduce the number.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_345_THE_TWO_PAWNSa" id="X_345_THE_TWO_PAWNSa"></a><a href="#X_345_THE_TWO_PAWNS"><b>345.&mdash;THE TWO PAWNS.&mdash;<i>solution</i></b></a></p>
+
+<p>Call one pawn A and the other B. Now, owing to that optional first
+move, either pawn may make either 5 or 6 moves in reaching the eighth
+square. There are, therefore, four cases to be considered: (1) A 6
+moves and B 6 moves; (2) A 6 moves and B 5 moves; (3) A 5 moves and B
+6 moves; (4) A 5 moves and B 5 moves. In case (1) there are 12 moves,
+and we may select any 6 of these for A. Therefore 7&nbsp;&times;&nbsp;8&nbsp;&times;&nbsp;9&nbsp;&times;&nbsp;10&nbsp;&times;&nbsp;11&nbsp;&times;&nbsp;12
+divided by 1&nbsp;&times;&nbsp;2&nbsp;&times;&nbsp;3&nbsp;&times;&nbsp;4&nbsp;&times;&nbsp;5&nbsp;&times;&nbsp;6 gives us the number of variations for this
+case&mdash;that is, 924. Similarly for case (2), 6 selections out of 11
+will be 462; in case (3), 5 selections out of 11 will also be 462; and
+in case (4), 5 selections out of 10 will be 252. Add these four
+numbers together and we get 2,100, which is the correct number of
+different ways in which the pawns may advance under the conditions.
+(See No. <a href="#X_270_THE_GLASS_BALLSa">270</a>, on p. <a href="#Page_204">204</a>.)</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_346_SETTING_THE_BOARDa" id="X_346_SETTING_THE_BOARDa"></a><a href="#X_346_SETTING_THE_BOARD"><b>346.&mdash;SETTING THE BOARD.&mdash;<i>solution</i></b></a></p>
+
+<p>The White pawns may be arranged in 40,320 ways, the White rooks in 2
+ways, the bishops in 2 ways, and the knights in 2 ways. Multiply these
+numbers together, and we find that the White pieces may be placed in
+322,560 different <span class='pagenum'>Pg 232<a name="Page_232" id="Page_232"></a></span>ways. The Black pieces may, of course, be placed in
+the same number of ways. Therefore the men may be set up in 322,560 &times;
+322,560&nbsp;=&nbsp;104,044,953,600 ways. But the point that nearly everybody
+overlooks is that the board may be placed in two different ways for
+every arrangement. Therefore the answer is doubled, and is
+208,089,907,200 different ways.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_347_COUNTING_THE_RECTANGLESa" id="X_347_COUNTING_THE_RECTANGLESa"></a><a href="#X_347_COUNTING_THE_RECTANGLES"><b>347.&mdash;COUNTING THE RECTANGLES.&mdash;<i>solution</i></b></a></p>
+
+<p>There are 1,296 different rectangles in all, 204 of which are squares,
+counting the square board itself as one, and 1,092 rectangles that are
+not squares. The general formula is that a board of <i>n</i><sup>2</sup> squares
+contains <span class='su2'>((<i>n</i><sup>2</sup>&nbsp;+&nbsp;<i>n</i>)<sup>2</sup>)</span>/<sub>4</sub> rectangles, of which <span class='su2'>(2<i>n</i><sup>3</sup>&nbsp;+&nbsp;3<i>n</i><sup>2</sup>&nbsp;+&nbsp;<i>n</i>)</span>/<sub>6</sub>
+are squares and <span class='su2'>(3<i>n</i><sup>4</sup>&nbsp;+&nbsp;2<i>n</i><sup>3</sup>&nbsp;-&nbsp;3<i>n</i><sup>2</sup>&nbsp;-&nbsp;2<i>n</i>)</span>/<sub>12</sub> are rectangles that are
+not squares. It is curious and interesting that the total number of
+rectangles is always the square of the triangular number whose side is
+<i>n</i>.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_348_THE_ROOKERYa" id="X_348_THE_ROOKERYa"></a><a href="#X_348_THE_ROOKERY"><b>348.&mdash;THE ROOKERY.&mdash;<i>solution</i></b></a></p>
+
+<p>The answer involves the little point that in the final position the
+numbered rooks must be in numerical order in the direction contrary to
+that in which they appear in the original diagram, otherwise it cannot
+be solved. Play the rooks in the following order of their numbers. As
+there is never more than one square to which a rook can move (except
+on the final move), the notation is obvious&mdash;5, 6, 7, 5, 6, 4, 3, 6,
+4, 7, 5, 4, 7, 3, 6, 7, 3, 5, 4, 3, 1, 8, 3, 4, 5, 6, 7, 1, 8, 2, 1,
+and rook takes bishop, checkmate. These are the fewest possible
+moves&mdash;thirty-two. The Black king's moves are all forced, and need not
+be given.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_349_STALEMATEa" id="X_349_STALEMATEa"></a><a href="#X_349_STALEMATE"><b>349.&mdash;STALEMATE.&mdash;<i>solution</i></b></a></p>
+
+<p>Working independently, the same position was arrived at by Messrs. S.
+Loyd, E.N. Frankenstein, W.H. Thompson, and myself. So the following
+may be accepted as the best solution possible to this curious problem
+:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>White.</td><td align='center'>Black.</td></tr>
+<tr><td align='left'>1. P&mdash;Q4</td><td align='left'>1. P&mdash;K4</td></tr>
+<tr><td align='left'>2. Q&mdash;Q3</td><td align='left'>2. Q&mdash;R5</td></tr>
+<tr><td align='left'>3. Q&mdash;KKt3</td><td align='left'>3. B&mdash;Kt5 ch</td></tr>
+<tr><td align='left'>4. Kt&mdash;Q2</td><td align='left'>4. P&mdash;QR4</td></tr>
+<tr><td align='left'>5. P&mdash;R4</td><td align='left'>5. P&mdash;Q3</td></tr>
+<tr><td align='left'>6. P&mdash;R3</td><td align='left'>6. B&mdash;K3</td></tr>
+<tr><td align='left'>7. R&mdash;R3</td><td align='left'>7. P&mdash;KB4</td></tr>
+<tr><td align='left'>8. Q&mdash;R2</td><td align='left'>8. P&mdash;B4</td></tr>
+<tr><td align='left'>9. R&mdash;KKt3</td><td align='left'>9. B&mdash;Kt6</td></tr>
+<tr><td align='left'>10. P&mdash;QB4</td><td align='left'>10. P&mdash;B5</td></tr>
+<tr><td align='left'>11. P&mdash;B3</td><td align='left'>11. P&mdash;K5</td></tr>
+<tr><td align='left'>12. P&mdash;Q5</td><td align='left'>12. P&mdash;K6</td></tr>
+</table></div>
+
+<p>And White is stalemated.</p>
+
+<p>We give a diagram of the curious position arrived at. It will be seen
+that not one of White's pieces may be moved.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a349.png" width="400" height="405" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_350_THE_FORSAKEN_KINGa" id="X_350_THE_FORSAKEN_KINGa"></a><a href="#X_350_THE_FORSAKEN_KING"><b>350.&mdash;THE FORSAKEN KING.&mdash;<i>solution</i></b></a></p>
+
+<p>Play as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>White.</td><td align='center'>Black.</td></tr>
+<tr><td align='left'>1. P to K 4th</td><td align='left'>1. Any move</td></tr>
+<tr><td align='left'>2. Q to Kt 4th</td><td align='left'>2. Any move except on KB file (a)</td></tr>
+<tr><td align='left'>3. Q to Kt 7th</td><td align='left'>3. K moves to royal row</td></tr>
+<tr><td align='left'>4. B to Kt 5th</td><td align='left'>4. Any move</td></tr>
+<tr><td align='left'>5. Mate in two moves</td></tr>
+<tr><td align='left'></td><td align='left'>If 3. K other than to royal row</td></tr>
+<tr><td align='left'>4. P to Q 4th</td><td align='left'>4. Any move</td></tr>
+<tr><td align='left'>5. Mate in two moves</td></tr>
+<tr><td align='left'></td><td align='left'>(a) If 2. Any move on KB file</td></tr>
+<tr><td align='left'>3. Q to Q 7th</td><td align='left'>3. K moves to royal row</td></tr>
+<tr><td align='left'>4. P to Q Kt 3rd</td><td align='left'>4. Any move</td></tr>
+<tr><td align='left'>5. Mate in two moves</td></tr>
+<tr><td align='left'></td><td align='left'>If 3. K other than to royal row</td></tr>
+<tr><td align='left'>4. P to Q 4th</td><td align='left'>4. Any move</td></tr>
+<tr><td align='left'>5. Mate in two moves</td></tr>
+</table></div>
+
+<p>Of course, by "royal row" is meant the row on which the king
+originally stands at the beginning of a game. Though, if Black plays
+badly, he may, in certain positions, be mated in fewer moves, the
+above provides for every variation he can possibly bring about.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_351_THE_CRUSADERa" id="X_351_THE_CRUSADERa"></a><a href="#X_351_THE_CRUSADER"><b>351.&mdash;THE CRUSADER.&mdash;<i>solution</i></b></a></p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>White.</td><td align='center'>Black.</td></tr>
+<tr><td align='left'>1. Kt to QB 3rd</td><td align='left'>1. P to Q 4th</td></tr>
+<tr><td align='left'>2. Kt takes QP</td><td align='left'>2. Kt to QB 3rd</td></tr>
+<tr><td align='left'>3. Kt takes KP</td><td align='left'>3. P to KKt 4th</td></tr>
+<tr><td align='left'>4. Kt takes B</td><td align='left'>4. Kt to KB 3rd</td></tr>
+<tr><td align='left'>5. Kt takes P</td><td align='left'>5. Kt to K 5th</td></tr>
+<tr><td align='left'>6. Kt takes Kt</td><td align='left'>6. Kt to B 6th</td></tr>
+<tr><td align='left'>7. Kt takes Q</td><td align='left'>7. R to KKt sq</td></tr>
+<tr><td align='left'>8. Kt takes BP</td><td align='left'>8. R to KKt 3rd</td></tr>
+<tr><td align='left'>9. Kt takes P</td><td align='left'>9. R to K 3rd</td></tr>
+<tr><td align='left'>10. Kt takes P</td><td align='left'>10. Kt to Kt 8th</td></tr>
+<tr><td align='left'>11. Kt takes B</td><td align='left'>11. R to R 6th</td></tr>
+<tr><td align='left'>12. Kt takes R</td><td align='left'>12. P to Kt 4th</td></tr>
+<tr><td align='left'>13. Kt takes P (ch)</td><td align='left'>13. K to B 2nd</td></tr>
+<tr><td align='left'>14. Kt takes P</td><td align='left'>14. K to Kt 3rd</td></tr>
+<tr><td align='left'>15. Kt takes R</td><td align='left'>15. K to R 4th</td></tr>
+<tr><td align='left'>16. Kt takes Kt</td><td align='left'>16. K to R 5th</td></tr>
+<tr><td align='center' colspan='2'>White now mates in three moves.</td></tr>
+<tr><td align='left'>17. P to Q 4th</td><td align='left'>17. K to R 4th</td></tr>
+<tr><td align='left'>18. Q to Q 3rd</td><td align='left'>18. K moves</td></tr>
+<tr><td align='left'>19. Q to KR 3rd (mate)</td></tr>
+<tr><td align='left'></td><td align='left'>If 17. K to Kt 5th</td></tr>
+<tr><td align='left'>18. P to K 4th (dis. ch)</td><td align='left'>18. K moves</td></tr>
+<tr><td align='left'>19. P to KKt 3rd (mate)</td></tr>
+</table></div>
+
+<p><span class='pagenum'>Pg 233<a name="Page_233" id="Page_233"></a></span></p>
+<p>The position after the sixteenth move, with the mate in three moves,
+was first given by S. Loyd in <i>Chess Nuts</i>.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_352_IMMOVABLE_PAWNSa" id="X_352_IMMOVABLE_PAWNSa"></a><a href="#X_352_IMMOVABLE_PAWNS"><b>352.&mdash;IMMOVABLE PAWNS.&mdash;<i>solution</i></b></a></p>
+
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='left'>1. Kt to KB 3</td></tr>
+<tr><td align='left'>2. Kt to KR 4</td></tr>
+<tr><td align='left'>3. Kt to Kt 6</td></tr>
+<tr><td align='left'>4. Kt takes R</td></tr>
+<tr><td align='left'>5. Kt to Kt 6</td></tr>
+<tr><td align='left'>6. Kt takes B</td></tr>
+<tr><td align='left'>7. K takes Kt</td></tr>
+<tr><td align='left'>8. Kt to QB 3</td></tr>
+<tr><td align='left'>9. Kt to R 4</td></tr>
+<tr><td align='left'>10. Kt to Kt 6</td></tr>
+<tr><td align='left'>11. Kt takes R</td></tr>
+<tr><td align='left'>12. Kt to Kt 6</td></tr>
+<tr><td align='left'>13. Kt takes B</td></tr>
+<tr><td align='left'>14. Kt to Q 6</td></tr>
+<tr><td align='left'>15. Q to K sq</td></tr>
+<tr><td align='left'>16. Kt takes Q</td></tr>
+<tr><td align='left'>17. K takes Kt, and the position is reached.</td></tr>
+</table></div>
+
+<p>Black plays precisely the same moves as White, and therefore we give
+one set of moves only. The above seventeen moves are the fewest
+possible.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_353_THIRTY-SIX_MATESa" id="X_353_THIRTY-SIX_MATESa"></a><a href="#X_353_THIRTY-SIX_MATES"><b>353.&mdash;THIRTY-SIX MATES.&mdash;<i>solution</i></b></a></p>
+
+<p>Place the remaining eight White pieces thus: K at KB 4th, Q at QKt
+6th, R at Q 6th, R at KKt 7th, B at Q 5th, B at KR 8th, Kt at QR 5th,
+and Kt at QB 5th. The following mates can then be given:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='left'>By discovery from Q</td><td align='right'>8</td></tr>
+<tr><td align='left'>By discovery from R at Q 6th</td><td align='right'>13</td></tr>
+<tr><td align='left'>By discovery from B at R 8th</td><td align='right'>11</td></tr>
+<tr><td align='left'>Given by Kt at R 5th</td><td align='right'>2</td></tr>
+<tr><td align='left'>Given by pawns</td><td align='right'>2</td></tr>
+<tr><td align='left'>Total</td><td align='right' class='bt'>36</td></tr>
+</table></div>
+
+<p>Is it possible to construct a position in which more than thirty-six
+different mates on the move can be given? So far as I know, nobody has
+yet beaten my arrangement.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_354_AN_AMAZING_DILEMMAa" id="X_354_AN_AMAZING_DILEMMAa"></a><a href="#X_354_AN_AMAZING_DILEMMA"><b>354.&mdash;AN AMAZING DILEMMA.&mdash;<i>solution</i></b></a></p>
+
+<p>Mr Black left his king on his queen's knight's 7th, and no matter what
+piece White chooses for his pawn, Black cannot be checkmated. As we
+said, the Black king takes no notice of checks and never moves. White
+may queen his pawn, capture the Black rook, and bring his three pieces
+up to the attack, but mate is quite impossible. The Black king cannot
+be left on any other square without a checkmate being possible.</p>
+
+<p>The late Sam Loyd first pointed out the peculiarity on which this
+puzzle is based.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_355_CHECKMATEa" id="X_355_CHECKMATEa"></a><a href="#X_355_CHECKMATE"><b>355.&mdash;CHECKMATE!&mdash;<i>solution</i></b></a></p>
+
+<p>Remove the White pawn from B 6th to K 4th and place a Black pawn on
+Black's KB 2nd. Now, White plays P to K 5th, check, and Black must
+play P to B 4th. Then White plays P takes P <i>en passant</i>, checkmate.
+This was therefore White's last move, and leaves the position given.
+It is the only possible solution.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_356_QUEER_CHESSa" id="X_356_QUEER_CHESSa"></a><a href="#X_356_QUEER_CHESS"><b>356.&mdash;QUEER CHESS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a356.png" width="400" height="163" alt="" title="" />
+</div>
+
+<p>If you place the pieces as follows (where only a portion of the board
+is given, to save space), the Black king is in check, with no possible
+move open to him. The reader will now see why I avoided the term
+"checkmate," apart from the fact that there is no White king. The
+position is impossible in the game of chess, because Black could not be
+given check by both rooks at the same time, nor could he have moved into
+check on his last move.</p>
+
+<p>I believe the position was first published by the late S. Loyd.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_357_ANCIENT_CHINESE_PUZZLEa" id="X_357_ANCIENT_CHINESE_PUZZLEa"></a><a href="#X_357_ANCIENT_CHINESE_PUZZLE"><b>357.&mdash;ANCIENT CHINESE PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Play as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='left'>1. R&mdash;Q 6</td></tr>
+<tr><td align='left'>2. K&mdash;R 7</td></tr>
+<tr><td align='left'>3. R (R 6)&mdash;B 6 (mate).</td></tr>
+</table></div>
+
+<p>Black's moves are forced, so need not be given.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_358_THE_SIX_PAWNSa" id="X_358_THE_SIX_PAWNSa"></a><a href="#X_358_THE_SIX_PAWNS"><b>358.&mdash;THE SIX PAWNS.&mdash;<i>solution</i></b></a></p>
+
+<p>The general formula for six pawns on all squares greater than 2<sup>2</sup> is
+this: Six times the square of the number of combinations of <i>n</i> things
+taken three at a time, where <i>n</i> represents the number of squares on
+the side of the board. Of course, where <i>n</i> is even the unoccupied
+squares in the rows and columns will be even, and where <i>n</i> is odd the
+number of squares will be odd. Here <i>n</i> is 8, so the answer is 18,816
+different ways. This is "The Dyer's Puzzle" (<i>Canterbury Puzzles</i>, No.
+27) in another form. I repeat it here in order to explain a method of
+solving that will be readily grasped by the novice. First of all, it
+is evident that if we put a pawn on any line, we must put a second one
+in that line in order that the remainder may be even in number. We
+cannot put four or six in any row without making it impossible to get
+an even number in all the columns interfered with. We have, therefore,
+to put two pawns in each of three rows and in each of three columns.
+Now, there are just six schemes or arrangements that fulfil these
+conditions, and these are shown in Diagrams A to F, inclusive, on next
+page.</p>
+
+<p><span class='pagenum'>Pg 234<a name="Page_234" id="Page_234"></a></span></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a358.png" width="400" height="218" alt="" title="" />
+</div>
+
+<p>I will just remark in passing that A and B are the only distinctive
+arrangements, because, if you give A a quarter-turn, you get F; and if
+you give B three quarter-turns in the direction that a clock hand
+moves, you will get successively C, D, and E. No matter how you may
+place your six pawns, if you have complied with the conditions of the
+puzzle they will fall under one of these arrangements. Of course it
+will be understood that mere expansions do not destroy the essential
+character of the arrangements. Thus G is only an expansion of form A.
+The solution therefore consists in finding the number of these
+expansions. Supposing we confine our operations to the first three
+rows, as in G, then with the pairs <i>a</i> and <i>b</i> placed in the first and
+second columns the pair <i>c</i> may be disposed in any one of the
+remaining six columns, and so give six solutions. Now slide pair <i>b</i>
+into the third column, and there are five possible positions for <i>c</i>.
+Slide <i>b</i> into the fourth column, and <i>c</i> may produce four new
+solutions. And so on, until (still leaving <i>a</i> in the first column)
+you have <i>b</i> in the seventh column, and there is only one place for
+<i>c</i>&mdash;in the eighth column. Then you may put <i>a</i> in the second column,
+<i>b</i> in the third, and <i>c</i> in the fourth, and start sliding <i>c</i> and <i>b</i>
+as before for another series of solutions.</p>
+
+<p>We find thus that, by using form A alone and confining our operations
+to the three top rows, we get as many answers as there are
+combinations of 8 things taken 3 at a time. This is <sup>(8&nbsp;&times;&nbsp;7&nbsp;&times;&nbsp;6)</sup>/<sub>(1 &times;
+2&nbsp;&times;&nbsp;3)</sub>&nbsp;=&nbsp;56. And it will at once strike the reader that if there are
+56 different ways of electing the columns, there must be for each of
+these ways just 56 ways of selecting the rows, for we may
+simultaneously work that "sliding" process downwards to the very
+bottom in exactly the same way as we have worked from left to right.
+Therefore the total number of ways in which form A may be applied is
+56&nbsp;&times;&nbsp;6&nbsp;=&nbsp;3,136. But there are, as we have seen, six arrangements, and
+we have only dealt with one of these, A. We must, therefore, multiply
+this result by 6, which gives us 3,136&nbsp;&times;&nbsp;6&nbsp;=&nbsp;18,816, which is the
+total number of ways, as we have already stated.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_359_COUNTER_SOLITAIREa" id="X_359_COUNTER_SOLITAIREa"></a><a href="#X_359_COUNTER_SOLITAIRE"><b>359.&mdash;COUNTER SOLITAIRE.&mdash;<i>solution</i></b></a></p>
+
+<p>Play as follows: 3&mdash;11, 9&mdash;10, 1&mdash;2, 7&mdash;15, 8&mdash;16, 8&mdash;7, 5&mdash;13, 1&mdash;4,
+8&mdash;5, 6&mdash;14, 3&mdash;8, 6&mdash;3, 6&mdash;12, 1&mdash;6, 1&mdash;9, and all the counters will
+have been removed, with the exception of No. 1, as required by the
+conditions.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_360_CHESSBOARD_SOLITAIREa" id="X_360_CHESSBOARD_SOLITAIREa"></a><a href="#X_360_CHESSBOARD_SOLITAIRE"><b>360.&mdash;CHESSBOARD SOLITAIRE.&mdash;<i>solution</i></b></a></p>
+
+<p>Play as follows: 7&mdash;15, 8&mdash;16, 8&mdash;7, 2&mdash;10, 1&mdash;9, 1&mdash;2, 5&mdash;13, 3&mdash;4,
+6&mdash;3, 11&mdash;1, 14&mdash;8, 6&mdash;12, 5&mdash;6, 5&mdash;11, 31&mdash;23, 32&mdash;24, 32&mdash;31,
+26&mdash;18, 25&mdash;17, 25&mdash;26, 22&mdash;32, 14&mdash;22, 29&mdash;21, 14&mdash;29, 27&mdash;28,
+30&mdash;27, 25&mdash;14, 30&mdash;20, 25&mdash;30, 25&mdash;5. The two counters left on the
+board are 25 and 19&mdash;both belonging to the same group, as
+stipulated&mdash;and 19 has never been moved from its original place.</p>
+
+<p>I do not think any solution is possible in which only one counter is
+left on the board.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_361_THE_MONSTROSITYa" id="X_361_THE_MONSTROSITYa"></a><a href="#X_361_THE_MONSTROSITY"><b>361.&mdash;THE MONSTROSITY.&mdash;<i>solution</i></b></a></p>
+
+
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'></td><td align='center'>White</td><td align='center'>Black,</td></tr>
+<tr><td align='right'>1.</td><td align='left'>P to KB 4</td><td align='left'>P to QB 3</td></tr>
+<tr><td align='right'>2.</td><td align='left'>K to B 2</td><td align='left'>Q to R 4</td></tr>
+<tr><td align='right'>3.</td><td align='left'>K to K 3</td><td align='left'>K to Q sq</td></tr>
+<tr><td align='right'>4.</td><td align='left'>P to B 5</td><td align='left'>K to B 2</td></tr>
+<tr><td align='right'>5.</td><td align='left'>Q to K sq</td><td align='left'>K to Kt 3</td></tr>
+<tr><td align='right'>6.</td><td align='left'>Q to Kt 3</td><td align='left'>Kt to QR 3</td></tr>
+<tr><td align='right'>7.</td><td align='left'>Q to Kt 8</td><td align='left'>P to KR 4</td></tr>
+<tr><td align='right'>8.</td><td align='left'>Kt to KB 3</td><td align='left'>R to R 3</td></tr>
+<tr><td align='right'>9.</td><td align='left'>Kt to K 5</td><td align='left'>R to Kt 3</td></tr>
+<tr><td align='right'>10.</td><td align='left'>Q takes B</td><td align='left'>R to Kt 6, ch</td></tr>
+<tr><td align='right'>11.</td><td align='left'>P takes R</td><td align='left'>K to Kt 4</td></tr>
+<tr><td align='right'>12.</td><td align='left'>R to R 4</td><td align='left'>P to B 3</td></tr>
+<tr><td align='right'>13.</td><td align='left'>R to Q 4</td><td align='left'>P takes Kt</td></tr>
+<tr><td align='right'>14.</td><td align='left'>P to QKt 4</td><td align='left'>P takes R, ch</td></tr>
+<tr><td align='right'>15.</td><td align='left'>K to B 4</td><td align='left'>P to R 5</td></tr>
+<tr><td align='right'>16.</td><td align='left'>Q to K 8</td><td align='left'>P to R 6</td></tr>
+<tr><td align='right'>17.</td><td align='left'>Kt to B 3, ch</td><td align='left'>P takes Kt</td></tr>
+<tr><td align='right'>18.</td><td align='left'>B to R 3</td><td align='left'>P to R 7</td></tr>
+<tr><td align='right'>19.</td><td align='left'>R to Kt sq</td><td align='left'>P to R 8 (Q)</td></tr>
+<tr><td align='right'>20.</td><td align='left'>R to Kt 2</td><td align='left'>P takes R</td></tr>
+<tr><td align='right'>21.</td><td align='left'>K to Kt 5</td><td align='left'>Q to KKt 8</td></tr>
+<tr><td align='right'>22.</td><td align='left'>Q to R 5</td><td align='left'>K to R 5</td></tr>
+<tr><td align='right'>23.</td><td align='left'>P to Kt 5</td><td align='left'>R to B sq</td></tr>
+<tr><td align='right'>24.</td><td align='left'>P to Kt 6</td><td align='left'>R to B 2</td></tr>
+<tr><td align='right'>25.</td><td align='left'>P takes R</td><td align='left'>P to Kt 8 (B)</td></tr>
+<tr><td align='right'>26.</td><td align='left'>P to B 8 (R)</td><td align='left'>Q to B 2</td></tr>
+<tr><td align='right'>27.</td><td align='left'>B to Q 6</td><td align='left'>Kt to Kt 5</td></tr>
+<tr><td align='right'>28.</td><td align='left'>K to Kt 6</td><td align='left'>K to R 6</td></tr>
+<tr><td align='right'>29.</td><td align='left'>R to R 8</td><td align='left'>K to Kt 7</td></tr>
+<tr><td align='right'>30.</td><td align='left'>P to R 4</td><td align='left'>Q (Kt 8) to Kt 3</td></tr>
+<tr><td align='right'>31.</td><td align='left'>P to R 5</td><td align='left'>K to B 8</td></tr>
+<tr><td align='right'>32.</td><td align='left'>P takes Q</td><td align='left'>K to Q 8</td></tr>
+<tr><td align='right'>33.</td><td align='left'>P takes Q</td><td align='left'>K to K 8</td></tr>
+<tr><td align='right'>34.</td><td align='left'>K to B 7</td><td align='left'>Kt to KR 3, ch</td></tr>
+<tr><td align='right'>35.</td><td align='left'>K to K 8</td><td align='left'>B to R 7</td></tr>
+<tr><td align='right'>36.</td><td align='left'>P to B 6</td><td align='left'>B to Kt sq</td></tr>
+<tr><td align='right'>37.</td><td align='left'>P to B 7</td><td align='left'>K takes B</td></tr>
+<tr><td align='right'>38.</td><td align='left'>P to B 8 (B)</td><td align='left'>Kt to Q 4</td></tr>
+<tr><td align='right'>39.</td><td align='left'>B to Kt 8</td><td align='left'>Kt to B 3, ch</td></tr>
+<tr><td align='right'>40.</td><td align='left'>K to Q 8</td><td align='left'>Kt to K sq</td></tr>
+<tr><td align='right'>41.</td><td align='left'>P takes Kt (R)</td><td align='left'>Kt to B 2, ch</td></tr>
+<tr><td align='right'>42.</td><td align='left'>K to B 7</td><td align='left'>Kt to Q sq</td></tr>
+<tr><td align='right'>43.</td><td align='left'>Q to B 7, ch</td><td align='left'>K to Kt 8</td></tr>
+</table></div>
+
+<p>And the position is reached.</p>
+
+<p>The order of the moves is immaterial, and this order may be greatly
+varied. But, al<span class='pagenum'>Pg 235<a name="Page_235" id="Page_235"></a></span>though many attempts have been made, nobody has
+succeeded in reducing the number of my moves.</p>
+
+
+<hr style="width: 30%;" />
+<p><a name="X_362_THE_WASSAIL_BOWLa" id="X_362_THE_WASSAIL_BOWLa"></a><a href="#X_362_THE_WASSAIL_BOWL"><b>362.&mdash;THE WASSAIL BOWL.&mdash;<i>solution</i></b></a></p>
+
+
+<p>The division of the twelve pints of ale can be made in eleven
+manipulations, as below. The six columns show at a glance the quantity
+of ale in the barrel, the five-pint jug, the three-pint jug, and the
+tramps X, Y, and Z respectively after each manipulation.</p>
+
+
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>Barrel.</td><td></td><td align='center'>5-pint.</td><td></td><td align='center'>3-pint.</td><td></td><td align='center'>X.</td><td></td><td align='center'>Y.</td><td></td><td align='center'>Z.</td></tr>
+<tr><td align='center'>7</td><td align='center'>..</td><td align='center'>5</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>0</td></tr>
+<tr><td align='center'>7</td><td align='center'>..</td><td align='center'>2</td><td align='center'>..</td><td align='center'>3</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>0</td></tr>
+<tr><td align='center'>7</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>3</td><td align='center'>..</td><td align='center'>2</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>0</td></tr>
+<tr><td align='center'>7</td><td align='center'>..</td><td align='center'>3</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>2</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>0</td></tr>
+<tr><td align='center'>4</td><td align='center'>..</td><td align='center'>3</td><td align='center'>..</td><td align='center'>3</td><td align='center'>..</td><td align='center'>2</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>0</td></tr>
+<tr><td align='center'>0</td><td align='center'>..</td><td align='center'>3</td><td align='center'>..</td><td align='center'>3</td><td align='center'>..</td><td align='center'>2</td><td align='center'>..</td><td align='center'>4</td><td align='center'>..</td><td align='center'>0</td></tr>
+<tr><td align='center'>0</td><td align='center'>..</td><td align='center'>5</td><td align='center'>..</td><td align='center'>1</td><td align='center'>..</td><td align='center'>2</td><td align='center'>..</td><td align='center'>4</td><td align='center'>..</td><td align='center'>0</td></tr>
+<tr><td align='center'>0</td><td align='center'>..</td><td align='center'>5</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>2</td><td align='center'>..</td><td align='center'>4</td><td align='center'>..</td><td align='center'>1</td></tr>
+<tr><td align='center'>0</td><td align='center'>..</td><td align='center'>2</td><td align='center'>..</td><td align='center'>3</td><td align='center'>..</td><td align='center'>2</td><td align='center'>..</td><td align='center'>4</td><td align='center'>..</td><td align='center'>1</td></tr>
+<tr><td align='center'>0</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>3</td><td align='center'>..</td><td align='center'>4</td><td align='center'>..</td><td align='center'>4</td><td align='center'>..</td><td align='center'>1</td></tr>
+<tr><td align='center'>0</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>4</td><td align='center'>..</td><td align='center'>4</td><td align='center'>..</td><td align='center'>4</td></tr>
+</table></div>
+
+
+<p>And each man has received his four pints of ale.</p>
+
+
+<hr style="width: 30%;" />
+<p><a name="X_363_THE_DOCTORS_QUERYa" id="X_363_THE_DOCTORS_QUERYa"></a><a href="#X_363_THE_DOCTORS_QUERY"><b>363.&mdash;THE DOCTOR'S QUERY.&mdash;<i>solution</i></b></a></p>
+
+
+<p>The mixture of spirits of wine and water is in the proportion of 40 to
+1, just as in the other bottle it was in the proportion of 1 to 40.</p>
+
+
+<hr style="width: 30%;" />
+<p><a name="X_364_THE_BARREL_PUZZLEa" id="X_364_THE_BARREL_PUZZLEa"></a><a href="#X_364_THE_BARREL_PUZZLE"><b>364.&mdash;THE BARREL PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a364.png" width="400" height="296" alt="" title="" />
+</div>
+
+<p>All that is necessary is to tilt the barrel as in Fig. 1, and if the
+edge of the surface of the water exactly touches the lip <i>a</i> at the
+same time that it touches the edge of the bottom <i>b</i>, it will be just
+half full. To be more exact, if the bottom is an inch or so from the
+ground, then we can allow for that, and the thickness of the bottom,
+at the top. If when the surface of the water reached the lip <i>a</i> it
+had risen to the point <i>c</i> in Fig. 2, then it would be more than half
+full. If, as in Fig. 3, some portion of the bottom were visible and
+the level of the water fell to the point <i>d</i>, then it would be less
+than half full.</p>
+
+<p>This method applies to all symmetrically constructed vessels.</p>
+
+
+<hr style="width: 30%;" />
+<p><a name="X_365_NEW_MEASURING_PUZZLEa" id="X_365_NEW_MEASURING_PUZZLEa"></a><a href="#X_365_NEW_MEASURING_PUZZLE"><b>365.&mdash;NEW MEASURING PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+
+<p>The following solution in eleven manipulations shows the contents of
+every vessel at the start and after every manipulation:&mdash;</p>
+
+
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>10-quart.</td><td></td><td align='center'>10-quart.</td><td></td><td align='center'>5-quart.</td><td></td><td align='center'>4-quart.</td></tr>
+<tr><td align='center'>10</td><td align='center'>..</td><td align='center'>10</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>0</td></tr>
+<tr><td align='center'>5</td><td align='center'>..</td><td align='center'>10</td><td align='center'>..</td><td align='center'>5</td><td align='center'>..</td><td align='center'>0</td></tr>
+<tr><td align='center'>5</td><td align='center'>..</td><td align='center'>10</td><td align='center'>..</td><td align='center'>1</td><td align='center'>..</td><td align='center'>4</td></tr>
+<tr><td align='center'>9</td><td align='center'>..</td><td align='center'>10</td><td align='center'>..</td><td align='center'>1</td><td align='center'>..</td><td align='center'>0</td></tr>
+<tr><td align='center'>9</td><td align='center'>..</td><td align='center'>6</td><td align='center'>..</td><td align='center'>1</td><td align='center'>..</td><td align='center'>4</td></tr>
+<tr><td align='center'>9</td><td align='center'>..</td><td align='center'>7</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>4</td></tr>
+<tr><td align='center'>9</td><td align='center'>..</td><td align='center'>7</td><td align='center'>..</td><td align='center'>4</td><td align='center'>..</td><td align='center'>0</td></tr>
+<tr><td align='center'>9</td><td align='center'>..</td><td align='center'>3</td><td align='center'>..</td><td align='center'>4</td><td align='center'>..</td><td align='center'>4</td></tr>
+<tr><td align='center'>9</td><td align='center'>..</td><td align='center'>3</td><td align='center'>..</td><td align='center'>5</td><td align='center'>..</td><td align='center'>3</td></tr>
+<tr><td align='center'>9</td><td align='center'>..</td><td align='center'>8</td><td align='center'>..</td><td align='center'>0</td><td align='center'>..</td><td align='center'>3</td></tr>
+<tr><td align='center'>4</td><td align='center'>..</td><td align='center'>8</td><td align='center'>..</td><td align='center'>5</td><td align='center'>..</td><td align='center'>3</td></tr>
+<tr><td align='center'>4</td><td align='center'>..</td><td align='center'>10</td><td align='center'>..</td><td align='center'>3</td><td align='center'>..</td><td align='center'>3</td></tr>
+</table></div>
+
+
+
+<hr style="width: 30%;" />
+<p><a name="X_366_THE_HONEST_DAIRYMANa" id="X_366_THE_HONEST_DAIRYMANa"></a><a href="#X_366_THE_HONEST_DAIRYMAN"><b>366.&mdash;THE HONEST DAIRYMAN.&mdash;<i>solution</i></b></a></p>
+
+
+<p>Whatever the respective quantities of milk and water, the relative
+proportion sent to London would always be three parts of water to one
+of milk. But there are one or two points to be observed. There must
+originally be more water than milk, or there will be no water in A to
+double in the second transaction. And the water must not be more than
+three times the quantity of milk, or there will not be enough liquid
+in B to effect the second transaction. The third transaction has no
+effect on A, as the relative proportions in it must be the same as
+after the second transaction. It was introduced to prevent a quibble
+if the quantity of milk and water were originally the same; for though
+double "nothing" would be "nothing," yet the third transaction in such
+a case could not take place.</p>
+
+
+<hr style="width: 30%;" />
+<p><a name="X_367_WINE_AND_WATERa" id="X_367_WINE_AND_WATERa"></a><a href="#X_367_WINE_AND_WATER"><b>367.&mdash;WINE AND WATER.&mdash;<i>solution</i></b></a></p>
+
+
+<p>The wine in small glass was one-sixth of the total liquid, and the
+wine in large glass two-ninths of total. Add these together, and we
+find that the wine was seven-eighteenths of total fluid, and therefore
+the water eleven-eighteenths.</p>
+
+
+<hr style="width: 30%;" />
+<p><a name="X_368_THE_KEG_OF_WINEa" id="X_368_THE_KEG_OF_WINEa"></a><a href="#X_368_THE_KEG_OF_WINE"><b>368.&mdash;THE KEG OF WINE.&mdash;<i>solution</i></b></a></p>
+
+
+<p>The capacity of the jug must have been a little less than three
+gallons. To be more exact, it was 2.93 gallons.</p>
+
+
+<hr style="width: 30%;" />
+<p><a name="X_369_MIXING_THE_TEAa" id="X_369_MIXING_THE_TEAa"></a><a href="#X_369_MIXING_THE_TEA"><b>369.&mdash;MIXING THE TEA.&mdash;<i>solution</i></b></a></p>
+
+
+<p>There are three ways of mixing the teas. Taking them in the order of
+quality, 2<i>s</i>. 6<i>d</i>., 2<i>s</i>. 3<i>d</i>., 1<i>s</i>. 9p., mix 16 lbs., 1 lb., 3 lbs.; or 14
+lbs., 4 lbs., 2 lbs.; or 12 lbs., 7 lbs., 1 lb. In every case the
+twenty pounds mixture should be worth 2<i>s</i>. 4�<i>d</i>. per pound; but the
+last case requires the smallest quantity of the best tea, therefore it
+is the correct answer.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 236<a name="Page_236" id="Page_236"></a></span><a name="X_370_A_PACKING_PUZZLEa" id="X_370_A_PACKING_PUZZLEa"></a><a href="#X_370_A_PACKING_PUZZLE"><b>370.&mdash;A PACKING PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>On the side of the box, 14 by 22<sup>4</sup>/<sub>5</sub>, we can arrange 13 rows
+containing alternately 7 and 6 balls, or 85 in all. Above this we can
+place another layer consisting of 12 rows of 7 and 6 alternately, or a
+total of 78. In the length of 24<sup>9</sup>/<sub>10</sub> inches 15 such layers may be
+packed, the alternate layers containing 85 and 78 balls. Thus 8 times
+85 added to 7 times 78 gives us 1,226 for the full contents of the
+box.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_371_GOLD_PACKING_IN_RUSSIAa" id="X_371_GOLD_PACKING_IN_RUSSIAa"></a><a href="#X_371_GOLD_PACKING_IN_RUSSIA"><b>371.&mdash;GOLD PACKING IN RUSSIA.&mdash;<i>solution</i></b></a></p>
+
+<p>The box should be 100 inches by 100 inches by 11 inches deep, internal
+dimensions. We can lay flat at the bottom a row of eight slabs,
+lengthways, end to end, which will just fill one side, and nine of
+these rows will dispose of seventy-two slabs (all on the bottom), with
+a space left over on the bottom measuring 100 inches by 1 inch by 1
+inch. Now make eleven depths of such seventy-two slabs, and we have
+packed 792, and have a space 100 inches by 1 inch by 11 inches deep.
+In this we may exactly pack the remaining eight slabs on edge, end to
+end.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_372_THE_BARRELS_OF_HONEYa" id="X_372_THE_BARRELS_OF_HONEYa"></a><a href="#X_372_THE_BARRELS_OF_HONEY"><b>372.&mdash;THE BARRELS OF HONEY.&mdash;<i>solution</i></b></a></p>
+
+<p>The only way in which the barrels could be equally divided among the
+three brothers, so that each should receive his 3� barrels of honey
+and his 7 barrels, is as follows:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'></td><td align='center'>Full.</td><td align='center'>Half-full.</td><td align='center'>Empty.</td></tr>
+<tr><td align='center'>A</td><td align='center'>3</td><td align='center'>1</td><td align='center'>3</td></tr>
+<tr><td align='center'>B</td><td align='center'>2</td><td align='center'>3</td><td align='center'>2</td></tr>
+<tr><td align='center'>C</td><td align='center'>2</td><td align='center'>3</td><td align='center'>2</td></tr>
+</table></div>
+
+<p>There is one other way in which the division could be made, were it
+not for the objection that all the brothers made to taking more than
+four barrels of the same description. Except for this difficulty, they
+might have given B his quantity in exactly the same way as A above,
+and then have left C one full barrel, five half-full barrels, and one
+empty barrel. It will thus be seen that in any case two brothers would
+have to receive their allowance in the same way.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_373_CROSSING_THE_STREAMa" id="X_373_CROSSING_THE_STREAMa"></a><a href="#X_373_CROSSING_THE_STREAM"><b>373.&mdash;CROSSING THE STREAM.&mdash;<i>solution</i></b></a></p>
+
+<p>First, the two sons cross, and one returns Then the man crosses and
+the other son returns. Then both sons cross and one returns. Then the
+lady crosses and the other son returns Then the two sons cross and one
+of them returns for the dog. Eleven crossings in all.</p>
+
+<p>It would appear that no general rule can be given for solving these
+river-crossing puzzles. A formula can be found for a particular case
+(say on No. <a href="#X_375_FIVE_JEALOUS_HUSBANDSa">375</a> or <a href="#X_376_THE_FOUR_ELOPEMENTSa">376</a>) that would apply to any number of individuals
+under the restricted conditions; but it is not of much use, for some
+little added stipulation will entirely upset it. As in the case of the
+measuring puzzles, we generally have to rely on individual ingenuity.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_374_CROSSING_THE_RIVER_AXEa" id="X_374_CROSSING_THE_RIVER_AXEa"></a><a href="#X_374_CROSSING_THE_RIVER_AXE"><b>374.&mdash;CROSSING THE RIVER AXE.&mdash;<i>solution</i></b></a></p>
+
+<p>Here is the solution:&mdash;</p>
+
+<div class='center'>
+<table border="1" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'></td><td align='center'>{J 5)</td><td align='center'>G T8 3</td></tr>
+<tr><td align='center'>5</td><td align='center'>( J }</td><td align='center'>G T8 3</td></tr>
+<tr><td align='center'>5</td><td align='center'>{G 3)</td><td align='center'>JT8</td></tr>
+<tr><td align='center'>53</td><td align='center'>( G }</td><td align='center'>JT8</td></tr>
+<tr><td align='center'>53</td><td align='center'>{J T)</td><td align='center'>G 8</td></tr>
+<tr><td align='center'>J 5</td><td align='center'>(T 3}</td><td align='center'>G 8</td></tr>
+<tr><td align='center'>J 5</td><td align='center'>{G 8)</td><td align='center'>T 3</td></tr>
+<tr><td align='center'>G 8</td><td align='center'>(J 5}</td><td align='center'>T</td></tr>
+<tr><td align='center'>G 8</td><td align='center'>{J T)</td><td align='center'>53</td></tr>
+<tr><td align='center'>JT8</td><td align='center'>( G }</td><td align='center'>53</td></tr>
+<tr><td align='center'>JT8</td><td align='center'>{G 3)</td><td align='center'>5</td></tr>
+<tr><td align='center'>G T8 3</td><td align='center'>( J }</td><td align='center'>5</td></tr>
+<tr><td align='center'>G T8 3</td><td align='center'>{J 5)</td><td></td></tr>
+</table></div>
+
+<p>G, J, and T stand for Giles, Jasper, and Timothy; and 8, 5, 3, for
+&pound;800, &pound;500, and &pound;300 respectively. The two side columns represent the
+left bank and the right bank, and the middle column the river.
+Thirteen crossings are necessary, and each line shows the position
+when the boat is in mid-stream during a crossing, the point of the
+bracket indicating the direction.</p>
+
+<p>It will be found that not only is no person left alone on the land or
+in the boat with more than his share of the spoil, but that also no
+two persons are left with more than their joint shares, though this
+last point was not insisted upon in the conditions.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_375_FIVE_JEALOUS_HUSBANDSa" id="X_375_FIVE_JEALOUS_HUSBANDSa"></a><a href="#X_375_FIVE_JEALOUS_HUSBANDS"><b>375.&mdash;FIVE JEALOUS HUSBANDS.&mdash;<i>solution</i></b></a></p>
+
+<p>It is obvious that there must be an odd number of crossings, and that
+if the five husbands had not been jealous of one another the party
+might have all got over in nine crossings. But no wife was to be in
+the company of a man or men unless her husband was present. This
+entails two more crossings, eleven in all.</p>
+
+<p>The following shows how it might have been done. The capital letters
+stand for the husbands, and the small letters for their respective
+wives. The position of affairs is shown at the start, and after each
+crossing between the left bank and the right, and the boat is
+represented by the asterisk. So you can see at a glance that a, b, and
+c went over at the first crossing, that b and c returned at the second
+crossing, and so on.<span class='pagenum'>Pg 237<a name="Page_237" id="Page_237"></a></span></p>
+
+<div class='center'>
+<table border="1" cellpadding="4" cellspacing="0" summary="">
+<tr><td></td><td align='center'>&nbsp; &nbsp;ABCDE&nbsp;abcde&nbsp;*</td><td align='center'>..</td><td></td></tr>
+<tr><td align='right'>1.</td><td align='center'>&nbsp;ABCDE&nbsp; &nbsp; de&nbsp;</td><td align='center'>..</td><td align='center'>*&nbsp; &nbsp; &nbsp; &nbsp;abc</td></tr>
+<tr><td align='right'>2.</td><td align='center'>&nbsp;ABCDE&nbsp; bcde&nbsp;*</td><td align='center'>..</td><td align='center'>&nbsp; &nbsp; &nbsp; &nbsp; a</td></tr>
+<tr><td align='right'>3.</td><td align='center'>&nbsp;ABCDE&nbsp; &nbsp; &nbsp;e&nbsp;</td><td align='center'>..</td><td align='center'>*&nbsp; &nbsp; &nbsp; &nbsp;abcd</td></tr>
+<tr><td align='right'>4.</td><td align='center'>&nbsp;ABCDE&nbsp; &nbsp; de&nbsp;*</td><td align='center'>..</td><td align='center'>&nbsp; &nbsp; &nbsp; &nbsp; abc</td></tr>
+<tr><td align='right'>5.</td><td align='center'>&nbsp; &nbsp; DE&nbsp; &nbsp; de&nbsp;</td><td align='center'>..</td><td align='center'>*&nbsp;ABC&nbsp; &nbsp;abc</td></tr>
+<tr><td align='right'>6.</td><td align='center'>&nbsp; &nbsp;CDE&nbsp; &nbsp;cde&nbsp;*</td><td align='center'>..</td><td align='center'>&nbsp; AB&nbsp; &nbsp; ab</td></tr>
+<tr><td align='right'>7.</td><td align='center'>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;cde&nbsp;</td><td align='center'>..</td><td align='center'>*&nbsp;ABCDE&nbsp;ab</td></tr>
+<tr><td align='right'>8.</td><td align='center'>&nbsp; &nbsp; &nbsp; &nbsp; bcde&nbsp;*</td><td align='center'>..</td><td align='center'>&nbsp; ABCDE&nbsp;a</td></tr>
+<tr><td align='right'>9.</td><td align='center'>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;e&nbsp;</td><td align='center'>..</td><td align='center'>*&nbsp;ABCDE&nbsp;abcd</td></tr>
+<tr><td align='right'>10.</td><td align='center'>&nbsp; &nbsp; &nbsp; &nbsp;bc&nbsp;e&nbsp;*</td><td align='center'>..</td><td align='center'>&nbsp; ABCDE&nbsp;a&nbsp; d</td></tr>
+<tr><td align='right'>11.</td><td align='center'>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</td><td align='center'>..</td><td align='center'>* ABCDE abcde</td></tr>
+</table></div>
+
+<p>There is a little subtlety concealed in the words "show the <i>quickest</i>
+way."</p>
+
+<p>Everybody correctly assumes that, as we are told nothing of the rowing
+capabilities of the party, we must take it that they all row equally
+well. But it is obvious that two such persons should row more quickly
+than one.</p>
+
+<p>Therefore in the second and third crossings two of the ladies should
+take back the boat to fetch d, not one of them only. This does not
+affect the number of landings, so no time is lost on that account. A
+similar opportunity occurs in crossings 10 and 11, where the party
+again had the option of sending over two ladies or one only.</p>
+
+<p>To those who think they have solved the puzzle in nine crossings I
+would say that in every case they will find that they are wrong. No
+such jealous husband would, in the circumstances, send his wife over
+to the other bank to a man or men, even if she assured him that she
+was coming back next time in the boat. If readers will have this fact
+in mind, they will at once discover their errors.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_376_THE_FOUR_ELOPEMENTSa" id="X_376_THE_FOUR_ELOPEMENTSa"></a><a href="#X_376_THE_FOUR_ELOPEMENTS"><b>376.&mdash;THE FOUR ELOPEMENTS.&mdash;<i>solution</i></b></a></p>
+
+<p>If there had been only three couples, the island might have been
+dispensed with, but with four or more couples it is absolutely
+necessary in order to cross under the conditions laid down. It can be
+done in seventeen passages from land to land (though French
+mathematicians have declared in their books that in such circumstances
+twenty-four are needed), and it cannot be done in fewer. I will give
+one way. A, B, C, and D are the young men, and a, b, c, and d are the
+girls to whom they are respectively engaged. The three columns show
+the positions of the different individuals on the lawn, the island,
+and the opposite shore before starting and after each passage, while
+the asterisk indicates the position of the boat on every occasion.</p>
+
+
+<div class='center'>
+<table border="1" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center'>Lawn.</td><td align='center'>Island.</td><td align='center'>Shore.</td></tr>
+<tr><td align='center'>ABCDabcd *</td><td align='center'></td><td></td></tr>
+<tr><td align='center'>ABCD cd</td><td align='center'></td><td align='center'>ab *</td></tr>
+<tr><td align='center'>ABCD bcd *</td><td align='center'></td><td align='center'>a</td></tr>
+<tr><td align='center'>ABCD d</td><td align='center'>bc *</td><td align='center'>a</td></tr>
+<tr><td align='center'>ABCD cd *</td><td align='center'>b</td><td align='center'>a</td></tr>
+<tr><td align='center'>CD cd</td><td align='center'>b</td><td align='center'>AB a *</td></tr>
+<tr><td align='center'>BCD cd *</td><td align='center'>b</td><td align='center'>A a</td></tr>
+<tr><td align='center'>BCD</td><td align='center'>bcd *</td><td align='center'>A a</td></tr>
+<tr><td align='center'>BCD d *</td><td align='center'>bc</td><td align='center'>A a</td></tr>
+<tr><td align='center'>D d</td><td align='center'>bc</td><td align='center'>ABC a *</td></tr>
+<tr><td align='center'>D d</td><td align='center'>abc *</td><td align='center'>ABC</td></tr>
+<tr><td align='center'>D d</td><td align='center'>b</td><td align='center'>ABC a c *</td></tr>
+<tr><td align='center'>B D d *</td><td align='center'>b</td><td align='center'>A C a c</td></tr>
+<tr><td align='center'>d</td><td align='center'>b</td><td align='center'>ABCD a c *</td></tr>
+<tr><td align='center'>d</td><td align='center'>bc *</td><td align='center'>ABCD a</td></tr>
+<tr><td align='center'>d</td><td align='center'></td><td align='center'>ABCD abc *</td></tr>
+<tr><td align='center'>cd *</td><td align='center'></td><td align='center'>ABCD ab</td></tr>
+<tr><td align='center'></td><td align='center'></td><td align='center'>ABCD abcd *</td></tr>
+</table></div>
+
+
+<p>Having found the fewest possible passages, we should consider two
+other points in deciding on the "quickest method": Which persons were
+the most expert in handling the oars, and which method entails the
+fewest possible delays in getting in and out of the boat? We have no
+data upon which to decide the first point, though it is probable that,
+as the boat belonged to the girls' household, they would be capable
+oarswomen. The other point, however, is important, and in the solution
+I have given (where the girls do 8-13ths of the rowing and A and D
+need not row at all) there are only sixteen gettings-in and sixteen
+gettings-out. A man and a girl are never in the boat together, and no
+man ever lands on the island. There are other methods that require
+several more exchanges of places.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_377_STEALING_THE_CASTLE_TREASUREa" id="X_377_STEALING_THE_CASTLE_TREASUREa"></a><a href="#X_377_STEALING_THE_CASTLE_TREASURE"><b>377.&mdash;STEALING THE CASTLE TREASURE.&mdash;<i>solution</i></b></a></p>
+
+<p>Here is the best answer, in eleven manipulations:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='left'>Treasure down.</td></tr>
+<tr><td align='left'>Boy down&mdash;treasure up.</td></tr>
+<tr><td align='left'>Youth down&mdash;boy up.</td></tr>
+<tr><td align='left'>Treasure down.</td></tr>
+<tr><td align='left'>Man down&mdash;youth and treasure up.</td></tr>
+<tr><td align='left'>Treasure down.</td></tr>
+<tr><td align='left'>Boy down&mdash;treasure up.</td></tr>
+<tr><td align='left'>Treasure down.</td></tr>
+<tr><td align='left'>Youth down&mdash;boy up.</td></tr>
+<tr><td align='left'>Boy down&mdash;treasure up.</td></tr>
+<tr><td align='left'>Treasure down.</td></tr>
+</table></div>
+
+<hr style="width: 30%;" />
+<p><a name="X_378_DOMINOES_IN_PROGRESSIONa" id="X_378_DOMINOES_IN_PROGRESSIONa"></a><a href="#X_378_DOMINOES_IN_PROGRESSION"><b>378.&mdash;DOMINOES IN PROGRESSION.&mdash;<i>solution</i></b></a></p>
+
+<p>There are twenty-three different ways. You may start with any domino,
+except the 4&mdash;4 and those that bear a 5 or 6, though only certain
+initial dominoes may be played either way round. If you are given the
+common difference and the first domino is played, you have no option
+as to the other dominoes. Therefore all I need do is to give the
+initial domino for all the twenty-three ways, and state the common
+difference. This I will do as follows:&mdash;</p>
+
+<p>With a common difference of 1, the first domino may be either of
+these: 0&mdash;0, 0&mdash;1, 1&mdash;0, 0&mdash;2, 1&mdash;1, 2&mdash;0, 0&mdash;3, 1&mdash;2, 2&mdash;1, 3&mdash;0,
+0&mdash;4, 1&mdash;3, 2&mdash;2, 3&mdash;1, 1&mdash;4, 2&mdash;3, 3&mdash;2, 2&mdash;4, 3&mdash;3, 3&mdash;4. With a
+difference of 2, the first domino may be 0&mdash;0, 0&mdash;2, or 0&mdash;1. Take the
+last case of all as an example. Having played the 0&mdash;1, and the
+difference being 2, we are <span class='pagenum'>Pg 238<a name="Page_238" id="Page_238"></a></span>compelled to continue with 1&mdash;2, 2&mdash;3,
+3&mdash;4. 4&mdash;5, 5&mdash;6. There are three dominoes that can never be used at
+all. These are 0&mdash;5, 0&mdash;6, and 1&mdash;6. If we used a box of dominoes
+extending to 9&mdash;9, there would be forty different ways.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_379_THE_FIVE_DOMINOESa" id="X_379_THE_FIVE_DOMINOESa"></a><a href="#X_379_THE_FIVE_DOMINOES"><b>379.&mdash;THE FIVE DOMINOES.&mdash;<i>solution</i></b></a></p>
+
+<p>There are just ten different ways of arranging the dominoes. Here is
+one of them:&mdash;</p>
+
+<p>(2&mdash;0) (0&mdash;0) (0&mdash;1) (1&mdash;4) (4&mdash;0).</p>
+
+<p>I will leave my readers to find the remaining nine for themselves.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_380_THE_DOMINO_FRAME_PUZZLEa" id="X_380_THE_DOMINO_FRAME_PUZZLEa"></a><a href="#X_380_THE_DOMINO_FRAME_PUZZLE"><b>380.&mdash;THE DOMINO FRAME PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a380.png" width="400" height="400" alt="" title="" />
+</div>
+
+<p>The illustration is a solution. It will be found that all four sides
+of the frame add up 44. The sum of the pips on all the dominoes is
+168, and if we wish to make the sides sum to 44, we must take care
+that the four corners sum to 8, because these corners are counted
+twice, and 168 added to 8 will equal 4 times 44, which is necessary.
+There are many different solutions. Even in the example given certain
+interchanges are possible to produce different arrangements. For
+example, on the left-hand side the string of dominoes from 2&mdash;2 down
+to 3&mdash;2 may be reversed, or from 2&mdash;6 to 3&mdash;2, or from 3&mdash;0 to 5&mdash;3.
+Also, on the right-hand side we may reverse from 4&mdash;3 to 1&mdash;4. These
+changes will not affect the correctness of the solution.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_381_THE_CARD_FRAME_PUZZLEa" id="X_381_THE_CARD_FRAME_PUZZLEa"></a><a href="#X_381_THE_CARD_FRAME_PUZZLE"><b>381.&mdash;THE CARD FRAME PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The sum of all the pips on the ten cards is 55. Suppose we are trying
+to get 14 pips on every side. Then 4 times 14 is 56. But each of the
+four corner cards is added in twice, so that 55 deducted from 56, or
+1, must represent the sum of the four corner cards. This is clearly
+impossible; therefore 14 is also impossible. But suppose we came to
+trying 18. Then 4 times 18 is 72, and if we deduct 55 we get 17 as the
+sum of the corners. We need then only try different arrangements with
+the four corners always summing to 17, and we soon discover the
+following solution:&mdash;</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a381.png" width="400" height="401" alt="" title="" />
+</div>
+
+<p>The final trials are very limited in number, and must with a little
+judgment either bring us to a correct solution or satisfy us that a
+solution is impossible under the conditions we are attempting. The two
+centre cards on the upright sides can, of course, always be
+interchanged, but I do not call these different solutions. If you
+reflect in a mirror you get another arrangement, which also is not
+considered different. In the answer given, however, we may exchange
+the 5 with the 8 and the 4 with the 1. This is a different solution.
+There are two solutions with 18, four with 19, two with 20, and two
+with 22&mdash;ten arrangements in all. Readers may like to find all these
+for themselves.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_382_THE_CROSS_OF_CARDSa" id="X_382_THE_CROSS_OF_CARDSa"></a><a href="#X_382_THE_CROSS_OF_CARDS"><b>382.&mdash;THE CROSS OF CARDS.&mdash;<i>solution</i></b></a></p>
+
+<p>There are eighteen fundamental arrangements, as follows, where I only
+give the numbers in the horizontal bar, since the remainder must
+naturally fall into their places.</p>
+
+<div class='center'>
+<table border="0" cellpadding="2" cellspacing="0" summary="">
+<tr><td align='center'>5 6 1 7 4</td></tr>
+<tr><td align='center'>3 5 1 6 8</td></tr>
+<tr><td align='center'>3 4 1 7 8</td></tr>
+<tr><td align='center'>2 5 1 7 8</td></tr>
+<tr><td align='center'>2 5 3 6 8</td></tr>
+<tr><td align='center'>1 5 3 7 8</td></tr>
+<tr><td align='center'>2 4 3 7 8</td></tr>
+<tr><td align='center'>1 4 5 7 8</td></tr>
+<tr><td align='center'>2 3 5 7 8</td></tr>
+<tr><td align='center'>2 4 5 6 8</td></tr>
+<tr><td align='center'>3 4 5 6 7</td></tr>
+<tr><td align='center'>1 4 7 6 8</td></tr>
+<tr><td align='center'>2 3 7 6 8</td></tr>
+<tr><td align='center'>2 4 7 5 8</td></tr>
+<tr><td align='center'>3 4 9 5 6</td></tr>
+<tr><td align='center'>2 4 9 5 7</td></tr>
+<tr><td align='center'>1 4 9 6 7</td></tr>
+<tr><td align='center'>2 3 9 6 7</td></tr>
+</table></div>
+
+<p>It will be noticed that there must always be an odd number in the
+centre, that there are four ways each of adding up 23, 25, and 27, but
+only three ways each of summing to 24 and 26.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 239<a name="Page_239" id="Page_239"></a></span><a name="X_383_THE_T_CARD_PUZZLEa" id="X_383_THE_T_CARD_PUZZLEa"></a><a href="#X_383_THE_T_CARD_PUZZLE"><b>383.&mdash;THE "T" CARD PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>If we remove the ace, the remaining cards may he divided into two
+groups (each adding up alike) in four ways; if we remove 3, there are
+three ways; if 5, there are four ways; if 7, there are three ways; and
+if we remove 9, there are four ways of making two equal groups. There
+are thus eighteen different ways of grouping, and if we take any one
+of these and keep the odd card (that I have called "removed") at the
+head of the column, then one set of numbers can be varied in order in
+twenty-four ways in the column and the other four twenty-four ways in
+the horizontal, or together they may be varied in 24&nbsp;&times;&nbsp;24&nbsp;=&nbsp;576 ways.
+And as there are eighteen such cases, we multiply this number by 18
+and get 10,368, the correct number of ways of placing the cards. As
+this number includes the reflections, we must divide by 2, but we have
+also to remember that every horizontal row can change places with a
+vertical row, necessitating our multiplying by 2; so one operation
+cancels the other.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_384_CARD_TRIANGLESa" id="X_384_CARD_TRIANGLESa"></a><a href="#X_384_CARD_TRIANGLES"><b>384.&mdash;CARD TRIANGLES.&mdash;<i>solution</i></b></a></p>
+
+<p>The following arrangements of the cards show (1) the smallest possible
+sum, 17; and (2) the largest possible, 23.</p>
+
+<div class="figcenter" style="width: 300px;">
+<img src="images/a384.png" width="300" height="100" alt="" title="" />
+</div>
+
+<p>It will be seen that the two cards in the middle of any side may
+always be interchanged without affecting the conditions. Thus there
+are eight ways of presenting every fundamental arrangement. The number
+of fundamentals is eighteen, as follows: two summing to 17, four
+summing to 19, six summing to 20, four summing to 21, and two summing
+to 23. These eighteen fundamentals, multiplied by eight (for the
+reason stated above), give 144 as the total number of different ways
+of placing the cards.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_385_STRAND_PATIENCEa" id="X_385_STRAND_PATIENCEa"></a><a href="#X_385_STRAND_PATIENCE"><b>385.&mdash;"STRAND" PATIENCE.&mdash;<i>solution</i></b></a></p>
+
+<p>The reader may find a solution quite easy in a little over 200 moves,
+but, surprising as it may at first appear, not more than 62 moves are
+required. Here is the play: By "4 C up" I mean a transfer of the 4 of
+clubs with all the cards that rest on it. 1 D on space, 2 S on space,
+3 D on space, 2 S on 3 D, 1 H on 2 S, 2 C on space, 1 D on 2 C, 4 S on
+space, 3 H on 4 S (9 moves so far), 2 S up on 3 H (3 moves), 5 H and 5
+D exchanged, and 4 C on 5 D (6 moves), 3 D on 4 C (1), 6 S (with 5 H)
+on space (3), 4 C up on 5 H (3), 2 C up on 3 D (3), 7 D on space (1),
+6 C up on 7 D (3), 8 S on space (1), 7 H on 8 S (1), 8 C on 9 D (1), 7
+H on 8 C (1), 8 S on 9 H (1), 7 H on 8 S (1), 7 D up on 8 C (5), 4 C
+up on 5 D (9), 6 S up on 7 H (3), 4 S up on 5 H (7)&nbsp;=&nbsp;62 moves in all.
+This is my record; perhaps the reader can beat it.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_386_A_TRICK_WITH_DICEa" id="X_386_A_TRICK_WITH_DICEa"></a><a href="#X_386_A_TRICK_WITH_DICE"><b>386.&mdash;A TRICK WITH DICE.&mdash;<i>solution</i></b></a></p>
+
+<p>All you have to do is to deduct 250 from the result given, and the
+three figures in the answer will be the three points thrown with the
+dice. Thus, in the throw we gave, the number given would be 386; and
+when we deduct 250 we get 136, from which we know that the throws were
+1, 3, and 6.</p>
+
+<p>The process merely consists in giving 100<i>a</i>&nbsp;+&nbsp;10<i>b</i>&nbsp;+&nbsp;<i>c</i>&nbsp;+&nbsp;250, where <i>a</i>,
+<i>b</i>, and <i>c</i> represent the three throws. The result is obvious.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_387_THE_VILLAGE_CRICKET_MATCHa" id="X_387_THE_VILLAGE_CRICKET_MATCHa"></a><a href="#X_387_THE_VILLAGE_CRICKET_MATCH"><b>387.&mdash;THE VILLAGE CRICKET MATCH.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a387.png" width="400" height="325" alt="" title="" />
+</div>
+
+<p>The diagram No. 1 will show that as neither Mr. Podder nor Mr. Dumkins
+can ever have been within the crease opposite to that from which he
+started, Mr. Dumkins would score nothing by his performance. Diagram
+No. 2 will, however, make it clear that since Mr. Luffey and Mr.
+Struggles have, notwithstanding their energetic but careless
+movements, contrived to change places, the man&#339;uvre must increase Mr.
+Struggles's total by one run.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_388_SLOW_CRICKETa" id="X_388_SLOW_CRICKETa"></a><a href="#X_388_SLOW_CRICKET"><b>388.&mdash;SLOW CRICKET.&mdash;<i>solution</i></b></a></p>
+
+<p>The captain must have been "not out" and scored 21. Thus:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='left'>2 men (each lbw)</td><td align='right'>19</td></tr>
+<tr><td align='left'>4 men (each caught)</td><td align='right'>17</td></tr>
+<tr><td align='left'>1 man (run out)</td><td align='right'>0</td></tr>
+<tr><td align='left'>3 men (each bowled)</td><td align='right'>9</td></tr>
+<tr><td align='left'>1 man (captain&mdash;not out)</td><td align='right'>21</td></tr>
+<tr><td align='left' class='bt'>11</td><td align='right' class='bt'>66</td></tr>
+</table></div>
+
+<p>The captain thus scored exactly 15 more than the average of the team.
+The "others" who were bowled could only refer to three men, as the
+eleventh man would be "not out." The reader can discover for himself
+why the captain must have been that eleventh man. It would not
+necessarily follow with any figures.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 240<a name="Page_240" id="Page_240"></a></span><a name="X_389_THE_FOOTBALL_PLAYERSa" id="X_389_THE_FOOTBALL_PLAYERSa"></a><a href="#X_389_THE_FOOTBALL_PLAYERS"><b>389.&mdash;THE FOOTBALL PLAYERS.&mdash;<i>solution</i></b></a></p>
+
+<p>The smallest possible number of men is seven. They could be accounted
+for in three different ways: 1. Two with both arms sound, one with
+broken right arm, and four with both arms broken. 2. One with both
+arms sound, one with broken left arm, two with broken right arm, and
+three with both arms broken. 3. Two with left arm broken, three with
+right arm broken, and two with both arms broken. But if every man was
+injured, the last case is the only one that would apply.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_390_THE_HORSE-RACE_PUZZLEa" id="X_390_THE_HORSE-RACE_PUZZLEa"></a><a href="#X_390_THE_HORSE-RACE_PUZZLE"><b>390.&mdash;THE HORSE-RACE PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The answer is: &pound;12 on Acorn, &pound;15 on Bluebottle, &pound;20 on Capsule.</p>
+<hr style="width: 30%;" />
+<p><a name="X_391_THE_MOTOR-CAR_RACEa" id="X_391_THE_MOTOR-CAR_RACEa"></a><a href="#X_391_THE_MOTOR-CAR_RACE"><b>391.&mdash;THE MOTOR-CAR RACE.&mdash;<i>solution</i></b></a></p>
+
+<p>The first point is to appreciate the fact that, in a race round a
+circular track, there are the same number of cars behind one as there
+are before. All the others are both behind and before. There were
+thirteen cars in the race, including Gogglesmith's car. Then one-third
+of twelve added to three-quarters of twelve will give us thirteen&mdash;the
+correct answer.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_392_THE_PEBBLE_GAMEa" id="X_392_THE_PEBBLE_GAMEa"></a><a href="#X_392_THE_PEBBLE_GAME"><b>392.&mdash;THE PEBBLE GAME.&mdash;<i>solution</i></b></a></p>
+
+<p>In the case of fifteen pebbles, the first player wins if he first
+takes two. Then when he holds an odd number and leaves 1, 8, or 9 he
+wins, and when he holds an even number and leaves 4, 5, or 12 he also
+wins. He can always do one or other of these things until the end of
+the game, and so defeat his opponent. In the case of thirteen pebbles
+the first player must lose if his opponent plays correctly. In fact,
+the only numbers with which the first player ought to lose are 5 and
+multiples of 8 added to 5, such as 13, 21, 29, etc.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_393_THE_TWO_ROOKSa" id="X_393_THE_TWO_ROOKSa"></a><a href="#X_393_THE_TWO_ROOKS"><b>393.&mdash;THE TWO ROOKS.&mdash;<i>solution</i></b></a></p>
+
+<p>The second player can always win, but to ensure his doing so he must
+always place his rook, at the start and on every subsequent move, on
+the same diagonal as his opponent's rook. He can then force his
+opponent into a corner and win. Supposing the diagram to represent the
+positions of the rooks at the start, then, if Black played first,
+White might have placed his rook at A and won next move. Any square on
+that diagonal from A to H will win, but the best play is always to
+restrict the moves of the opposing rook as much as possible. If White
+played first, then Black should have placed his rook at B (F would not
+be so good, as it gives White more scope); then if White goes to C,
+Black moves to D; White to E, Black to F; White to G, Black to C;
+White to H, Black to I; and Black must win next move. If at any time
+Black had failed to move on to the same diagonal as White, then White
+could take Black's diagonal and win.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a393.png" width="400" height="400" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_394_PUSS_IN_THE_CORNERa" id="X_394_PUSS_IN_THE_CORNERa"></a><a href="#X_394_PUSS_IN_THE_CORNER"><b>394.&mdash;PUSS IN THE CORNER.&mdash;<i>solution</i></b></a></p>
+
+<p>No matter whether he plays first or second, the player A, who starts
+the game at 55, must win. Assuming that B adopts the very best lines
+of play in order to prolong as much as possible his existence, A, if
+he has first move, can always on his 12th move capture B; and if he
+has the second move, A can always on his 14th move make the capture.
+His point is always to get diagonally in line with his opponent, and
+by going to 33, if he has first move, he prevents B getting diagonally
+in line with himself. Here are two good games. The number in front of
+the hyphen is always A's move; that after the hyphen is B's:&mdash;</p>
+
+<p>33-8, 32-15, 31-22, 30-21, 29-14, 22-7, 15-6, 14-2, 7-3, 6-4, 11-, and
+A must capture on his next (12th) move, -13, 54-20, 53-27, 52-34,
+51-41, 50-34, 42-27, 35-20, 28-13, 21-6, 14-2, 7-3, 6-4, 11-, and A
+must capture on his next (14th) move.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_395_A_WAR_PUZZLE_GAMEa" id="X_395_A_WAR_PUZZLE_GAMEa"></a><a href="#X_395_A_WAR_PUZZLE_GAME"><b>395.&mdash;A WAR PUZZLE GAME.&mdash;<i>solution</i></b></a></p>
+
+<p>The Britisher can always catch the enemy, no matter how clever and
+elusive that astute individual may be; but curious though it may seem,
+the British general can only do so after he has paid a somewhat
+mysterious visit to the particular town marked "1" in the map, going
+in by 3 and leaving by 2, or entering by 2 and leaving by 3. The three
+towns that are shaded and have no numbers do not really come into the
+question, as some may suppose, for the simple reason that the
+Britisher never needs to enter any one of them, while the enemy cannot
+be forced to go into them, and would be clearly ill-advised to do so
+voluntarily. We may therefore leave these out of consideration
+altogether. No matter what the enemy may do, the Britisher should make
+the follow<span class='pagenum'>Pg 241<a name="Page_241" id="Page_241"></a></span>ing first nine moves: He should visit towns 24, 20, 19,
+15, 11, 7, 3, 1, 2. If the enemy takes it into his head also to go to
+town 1, it will be found that he will have to beat a precipitate
+retreat <i>the same way that he went in</i>, or the Britisher will
+infallibly catch him in towns 2 or 3, as the case may be. So the enemy
+will be wise to avoid that north-west corner of the map altogether.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a395.png" width="400" height="396" alt="" title="" />
+</div>
+
+<p>Now, when the British general has made the nine moves that I have
+given, the enemy will be, after his own ninth move, in one of the
+towns marked 5, 8, 11, 13, 14, 16, 19, 21, 24, or 27. Of course, if he
+imprudently goes to 3 or 6 at this point he will be caught at once.
+Wherever he may happen to be, the Britisher "goes for him," and has no
+longer any difficulty in catching him in eight more moves at most
+(seventeen in all) in one of the following ways. The Britisher will
+get to 8 when the enemy is at 5, and win next move; or he will get to
+19 when the enemy is at 22, and win next move; or he will get to 24
+when the enemy is at 27, and so win next move. It will be found that
+he can be forced into one or other of these fatal positions.</p>
+
+<p>In short, the strategy really amounts to this: the Britisher plays the
+first nine moves that I have given, and although the enemy does his
+very best to escape, our general goes after his antagonist and always
+driving him away from that north-west corner ultimately closes in with
+him, and wins. As I have said, the Britisher never need make more than
+seventeen moves in all, and may win in fewer moves if the enemy plays
+badly. But after playing those first nine moves it does not matter
+even if the Britisher makes a few bad ones. He may lose time, but
+cannot lose his advantage so long as he now keeps the enemy from town
+1, and must eventually catch him.</p>
+
+<p>This is a complete explanation of the puzzle. It may seem a little
+complex in print, but in practice the winning play will now be quite
+easy to the reader. Make those nine moves, and there ought to be no
+difficulty whatever in finding the concluding line of play. Indeed, it
+might almost be said that then it is difficult for the British general
+<i>not</i> to catch the enemy. It is a question of what in chess we call
+the "opposition," and the visit by the Britisher to town 1 "gives him
+the jump" on the enemy, as the man in the street would say.</p>
+
+<p>Here is an illustrative example in which the enemy avoids capture as
+long as it is possible for him to do so. The Britisher's moves are
+above the line and the enemy's below it. Play them alternately.</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right' class='bb'>24</td><td align='right' class='bb'>20</td><td align='right' class='bb'>19</td><td align='right' class='bb'>15</td><td align='right' class='bb'>11</td><td align='right' class='bb'>7</td><td align='right' class='bb'>3</td><td align='right' class='bb'>1</td><td align='right' class='bb'>2</td><td align='right' class='bb'>6</td><td align='right' class='bb'>10</td><td align='right' class='bb'>14</td><td align='right' class='bb'>18</td><td align='right' class='bb'>19</td><td align='right' class='bb'>20</td><td align='right' class='bb'>24</td></tr>
+<tr><td align='right'>13</td><td align='right'>9</td><td align='right'>13</td><td align='right'>17</td><td align='right'>21</td><td align='right'>20</td><td align='right'>24</td><td align='right'>23</td><td align='right'>19</td><td align='right'>15</td><td align='right'>19</td><td align='right'>23</td><td align='right'>24</td><td align='right'>25</td><td align='right'>27</td></tr>
+</table></div>
+
+
+<p>The enemy must now go to 25 or B, in either of which towns he is
+immediately captured.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_396_A_MATCH_MYSTERYa" id="X_396_A_MATCH_MYSTERYa"></a><a href="#X_396_A_MATCH_MYSTERY"><b>396.&mdash;A MATCH MYSTERY.&mdash;<i>solution</i></b></a></p>
+
+<p>If you form the three heaps (and are therefore the second to draw),
+any one of the following thirteen groupings will give you a win if you
+play correctly: 15, 14, 1; 15, 13, 2; 15, 12, 3; 15, 11, 4; 15, 10, 5;
+15, 9, 6; 15, 8, 7; 14, 13, 3; 14, 11, 5; 14, 9, 7; 13, 11, 6; 13, 10,
+7; 12, 11, 7.</p>
+
+<p>The beautiful general solution of this problem is as follows. Express
+the number in every heap in powers of 2, avoiding repetitions and
+remembering that 2<sup>0</sup>&nbsp;=&nbsp;1. Then if you so leave the matches to your
+opponent that there is an even number of every power, you can win. And
+if at the start you leave the powers even, you can always continue to
+do so throughout the game. Take, as example, the last grouping given
+above&mdash;12, 11, 7. Expressed in powers of 2 we have&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>12</td><td align='center'>=</td><td align='center'>8</td><td align='center'>4</td><td align='center'>-</td><td align='center'>-</td></tr>
+<tr><td align='right'>11</td><td align='center'>=</td><td align='center'>8</td><td align='center'>-</td><td align='center'>2</td><td align='center'>1</td></tr>
+<tr><td align='right'>7</td><td align='center'>=</td><td align='center'>-</td><td align='center'>4</td><td align='center'>2</td><td align='center'>1</td></tr>
+<tr><td align='right'></td><td></td><td align='center' class='bt'>2</td><td align='center' class='bt'>2</td><td align='center' class='bt'>2</td><td align='center' class='bt'>2</td></tr>
+</table></div>
+
+<p>As there are thus two of every power, you must win. Say your opponent
+takes 7 from the 12 heap. He then leaves&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>5</td><td align='center'>=</td><td align='center'>-</td><td align='center'>4</td><td align='center'>-</td><td align='center'>1</td></tr>
+<tr><td align='right'>11</td><td align='center'>=</td><td align='center'>8</td><td align='center'>-</td><td align='center'>2</td><td align='center'>1</td></tr>
+<tr><td align='right'>7</td><td align='center'>=</td><td align='center'>-</td><td align='center'>4</td><td align='center'>2</td><td align='center'>1</td></tr>
+<tr><td align='right'></td><td></td><td align='center' class='bt'>1</td><td align='center' class='bt'>2</td><td align='center' class='bt'>2</td><td align='center' class='bt'>3</td></tr>
+</table></div>
+
+<p>Here the powers are not all even in number, but by taking 9 from the
+11 heap you immediately restore your winning position, thus&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>5</td><td align='center'>=</td><td align='center'>-</td><td align='center'>4</td><td align='center'>-</td><td align='center'>1</td></tr>
+<tr><td align='right'>2</td><td align='center'>=</td><td align='center'>-</td><td align='center'>-</td><td align='center'>2</td><td align='center'>-</td></tr>
+<tr><td align='right'>7</td><td align='center'>=</td><td align='center'>-</td><td align='center'>4</td><td align='center'>2</td><td align='center'>1</td></tr>
+<tr><td align='right'></td><td></td><td align='center' class='bt'>-</td><td align='center' class='bt'>2</td><td align='center' class='bt'>2</td><td align='center' class='bt'>2</td></tr>
+</table></div>
+
+<p>And so on to the end. This solution is quite <span class='pagenum'>Pg 242<a name="Page_242" id="Page_242"></a></span>general, and applies to
+any number of matches and any number of heaps. A correspondent informs
+me that this puzzle game was first propounded by Mr. W.M.F. Mellor,
+but when or where it was published I have not been able to ascertain.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_397_THE_MONTENEGRIN_DICE_GAMEa" id="X_397_THE_MONTENEGRIN_DICE_GAMEa"></a><a href="#X_397_THE_MONTENEGRIN_DICE_GAME"><b>397.&mdash;THE MONTENEGRIN DICE GAME.&mdash;<i>solution</i></b></a></p>
+
+<p>The players should select the pairs 5 and 9, and 13 and 15, if the
+chances of winning are to be quite equal. There are 216 different ways
+in which the three dice may fall. They may add up 5 in 6 different
+ways and 9 in 25 different ways, making 31 chances out of 216 for the
+player who selects these numbers. Also the dice may add up 13 in 21
+different ways, and 15 in 10 different ways, thus giving the other
+player also 31 chances in 216.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_398_THE_CIGAR_PUZZLEa" id="X_398_THE_CIGAR_PUZZLEa"></a><a href="#X_398_THE_CIGAR_PUZZLE"><b>398.&mdash;THE CIGAR PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>Not a single member of the club mastered this puzzle, and yet I shall
+show that it is so simple that the merest child can understand its
+solution&mdash;when it is pointed out to him! The large majority of my
+friends expressed their entire bewilderment. Many considered that "the
+theoretical result, in any case, is determined by the relationship
+between the table and the cigars;" others, regarding it as a problem
+in the theory of Probabilities, arrived at the conclusion that the
+chances are slightly in favour of the first or second player, as the
+case may be. One man took a table and a cigar of particular
+dimensions, divided the table into equal sections, and proceeded to
+make the two players fill up these sections so that the second player
+should win. But why should the first player be so accommodating? At
+any stage he has only to throw down a cigar obliquely across several
+of these sections entirely to upset Mr. 2's calculations! We have to
+assume that each player plays the best possible; not that one
+accommodates the other.</p>
+
+<p>The theories of some other friends would be quite sound if the shape
+of the cigar were that of a torpedo&mdash;perfectly symmetrical and pointed
+at both ends.</p>
+
+<p>I will show that the first player should infallibly win, if he always
+plays in the best possible manner. Examine carefully the following
+diagram, No. 1, and all will be clear.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a398.png" width="400" height="208" alt="" title="" />
+</div>
+
+<p>The first player must place his first cigar <i>on end</i> in the exact
+centre of the table, as indicated by the little circle. Now, whatever
+the second player may do throughout, the first player must always
+repeat it in an exactly diametrically opposite position. Thus, if the
+second player places a cigar at A, I put one at AA; he places one at
+B, I put one at BB; he places one at C, I put one at CC; he places one
+at D, I put one at DD; he places one at E, I put one at EE; and so on
+until no more cigars can be placed without touching. As the cigars are
+supposed to be exactly alike in every respect, it is perfectly clear
+that for every move that the second player may choose to make, it is
+possible exactly to repeat it on a line drawn through the centre of
+the table. The second player can always duplicate the first player's
+move, no matter where he may place a cigar, or whether he places it on
+end or on its side. As the cigars are all alike in every respect, one
+will obviously balance over the edge of the table at precisely the
+same point as another. Of course, as each player is supposed to play
+in the best possible manner, it becomes a matter of theory. It is no
+valid objection to say that in actual practice one would not be
+sufficiently exact to be sure of winning. If as the first player you
+did not win, it would be in consequence of your <i>not</i> having played
+the best possible.</p>
+
+<p>The second diagram will serve to show why the first cigar must be
+placed on end. (And here I will say that the first cigar that I
+selected from a box I was able so to stand on end, and I am allowed to
+assume that all the other cigars would do the same.) If the first
+cigar were placed on its side, as at F, then the second player could
+place a cigar as at G&mdash;as near as possible, but not actually touching
+F. Now, in this position you cannot repeat his play on the opposite
+side, because the two ends of the cigar are not alike. It will be seen
+that GG, when placed on the opposite side in the same relation to the
+centre, intersects, or lies on top of, F, whereas the cigars are not
+allowed to touch. You must therefore put the cigar farther away from
+the centre, which would result in your having insufficient room
+between the centre and the bottom left-hand corner to repeat
+everything that the other player would do between G and the top
+right-hand corner. Therefore the result would not be a certain win for
+the first player.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_399_THE_TROUBLESOME_EIGHTa" id="X_399_THE_TROUBLESOME_EIGHTa"></a><a href="#X_399_THE_TROUBLESOME_EIGHT"><b>399.&mdash;THE TROUBLESOME EIGHT.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a399.png" width="400" height="407" alt="" title="" />
+</div>
+
+<p>The conditions were to place a different number in each of the nine
+cells so that the three rows, <span class='pagenum'>Pg 243<a name="Page_243" id="Page_243"></a></span>three columns, and two diagonals should
+each add up 15. Probably the reader at first set himself an impossible
+task through reading into these conditions something which is not
+there&mdash;a common error in puzzle-solving. If I had said "a different
+figure," instead of "a different number," it would have been quite
+impossible with the 8 placed anywhere but in a corner. And it would
+have been equally impossible if I had said "a different whole number."
+But a number may, of course, be fractional, and therein lies the
+secret of the puzzle. The arrangement shown in the figure will be
+found to comply exactly with the conditions: all the numbers are
+different, and the square adds up 15 in all the required eight ways.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_400_THE_MAGIC_STRIPSa" id="X_400_THE_MAGIC_STRIPSa"></a><a href="#X_400_THE_MAGIC_STRIPS"><b>400.&mdash;THE MAGIC STRIPS.&mdash;<i>solution</i></b></a></p>
+
+<p>There are of course six different places between the seven figures in
+which a cut may be made, and the secret lies in keeping one strip
+intact and cutting each of the other six in a different place. After
+the cuts have been made there are a large number of ways in which the
+thirteen pieces may be placed together so as to form a magic square.
+Here is one of them:&mdash;</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a400.png" width="400" height="387" alt="" title="" />
+</div>
+
+<p>The arrangement has some rather interesting features. It will be seen
+that the uncut strip is at the top, but it will be found that if the
+bottom row of figures be placed at the top the numbers will still form
+a magic square, and that every successive removal from the bottom to
+the top (carrying the uncut strip stage by stage to the bottom) will
+produce the same result. If we imagine the numbers to be on seven
+complete <i>perpendicular</i> strips, it will be found that these columns
+could also be moved in succession from left to right or from right to
+left, each time producing a magic square.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_401_EIGHT_JOLLY_GAOL_BIRDSa" id="X_401_EIGHT_JOLLY_GAOL_BIRDSa"></a><a href="#X_401_EIGHT_JOLLY_GAOL_BIRDS"><b>401.&mdash;EIGHT JOLLY GAOL BIRDS.&mdash;<i>solution</i></b></a></p>
+
+<p>There are eight ways of forming the magic square&mdash;all merely different
+aspects of one fundamental arrangement. Thus, if you give our first
+square a quarter turn you will get the second square; and as the four
+sides may be in turn brought to the top, there are four aspects. These
+four in turn reflected in a mirror produce the remaining four aspects.
+Now, of these eight arrangements only four can possibly be reached
+under the conditions, and only two of these four can be reached in the
+fewest possible moves, which is nineteen. These two arrangements are
+shown. Move the men in the following order: 5, 3, 2, 5, 7, 6, 4, 1, 5,
+7, 6, 4, 1, 6, 4, 8, 3, 2, 7, and you get the first square. Move them
+thus: 4, 1, 2, 4, 1, 6, 7, 1, 5, 8, 1, 5, 6, 7, 5, 6, 4, 2, 7, and you
+have the arrangement in the second square. In the first case every man
+has moved, but in the second case the man numbered 3 has never left
+his cell. Therefore No. 3 must be the obstinate prisoner, and the
+second square must be the required arrangement.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a401.png" width="400" height="176" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_402_NINE_JOLLY_GAOL_BIRDSa" id="X_402_NINE_JOLLY_GAOL_BIRDSa"></a><a href="#X_402_NINE_JOLLY_GAOL_BIRDS"><b>402.&mdash;NINE JOLLY GAOL BIRDS.&mdash;<i>solution</i></b></a></p>
+
+<p>There is a pitfall set for the unwary in this little puzzle. At the
+start one man is allowed to be placed on the shoulders of another, so
+as to give always one empty cell to enable the prisoners to move about
+without any two ever being in a cell together. The two united
+prisoners are allowed to add their numbers together, and are, of
+course, permitted to remain together at the completion of the magic
+square. But they are obviously not compelled so to remain together,
+provided that one of the pair on his final move does not break the
+condition of entering a cell already occupied. After the acute solver
+has noticed this point, it is for him to determine which method is the
+better one&mdash;for the two to be together at the count or to separate. As
+a matter of fact, the puzzle can be solved in seventeen moves if the
+men are to remain together; but if they separate at the<span class='pagenum'>Pg 244<a name="Page_244" id="Page_244"></a></span>
+end, they may actually save a move and perform the feat in sixteen!
+The trick consists in placing the man in the centre on the back of one
+of the corner men, and then working the pair into the centre before
+their final separation.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a402.png" width="400" height="227" alt="" title="" />
+</div>
+
+<p>Here are the moves for getting the men into one or other of the above
+two positions. The numbers are those of the men in the order in which
+they move into the cell that is for the time being vacant. The pair is
+shown in brackets:&mdash;</p>
+
+<p>Place 5 on 1. Then, 6, 9, 8, 6, 4, (6), 2, 4, 9, 3, 4, 9, (6), 7, 6,
+1.</p>
+
+<p>Place 5 on 9. Then, 4, 1, 2, 4, 6, (14), 8, 6, 1, 7, 6, 1, (14), 3, 4,
+9.</p>
+
+<p>Place 5 on 3. Then, 6, (8), 2, 6, 4, 7, 8, 4, 7, 1, 6, 7, (8), 9, 4,
+3.</p>
+
+<p>Place 5 on 7. Then, 4, (12), 8, 4, 6, 3, 2, 6, 3, 9, 4, 3, (12), 1, 6,
+7.</p>
+
+<p>The first and second solutions produce Diagram A; the second and third
+produce Diagram B. There are only sixteen moves in every case. Having
+found the fewest moves, we had to consider how we were to make the
+burdened man do as little work as possible. It will at once be seen
+that as the pair have to go into the centre before separating they
+must take at fewest two moves. The labour of the burdened man can only
+be reduced by adopting the other method of solution, which, however,
+forces us to take another move.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_403_THE_SPANISH_DUNGEONa" id="X_403_THE_SPANISH_DUNGEONa"></a><a href="#X_403_THE_SPANISH_DUNGEON"><b>403.&mdash;THE SPANISH DUNGEON.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a403.png" width="400" height="183" alt="" title="" />
+</div>
+
+<p>This can best be solved by working backwards&mdash;that is to say, you must
+first catch your square, and then work back to the original position.
+We must first construct those squares which are found to require the
+least amount of readjustment of the numbers. Many of these we know
+cannot possibly be reached. When we have before us the most favourable
+possible arrangements, it then becomes a question of careful analysis
+to discover which position can be reached in the fewest moves. I am
+afraid, however, it is only after considerable study and experience
+that the solver is able to get such a grasp of the various "areas of
+disturbance" and methods of circulation that his judgment is of much
+value to him.</p>
+
+<p>The second diagram is a most favourable magic square position. It will
+be seen that prisoners 4, 8, 13, and 14 are left in their original
+cells. This position may be reached in as few as thirty-seven moves.
+Here are the moves: 15, 14, 10, 6, 7, 3, 2, 7, 6, 11, 3, 2, 7, 6, 11,
+10, 14, 3, 2, 11, 10, 9, 5, 1, 6, 10, 9, 5, 1, 6, 10, 9, 5, 2, 12, 15,
+3. This short solution will probably surprise many readers who may not
+find a way under from sixty to a hundred moves. The clever prisoner
+was No. 6, who in the original illustration will be seen with his arms
+extended calling out the moves. He and No. 10 did most of the work,
+each changing his cell five times. No. 12, the man with the crooked
+leg, was lame, and therefore fortunately had only to pass from his
+cell into the next one when his time came round.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_404_THE_SIBERIAN_DUNGEONSa" id="X_404_THE_SIBERIAN_DUNGEONSa"></a><a href="#X_404_THE_SIBERIAN_DUNGEONS"><b>404.&mdash;THE SIBERIAN DUNGEONS.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a404.png" width="400" height="389" alt="" title="" />
+</div>
+
+<p>In attempting to solve this puzzle it is clearly necessary to seek
+such magic squares as seem the most favourable for our purpose, and
+then carefully examine and try them for "fewest moves." Of course it
+at once occurs to us that if we can adopt a square in which a certain
+number of men need not leave their original cells, we may save moves
+on the one hand, but we may obstruct our movements on the other. For
+example, a magic square may be formed with the 6, 7, 13, and 16
+unmoved; but in such case it is obvious that a solution is impossible,
+since cells 14 and 15 can neither be left nor entered without breaking
+the condition of no two men ever being in the same cell together.</p>
+
+<p>The following solution in fourteen moves was found by Mr. G.
+Wotherspoon: 8-17, 16-21, 6-16, 14-8, 5-18, 4-14, 3-24, 11-20, 10-19,
+2-23, 13-22, 12-6, 1-5, 9-13. As this solution is in what I consider
+the theoretical minimum number of moves, I am confident that it cannot
+be improved upon, and on this point Mr. Wotherspoon is of the same
+opinion.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_405_CARD_MAGIC_SQUARESa" id="X_405_CARD_MAGIC_SQUARESa"></a><a href="#X_405_CARD_MAGIC_SQUARES"><b>405.&mdash;CARD MAGIC SQUARES.&mdash;<i>solution</i></b></a></p>
+
+<p>Arrange the cards as follows for the three new squares:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>3</td><td align='right'>2</td><td align='right'>4</td></tr>
+<tr><td align='right'>4</td><td align='right'>3</td><td align='right'>2</td></tr>
+<tr><td align='right'>2</td><td align='right'>4</td><td align='right'>3</td></tr>
+</table></div>
+<p><br /></p>
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>6</td><td align='right'>5</td><td align='right'>7</td></tr>
+<tr><td align='right'>7</td><td align='right'>6</td><td align='right'>5</td></tr>
+<tr><td align='right'>5</td><td align='right'>7</td><td align='right'>6</td></tr>
+</table></div>
+<p><br /></p>
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>9</td><td align='right'>8</td><td align='right'>10</td></tr>
+<tr><td align='right'>10</td><td align='right'>9</td><td align='right'>8</td></tr>
+<tr><td align='right'>8</td><td align='right'>10</td><td align='right'>9</td></tr>
+</table></div>
+
+<p>Three aces and one ten are not used. The summations of the four
+squares are thus: 9, 15, 18, and 27&mdash;all different, as required.</p>
+
+<hr style="width: 30%;" />
+<p><span class='pagenum'>Pg 245<a name="Page_245" id="Page_245"></a></span><a name="X_406_THE_EIGHTEEN_DOMINOESa" id="X_406_THE_EIGHTEEN_DOMINOESa"></a><a href="#X_406_THE_EIGHTEEN_DOMINOES"><b>406.&mdash;THE EIGHTEEN DOMINOES.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a406.png" width="400" height="401" alt="" title="" />
+</div>
+
+<p>The illustration explains itself. It will be found that the pips in
+every column, row, and long diagonal add up 18, as required.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_407_TWO_NEW_MAGIC_SQUARESa" id="X_407_TWO_NEW_MAGIC_SQUARESa"></a><a href="#X_407_TWO_NEW_MAGIC_SQUARES"><b>407.&mdash;TWO NEW MAGIC SQUARES.&mdash;<i>solution</i></b></a></p>
+
+<p>Here are two solutions that fulfil the conditions:&mdash;</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a407.png" width="400" height="208" alt="" title="" />
+</div>
+
+<p>The first, by subtracting, has a constant 8, and the associated pairs
+all have a difference of 4. The second square, by dividing, has a
+constant 9, and all the associated pairs produce 3 by division. These
+are two remarkable and instructive squares.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_408_MAGIC_SQUARES_OF_TWO_DEGREESa" id="X_408_MAGIC_SQUARES_OF_TWO_DEGREESa"></a><a href="#X_408_MAGIC_SQUARES_OF_TWO_DEGREES"><b>408.&mdash;MAGIC SQUARES OF TWO DEGREES.&mdash;<i>solution</i></b></a></p>
+
+<p>The following is the square that I constructed. As it stands the
+constant is 260. If for every number you substitute, in its allotted
+place, its square, then the constant will be 11,180. Readers can write
+out for themselves the second degree square.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a408.png" width="400" height="398" alt="" title="" />
+</div>
+
+<p>The main key to the solution is the pretty law that if eight numbers
+sum to 260 and their squares to 11,180, then the same will happen in
+the case of the eight numbers that are complementary to 65. Thus 1 +
+18&nbsp;+&nbsp;23&nbsp;+&nbsp;26&nbsp;+&nbsp;31&nbsp;+&nbsp;48&nbsp;+&nbsp;56&nbsp;+&nbsp;57&nbsp;=&nbsp;260, and the sum of their squares
+is 11,180. Therefore 64&nbsp;+&nbsp;47&nbsp;+&nbsp;42&nbsp;+&nbsp;39&nbsp;+&nbsp;34&nbsp;+&nbsp;17&nbsp;+&nbsp;9&nbsp;+&nbsp;8 (obtained by
+subtracting each of the above numbers from 65) will sum to 260 and
+their squares to 11,180. Note that in every one of the sixteen smaller
+squares the two diagonals sum to 65. There are four columns and four
+rows with their complementary columns and rows. Let us pick out the
+numbers found in the 2nd, 1st, 4th, and 3rd rows and arrange them thus
+:&mdash;</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='right'>1</td><td align='right'>8</td><td align='right'>28</td><td align='right'>29</td><td align='right'>42</td><td align='right'>47</td><td align='right'>51</td><td align='right'>54</td></tr>
+<tr><td align='right'>2</td><td align='right'>7</td><td align='right'>27</td><td align='right'>30</td><td align='right'>41</td><td align='right'>48</td><td align='right'>52</td><td align='right'>53</td></tr>
+<tr><td align='right'>3</td><td align='right'>6</td><td align='right'>26</td><td align='right'>31</td><td align='right'>44</td><td align='right'>45</td><td align='right'>49</td><td align='right'>56</td></tr>
+<tr><td align='right'>4</td><td align='right'>5</td><td align='right'>25</td><td align='right'>32</td><td align='right'>43</td><td align='right'>46</td><td align='right'>50</td><td align='right'>55</td></tr>
+</table></div>
+
+<p>Here each column contains four consecutive numbers cyclically
+arranged, four running in one direction and four in the other. The
+numbers in the 2nd, 5th, 3rd, and 8th columns of the square may be
+similarly grouped. The great difficulty lies in discovering the
+conditions governing these groups of numbers, the pairing of the
+complementaries in the squares of four and the formation of the
+diagonals. But when a correct solution is shown, as above, it
+discloses all the more important keys to the mystery. I am inclined to
+think this square of two degrees the most elegant thing that exists in
+magics. I believe such a magic square cannot be constructed in the
+case of any order lower than 8.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_409_THE_BASKETS_OF_PLUMSa" id="X_409_THE_BASKETS_OF_PLUMSa"></a><a href="#X_409_THE_BASKETS_OF_PLUMS"><b>409.&mdash;THE BASKETS OF PLUMS.&mdash;<i>solution</i></b></a></p>
+
+<p>As the merchant told his man to distribute the contents of one of the
+baskets of plums "among some children," it would not be permissible to
+give the complete basketful to one child; and as it was also directed
+that the man was to give "plums to every child, so that each should
+receive an equal number," it would also not be allowed to select just
+as many children as there were plums in a basket and give each child a
+single plum. Consequently, if the number of <span class='pagenum'>Pg 246<a name="Page_246" id="Page_246"></a></span>plums in every basket was
+a prime number, then the man would be correct in saying that the
+proposed distribution was quite impossible. Our puzzle, therefore,
+resolves itself into forming a magic square with nine different prime
+numbers.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a409.png" width="400" height="428" alt="" title="" />
+</div>
+
+<p>In Diagram A we have a magic square in prime numbers, and it is the
+one giving the smallest constant sum that is possible. As to the
+little trap I mentioned, it is clear that Diagram A is barred out by
+the words "every basket contained plums," for one plum is not plums.
+And as we were referred to the baskets, "as shown in the
+illustration," it is perfectly evident, without actually attempting to
+count the plums, that there are at any rate more than 7 plums in every
+basket. Therefore C is also, strictly speaking, barred. Numbers over
+20 and under, say, 250 would certainly come well within the range of
+possibility, and a large number of arrangements would come within
+these limits. Diagram B is one of them. Of course we can allow for the
+false bottoms that are so frequently used in the baskets of
+fruitsellers to make the basket appear to contain more fruit than it
+really does.</p>
+
+<p>Several correspondents assumed (on what grounds I cannot think) that
+in the case of this problem the numbers cannot be in consecutive
+arithmetical progression, so I give Diagram D to show that they were
+mistaken. The numbers are 199, 409, 619, 829, 1,039, 1,249, 1,459,
+1,669, and 1,879&mdash;all primes with a common difference of 210.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_410_THE_MANDARINS_T_PUZZLEa" id="X_410_THE_MANDARINS_T_PUZZLEa"></a><a href="#X_410_THE_MANDARINS_T_PUZZLE"><b>410.&mdash;THE MANDARIN'S "T" PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>There are many different ways of arranging the numbers, and either the
+2 or the 3 may be omitted from the "T" enclosure. The arrangement that
+I give is a "nasik" square. Out of the total of 28,800 nasik squares
+of the fifth order this is the only one (with its one reflection) that
+fulfils the "T" condition. This puzzle was suggested to me by Dr. C.
+Planck.</p>
+
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a410.png" width="400" height="400" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_411_A_MAGIC_SQUARE_OF_COMPOSITESa" id="X_411_A_MAGIC_SQUARE_OF_COMPOSITESa"></a><a href="#X_411_A_MAGIC_SQUARE_OF_COMPOSITES"><b>411.&mdash;A MAGIC SQUARE OF COMPOSITES.&mdash;<i>solution</i></b></a></p>
+
+<p>The problem really amounts to finding the smallest prime such that the
+next higher prime shall exceed it by 10 at least. If we write out a
+little list of primes, we shall not need to exceed 150 to discover
+what we require, for after 113 the next prime is 127. We can then form
+the square in the diagram, where every number is composite. This is
+the solution in the smallest numbers. We thus see that the answer is
+arrived at quite easily, in a square of the third order, by trial. But
+I propose to show how we may get an answer (not, it is true, the one
+in smallest numbers) without any tables or trials, but in a very
+direct and rapid manner.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a411.png" width="400" height="404" alt="" title="" />
+</div>
+
+<p>First write down any consecutive numbers, the smallest being greater
+than 1&mdash;say, 2, 3, 4, 5, 6, 7, 8, 9, 10. The only factors in these
+numbers are 2, 3, 5, and 7. We therefore mul<span class='pagenum'>Pg 247<a name="Page_247" id="Page_247"></a></span>tiply these four numbers
+together and add the product, 210, to each of the nine numbers. The
+result is the nine consecutive composite numbers, 212 to 220
+inclusive, with which we can form the required square. Every number
+will necessarily be divisible by its difference from 210. It will be
+very obvious that by this method we may find as many consecutive
+composites as ever we please. Suppose, for example, we wish to form a
+magic square of sixteen such numbers; then the numbers 2 to 17 contain
+the factors 2, 3, 5, 7, 11, 13, and 17, which, multiplied together,
+make 510510 to be added to produce the sixteen numbers 510512 to
+510527 inclusive, all of which are composite as before.</p>
+
+<p>But, as I have said, these are not the answers in the smallest
+numbers: for if we add 523 to the numbers 1 to 16, we get sixteen
+consecutive composites; and if we add 1,327 to the numbers 1 to 25, we
+get twenty-five consecutive composites, in each case the smallest
+numbers possible. Yet if we required to form a magic square of a
+hundred such numbers, we should find it a big task by means of tables,
+though by the process I have shown it is quite a simple matter. Even
+to find thirty-six such numbers you will search the tables up to
+10,000 without success, and the difficulty increases in an
+accelerating ratio with each square of a larger order.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_412_THE_MAGIC_KNIGHTS_TOURa" id="X_412_THE_MAGIC_KNIGHTS_TOURa"></a><a href="#X_412_THE_MAGIC_KNIGHTS_TOUR"><b>412.&mdash;THE MAGIC KNIGHT'S TOUR.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a412.png" width="400" height="394" alt="" title="" />
+</div>
+
+<p>Here each successive number (in numerical order) is a knight's move
+from the preceding number, and as 64 is a knight's move from 1, the
+tour is "re-entrant." All the columns and rows add up 260.
+Unfortunately, it is not a perfect magic square, because the diagonals
+are incorrect, one adding up 264 and the other 256&mdash;requiring only the
+transfer of 4 from one diagonal to the other. I think this is the best
+result that has ever been obtained (either re-entrant or not), and
+nobody can yet say whether a perfect solution is possible or
+impossible.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_413_A_CHESSBOARD_FALLACYa" id="X_413_A_CHESSBOARD_FALLACYa"></a><a href="#X_413_A_CHESSBOARD_FALLACY"><b>413.&mdash;A CHESSBOARD FALLACY.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a413.png" width="400" height="240" alt="" title="" />
+</div>
+
+<p>The explanation of this little fallacy is as follows. The error lies
+in assuming that the little triangular piece, marked C, is exactly the
+same height as one of the little squares of the board. As a matter of
+fact, its height (if we make the sixty-four squares each a square
+inch) will be 1<sup>1</sup>/<sub>7</sub> in. Consequently the rectangle is really 9<sup>1</sup>/<sub>7</sub> in.
+by 7 in., so that the area is sixty-four square inches in either case.
+Now, although the pieces do fit together exactly to form the perfect
+rectangle, yet the directions of the horizontal lines in the pieces
+will not coincide. The new diagram above will make everything quite
+clear to the reader.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_414_WHO_WAS_FIRSTa" id="X_414_WHO_WAS_FIRSTa"></a><a href="#X_414_WHO_WAS_FIRST"><b>414.&mdash;WHO WAS FIRST?&mdash;<i>solution</i></b></a></p>
+
+<p>Biggs, who saw the smoke, would be first; Carpenter, who saw the
+bullet strike the water, would be second; and Anderson, who heard the
+report, would be last of all.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_415_A_WONDERFUL_VILLAGEa" id="X_415_A_WONDERFUL_VILLAGEa"></a><a href="#X_415_A_WONDERFUL_VILLAGE"><b>415.&mdash;A WONDERFUL VILLAGE.&mdash;<i>solution</i></b></a></p>
+
+<p>When the sun is in the horizon of any place (whether in Japan or
+elsewhere), he is the length of half the earth's diameter more distant
+from that place than in his meridian at noon. As the earth's
+semi-diameter is nearly 4,000 miles, the sun must be considerably more
+than 3,000 miles nearer at noon than at his rising, there being no
+valley even the hundredth part of 1,000 miles deep.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_416_A_CALENDAR_PUZZLEa" id="X_416_A_CALENDAR_PUZZLEa"></a><a href="#X_416_A_CALENDAR_PUZZLE"><b>416.&mdash;A CALENDAR PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The first day of a century can never fall on a Sunday; nor on a
+Wednesday or a Friday.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_417_THE_TIRING_IRONSa" id="X_417_THE_TIRING_IRONSa"></a><a href="#X_417_THE_TIRING_IRONS"><b>417.&mdash;THE TIRING IRONS.&mdash;<i>solution</i></b></a></p>
+
+<p>I will give my complete working of the solution, so that readers may
+see how easy it is when you know how to proceed. And first of all, as
+there is an even number of rings, I will say that they may all be
+taken off in one-third of (2<sup>(n&nbsp;+&nbsp;1)</sup>&nbsp;-&nbsp;2) moves; and since <i>n</i> in our
+case is 14, all the rings may be taken off in 10,922 moves. Then I say
+10,922&nbsp;-&nbsp;9,999&nbsp;=&nbsp;923, and proceed to find the position when only 923
+out of the 10,922 moves remain to be made. Here is the curious method
+of doing this. It is based on <span class='pagenum'>Pg 248<a name="Page_248" id="Page_248"></a></span>the binary scale method used by
+Monsieur L. Gros, for an account of which see W.W. Rouse Ball's
+<i>Mathematical Recreations</i>.</p>
+
+<p>Divide 923 by 2, and we get 461 and the remainder 1; divide 461 by 2,
+and we get 230 and the remainder 1; divide 230 by 2, and we get 115
+and the remainder nought. Keep on dividing by 2 in this way as long as
+possible, and all the remainders will be found to be 1, 1, 1, 0, 0, 1,
+1, 0, 1, 1, the last remainder being to the left and the first
+remainder to the right. As there are fourteen rings and only ten
+figures, we place the difference, in the form of four noughts, in
+brackets to the left, and bracket all those figures that repeat a
+figure on their left. Then we get the following arrangement: (0 0 0 0) 1
+(1 1) 0 (0) 1 (1) 0 1 (1). This is the correct answer to the puzzle, for
+if we now place rings below the line to represent the figures in
+brackets and rings on the line for the other figures, we get the
+solution in the required form, as below:&mdash;</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a417.png" width="400" height="65" alt="" title="" />
+</div>
+
+<p>This is the exact position of the rings after the 9,999th move has
+been made, and the reader will find that the method shown will solve
+any similar question, no matter how many rings are on the
+tiring-irons. But in working the inverse process, where you are
+required to ascertain the number of moves necessary in order to reach
+a given position of the rings, the rule will require a little
+modification, because it does not necessarily follow that the position
+is one that is actually reached in course of taking off all the rings
+on the irons, as the reader will presently see. I will here state that
+where the total number of rings is odd the number of moves required to
+take them all off is one-third of (2<sup>(<i>n</i>&nbsp;+&nbsp;1)</sup>&nbsp;-&nbsp;1).</p>
+
+<p>With <i>n</i> rings (where <i>n</i> is <i>odd</i>) there are 2<sup><i>n</i></sup> positions counting
+all on and all off. In <sup>1</sup>/<sub>3</sub> (2<sup>(n+1)</sup>&nbsp;+&nbsp;2) positions they are all
+removed. The number of positions not used is (<sup>1</sup>/<sub>3</sub>)(2<sup><i>n</i></sup>&nbsp;-&nbsp;2).</p>
+
+<p>With <i>n</i> rings (where <i>n</i> is <i>even</i>) there are 2<sup>n</sup> positions counting
+all on and all off. In (2<sup>(<i>n</i>&nbsp;+&nbsp;1)</sup>&nbsp;+&nbsp;1) positions they are all removed.
+The number of positions not used is here (<sup>1</sup>/<sub>3</sub>)(2<sup><i>n</i></sup>&nbsp;-&nbsp;1).</p>
+
+<p>It will be convenient to tabulate a few cases.</p>
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='center' class='bb'>No. of <br />Rings.</td><td align='center' class='bb'>Total <br />Positions.</td><td align='center' class='bb'>Positions <br />used.</td><td align='center' class='bb'>Positions <br />not used.</td></tr>
+<tr><td align='right'>1</td><td align='right'>2</td><td align='right'>2</td><td align='right'>0</td></tr>
+<tr><td align='right'>3</td><td align='right'>8</td><td align='right'>6</td><td align='right'>2</td></tr>
+<tr><td align='right'>5</td><td align='right'>32</td><td align='right'>22</td><td align='right'>10</td></tr>
+<tr><td align='right'>7</td><td align='right'>128</td><td align='right'>86</td><td align='right'>42</td></tr>
+<tr><td align='right'>9</td><td align='right'>512</td><td align='right'>342</td><td align='right'>170</td></tr>
+<tr><td>&nbsp;</td></tr>
+<tr><td align='right'>2</td><td align='right'>4</td><td align='right'>3</td><td align='right'>1</td></tr>
+<tr><td align='right'>4</td><td align='right'>16</td><td align='right'>11</td><td align='right'>5</td></tr>
+<tr><td align='right'>6</td><td align='right'>64</td><td align='right'>43</td><td align='right'>21</td></tr>
+<tr><td align='right'>8</td><td align='right'>256</td><td align='right'>171</td><td align='right'>85</td></tr>
+<tr><td align='right'>10</td><td align='right'>1024</td><td align='right'>683</td><td align='right'>341</td></tr>
+</table></div>
+
+<p>Note first that the number of <i>positions used</i> is one more than the
+number of <i>moves</i> required to take all the rings off, because we are
+including "all on" which is a position but not a move. Then note that
+the number of <i>positions not used</i> is the same as the number of <i>moves
+used</i> to take off a set that has one ring fewer. For example, it takes
+85 moves to remove 7 rings, and the 42 positions not used are exactly
+the number of moves required to take off a set of 6 rings. The fact is
+that if there are 7 rings and you take off the first 6, and then wish
+to remove the 7th ring, there is no course open to you but to reverse
+all those 42 moves that never ought to have been made. In other words,
+you must replace all the 7 rings on the loop and start afresh! You
+ought first to have taken off 5 rings, to do which you should have
+taken off 3 rings, and previously to that 1 ring. To take off 6 you
+first remove 2 and then 4 rings.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_418_SUCH_A_GETTING_UPSTAIRSa" id="X_418_SUCH_A_GETTING_UPSTAIRSa"></a><a href="#X_418_SUCH_A_GETTING_UPSTAIRS"><b>418.&mdash;SUCH A GETTING UPSTAIRS.&mdash;<i>solution</i></b></a></p>
+
+<p>Number the treads in regular order upwards, 1 to 8. Then proceed as
+follows: 1 (step back to floor), 1, 2, 3 (2), 3, 4, 5 (4), 5, 6, 7
+(6), 7, 8, landing (8), landing. The steps in brackets are taken in a
+backward direction. It will thus be seen that by returning to the
+floor after the first step, and then always going three steps forward
+for one step backward, we perform the required feat in nineteen steps.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_419_THE_FIVE_PENNIESa" id="X_419_THE_FIVE_PENNIESa"></a><a href="#X_419_THE_FIVE_PENNIES"><b>419.&mdash;THE FIVE PENNIES.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a419.png" width="400" height="349" alt="" title="" />
+</div>
+
+<p>First lay three of the pennies in the way shown in Fig. 1. Now hold
+the remaining two pennies in the position shown in Fig. 2, so that
+they touch one another at the top, and at the base are in contact with
+the three horizontally placed coins. Then the five pennies will be
+equidistant, for every penny will touch every other penny.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_420_THE_INDUSTRIOUS_BOOKWORMa" id="X_420_THE_INDUSTRIOUS_BOOKWORMa"></a><a href="#X_420_THE_INDUSTRIOUS_BOOKWORM"><b>420.&mdash;THE INDUSTRIOUS BOOKWORM.&mdash;<i>solution</i></b></a></p>
+
+<p>The hasty reader will assume that the bookworm, in boring from the
+first to the last page <span class='pagenum'>Pg 249<a name="Page_249" id="Page_249"></a></span>of a book in three volumes, standing in their
+proper order on the shelves, has to go through all three volumes and
+four covers. This, in our case, would mean a distance of 9� in.,
+which is a long way from the correct answer. You will find, on
+examining any three consecutive volumes on your shelves, that the
+first page of Vol. I. and the last page of Vol. III. are actually the
+pages that are nearest to Vol. II., so that the worm would only have
+to penetrate four covers (together, � in.) and the leaves in the
+second volume (3 in.), or a distance of 3� inches, in order to
+tunnel from the first page to the last.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_421_A_CHAIN_PUZZLEa" id="X_421_A_CHAIN_PUZZLEa"></a><a href="#X_421_A_CHAIN_PUZZLE"><b>421.&mdash;A CHAIN PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>To open and rejoin a link costs threepence. Therefore to join the nine
+pieces into an endless chain would cost 2<i>s</i>. 3<i>d</i>., whereas a new chain
+would cost 2<i>s</i>. 2<i>d</i>. But if we break up the piece of eight links, these
+eight will join together the remaining eight pieces at a cost of 2<i>s</i>.
+But there is a subtle way of even improving on this. Break up the two
+pieces containing three and four links respectively, and these seven
+will join together the remaining seven pieces at a cost of only 1<i>s</i>.
+9<i>d</i>.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_422_THE_SABBATH_PUZZLEa" id="X_422_THE_SABBATH_PUZZLEa"></a><a href="#X_422_THE_SABBATH_PUZZLE"><b>422.&mdash;THE SABBATH PUZZLE.&mdash;<i>solution</i></b></a></p>
+
+<p>The way the author of the old poser proposed to solve the difficulty
+was as follows: From the Jew's abode let the Christian and the Turk
+set out on a tour round the globe, the Christian going due east and
+the Turk due west. Readers of Edgar Allan Poe's story, <i>Three Sundays
+in a Week</i>, or of Jules Verne's <i>Round the World in Eighty Days</i>, will
+know that such a proceeding will result in the Christian's gaining a
+day and in the Turk's losing a day, so that when they meet again at
+the house of the Jew their reckoning will agree with his, and all
+three may keep their Sabbath on the same day. The correctness of this
+answer, of course, depends on the popular notion as to the definition
+of a day&mdash;the average duration between successive sun-rises. It is an
+old quibble, and quite sound enough for puzzle purposes. Strictly
+speaking, the two travellers ought to change their reckonings on
+passing the 180th meridian; otherwise we have to admit that at the
+North or South Pole there would only be one Sabbath in seven years.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_423_THE_RUBY_BROOCHa" id="X_423_THE_RUBY_BROOCHa"></a><a href="#X_423_THE_RUBY_BROOCH"><b>423.&mdash;THE RUBY BROOCH.&mdash;<i>solution</i></b></a></p>
+
+<p>In this case we were shown a sketch of the brooch exactly as it
+appeared after the four rubies had been stolen from it. The reader was
+asked to show the positions from which the stones "may have been
+taken;" for it is not possible to show precisely how the gems were
+originally placed, because there are many such ways. But an important
+point was the statement by Lady Littlewood's brother: "I know the
+brooch well. It originally contained forty-five stones, and there are
+now only forty-one. Somebody has stolen four rubies, and then reset as
+small a number as possible in such a way that there shall always be
+eight stones in any of the directions you have mentioned."</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a423.png" width="400" height="410" alt="" title="" />
+</div>
+
+<p>The diagram shows the arrangement before the robbery. It will be seen
+that it was only necessary to reset one ruby&mdash;the one in the centre.
+Any solution involving the resetting of more than one stone is not in
+accordance with the brother's statement, and must therefore be wrong.
+The original arrangement was, of course, a little unsymmetrical, and
+for this reason the brooch was described as "rather eccentric."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_424_THE_DOVETAILED_BLOCKa" id="X_424_THE_DOVETAILED_BLOCKa"></a><a href="#X_424_THE_DOVETAILED_BLOCK"><b>424.&mdash;THE DOVETAILED BLOCK.&mdash;<i>solution</i></b></a></p>
+
+<div class="figcenter" style="width: 300px;">
+<img src="images/a424.png" width="300" height="519" alt="" title="" />
+</div>
+
+<p>The mystery is made clear by the illustration. It will be seen at once
+how the two pieces slide together in a diagonal direction.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_425_JACK_AND_THE_BEANSTALKa" id="X_425_JACK_AND_THE_BEANSTALKa"></a><a href="#X_425_JACK_AND_THE_BEANSTALK"><b>425.&mdash;JACK AND THE BEANSTALK.&mdash;<i>solution</i></b></a></p>
+
+<p>The serious blunder that the artist made in this drawing was in
+depicting the tendrils of</p>
+
+<p><span class='pagenum'>Pg 250<a name="Page_250" id="Page_250"></a></span></p><div class="figcenter" style="width: 400px;">
+<img src="images/a425.png" width="400" height="402" alt="" title="" />
+</div>
+
+<p>the bean climbing spirally as at A above, whereas the French bean, or
+scarlet runner, the variety clearly selected by the artist in the
+absence of any authoritative information on the point, always climbs
+as shown at B. Very few seem to be aware of this curious little fact.
+Though the bean always insists on a sinistrorsal growth, as B, the hop
+prefers to climb in a dextrorsal manner, as A. Why, is one of the
+mysteries that Nature has not yet unfolded.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_426_THE_HYMN-BOARD_POSERa" id="X_426_THE_HYMN-BOARD_POSERa"></a><a href="#X_426_THE_HYMN-BOARD_POSER"><b>426.&mdash;THE HYMN-BOARD POSER.&mdash;<i>solution</i></b></a></p>
+
+<p>This puzzle is not nearly so easy as it looks at first sight. It was
+required to find the smallest possible number of plates that would be
+necessary to form a set for three hymn-boards, each of which would
+show the five hymns sung at any particular service, and then to
+discover the lowest possible cost for the same. The hymn-book contains
+700 hymns, and therefore no higher number than 700 could possibly be
+needed.</p>
+
+<p>Now, as we are required to use every legitimate and practical method
+of economy, it should at once occur to us that the plates must be
+painted on both sides; indeed, this is such a common practice in cases
+of this kind that it would readily occur to most solvers. We should
+also remember that some of the figures may possibly be reversed to
+form other figures; but as we were given a sketch of the actual shapes
+of these figures when painted on the plates, it would be seen that
+though the 6's may be turned upside down to make 9's, none of the
+other figures can be so treated.</p>
+
+<p>It will be found that in the case of the figures 1, 2, 3, 4, and 5,
+thirty-three of each will be required in order to provide for every
+possible emergency; in the case of 7, 8, and 0, we can only need
+thirty of each; while in the case of the figure 6 (which may be
+reversed for the figure 9) it is necessary to provide exactly
+forty-two.</p>
+
+<p>It is therefore clear that the total number of figures necessary is
+297; but as the figures are painted on both sides of the plates, only
+149 such plates are required. At first it would appear as if one of
+the plates need only have a number on one side, the other side being
+left blank. But here we come to a rather subtle point in the problem.</p>
+
+<p>Readers may have remarked that in real life it is sometimes cheaper
+when making a purchase to buy more articles than we require, on the
+principle of a reduction on taking a quantity: we get more articles
+and we pay less. Thus, if we want to buy ten apples, and the price
+asked is a penny each if bought singly, or ninepence a dozen, we
+should both save a penny and get two apples more than we wanted by
+buying the full twelve. In the same way, since there is a regular
+scale of reduction for plates painted alike, we actually save by
+having two figures painted on that odd plate. Supposing, for example,
+that we have thirty plates painted alike with 5 on one side and 6 on
+the other. The rate would be 4�<i>d</i>., and the cost 11<i>s</i>. 10�<i>d</i>. But
+if the odd plate with, say, only a 5 on one side of it have a 6
+painted on the other side, we get thirty-one plates at the reduced
+rate of 4�<i>d</i>., thus saving a farthing on each of the previous
+thirty, and reducing the cost of the last one from 1<i>s</i>. to 4�<i>d</i>.</p>
+
+<p>But even after these points are all seen there comes in a new
+difficulty: for although it will be found that all the 8's may be on
+the backs of the 7's, we cannot have all the 2's on the backs of the
+1's, nor all the 4 on the backs of the 3's, etc. There is a great
+danger, in our attempts to get as many as possible painted alike, of
+our so adjusting the figures that some particular combination of hymns
+cannot be represented.</p>
+
+<p>Here is the solution of the difficulty that was sent to the vicar of
+Chumpley St. Winifred. Where the sign  is placed between two figures,
+it implies that one of these figures is on one side of the plate and
+the other on the other side.</p>
+
+
+<div class='center'>
+<table border="0" cellpadding="4" cellspacing="0" summary="">
+<tr><td align='left'></td><td></td><td></td><td></td><td align='center'><i>d</i>.</td><td></td><td align='center'>&pound;</td><td align='center'><i>s</i>.</td><td align='right'><i>d</i>.</td></tr>
+<tr><td align='right'>31</td><td align='center'>plates painted</td><td align='center'>5 X 9</td><td align='center'>@</td><td align='right'>4�</td><td align='center'>=</td><td align='right'>0</td><td align='right'>11</td><td align='right'>7�</td></tr>
+<tr><td align='right'>30</td><td align='center'>"</td><td align='center'>7 X 8</td><td align='center'>@</td><td align='right'>4�</td><td align='center'>=</td><td align='right'>0</td><td align='right'>11</td><td align='right'>10�</td></tr>
+<tr><td align='right'>21</td><td align='center'>"</td><td align='center'>1 X 2</td><td align='center'>@</td><td align='right'>7</td><td align='center'>=</td><td align='right'>0</td><td align='right'>12</td><td align='right'>3</td></tr>
+<tr><td align='right'>21</td><td align='center'>"</td><td align='center'>3 X 0</td><td align='center'>@</td><td align='right'>7</td><td align='center'>=</td><td align='right'>0</td><td align='right'>12</td><td align='right'>3</td></tr>
+<tr><td align='right'>12</td><td align='center'>"</td><td align='center'>1 X 3</td><td align='center'>@</td><td align='right'>9�</td><td align='center'>=</td><td align='right'>0</td><td align='right'>9</td><td align='right'>3</td></tr>
+<tr><td align='right'>12</td><td align='center'>"</td><td align='center'>2 X 4</td><td align='center'>@</td><td align='right'>9�</td><td align='center'>=</td><td align='right'>0</td><td align='right'>9</td><td align='right'>3</td></tr>
+<tr><td align='right'>12</td><td align='center'>"</td><td align='center'>9 X 4</td><td align='center'>@</td><td align='right'>9�</td><td align='center'>=</td><td align='right'>0</td><td align='right'>9</td><td align='right'>3</td></tr>
+<tr><td align='right'>8</td><td align='center'>"</td><td align='center'>4 X 0</td><td align='center'>@</td><td align='right'>10�</td><td align='right'>=</td><td align='center'>0</td><td align='right'>6</td><td align='right'>10</td></tr>
+<tr><td align='right'>1</td><td align='center'>"</td><td align='center'>5 X 4</td><td align='center'>@</td><td align='right'>12</td><td align='right'>=</td><td align='center'>0</td><td align='right'>1</td><td align='right'>0</td></tr>
+<tr><td align='right'>1</td><td align='center'>"</td><td align='center'>5 X 0</td><td align='center'>@</td><td align='right'>12</td><td align='right'>=</td><td align='center'>0</td><td align='right'>1</td><td align='right'>0</td></tr>
+<tr><td align='right' class='bt'>149</td><td  align='left' colspan='4'>plates @ 6<i>d</i>. each</td><td align='center'>=</td><td align='right'>3</td><td align='center'>14</td><td align='right'>6</td></tr>
+<tr><td align='right'></td><td></td><td></td><td></td><td></td><td></td><td align='right' class='bt'>&pound;7</td><td align='right' class='bt'>19</td><td align='right' class='bt'>1</td></tr>
+</table></div>
+
+
+<p>Of course, if we could increase the number of plates, we might get the
+painting done for nothing, but such a contingency is prevented by the
+condition that the fewest possible plates must be provided.</p>
+
+<p><span class='pagenum'>Pg 251<a name="Page_251" id="Page_251"></a></span>This puzzle appeared in <i>Tit-Bits</i>, and the following remarks, made
+by me in the issue for 11th December 1897, may be of interest.</p>
+
+<p>The "Hymn-Board Poser" seems to have created extraordinary interest.
+The immense number of attempts at its solution sent to me from all
+parts of the United Kingdom and from several Continental countries
+show a very kind disposition amongst our readers to help the worthy
+vicar of Chumpley St. Winifred over his parochial difficulty. Every
+conceivable estimate, from a few shillings up to as high a sum as
+&pound;1,347, 10<i>s</i>., seems to have come to hand. But the astonishing part of
+it is that, after going carefully through the tremendous pile of
+correspondence, I find that only one competitor has succeeded in
+maintaining the reputation of the <i>Tit-Bits</i> solvers for their
+capacity to solve anything, and his solution is substantially the same
+as the one given above, the cost being identical. Some of his figures
+are differently combined, but his grouping of the plates, as shown in
+the first column, is exactly the same. Though a large majority of
+competitors clearly hit upon all the essential points of the puzzle,
+they completely collapsed in the actual arrangement of the figures.
+According to their methods, some possible selection of hymns, such as
+111, 112, 121, 122,211, cannot be set up. A few correspondents
+suggested that it might be possible so to paint the 7's that upside
+down they would appear as 2's or 4's; but this would, of course, be
+barred out by the fact that a representation of the actual figures to
+be used was given.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_427_PHEASANT-SHOOTINGa" id="X_427_PHEASANT-SHOOTINGa"></a><a href="#X_427_PHEASANT-SHOOTING"><b>427.&mdash;PHEASANT-SHOOTING.&mdash;<i>solution</i></b></a></p>
+
+<p>The arithmetic of this puzzle is very easy indeed. There were clearly
+24 pheasants at the start. Of these 16 were shot dead, 1 was wounded
+in the wing, and 7 got away. The reader may have concluded that the
+answer is, therefore, that "seven remained." But as they flew away it
+is clearly absurd to say that they "remained." Had they done so they
+would certainly have been killed. Must we then conclude that the 17
+that were shot remained, because the others flew away? No; because the
+question was not "how many remained?" but "how many still remained?"
+Now the poor bird that was wounded in the wing, though unable to fly,
+was very active in its painful struggles to run away. The answer is,
+therefore, that the 16 birds that were shot dead "still remained," or
+"remained still."</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_428_THE_GARDENER_AND_THE_COOKa" id="X_428_THE_GARDENER_AND_THE_COOKa"></a><a href="#X_428_THE_GARDENER_AND_THE_COOK"><b>428.&mdash;THE GARDENER AND THE COOK.&mdash;<i>solution</i></b></a></p>
+
+<p>Nobody succeeded in solving the puzzle, so I had to let the cat out of
+the bag&mdash;an operation that was dimly foreshadowed by the puss in the
+original illustration. But I first reminded the reader that this
+puzzle appeared on April 1, a day on which none of us ever resents
+being made an "April Fool;" though, as I practically "gave the thing
+away" by specially drawing attention to the fact that it was All
+Fools' Day, it was quite remarkable that my correspondents, without a
+single exception, fell into the trap.</p>
+
+<p>One large body of correspondents held that what the cook loses in
+stride is exactly made up in greater speed; consequently both advance
+at the same rate, and the result must be a tie. But another
+considerable section saw that, though this might be so in a race 200
+ft. straight away, it could not really be, because they each go a
+stated distance at "every bound," and as 100 is not an exact multiple
+of 3, the gardener at his thirty-fourth bound will go 2 ft. beyond the
+mark. The gardener will, therefore, run to a point 102 ft. straight
+away and return (204 ft. in all), and so lose by 4 ft. This point
+certainly comes into the puzzle. But the most important fact of all is
+this, that it so happens that the gardener was a pupil from the
+Horticultural College for Lady Gardeners at, if I remember aright,
+Swanley; while the cook was a very accomplished French chef of the
+hemale persuasion! Therefore "she (the gardener) made three bounds to
+his (the cook's) two." It will now be found that while the gardener is
+running her 204 ft. in 68 bounds of 3 ft., the somewhat infirm old
+cook can only make 45<sup>1</sup>/<sub>3</sub> of his 2 ft. bounds, which equals 90 ft. 8
+in. The result is that the lady gardener wins the race by 109 ft. 4
+in. at a moment when the cook is in the air, one-third through his
+46th bound.</p>
+
+<p>The moral of this puzzle is twofold: (1) Never take things for granted
+in attempting to solve puzzles; (2) always remember All Fools' Day
+when it comes round. I was not writing of <i>any</i> gardener and cook, but
+of a <i>particular</i> couple, in "a race that I witnessed." The statement
+of the eye-witness must therefore be accepted: as the reader was not
+there, he cannot contradict it. Of course the information supplied was
+insufficient, but the correct reply was: "Assuming the gardener to be
+the 'he,' the cook wins by 4 ft.; but if the gardener is the 'she,'
+then the gardener wins by 109 ft. 4 in." This would have won the
+prize. Curiously enough, one solitary competitor got on to the right
+track, but failed to follow it up. He said: "Is this a regular April 1
+catch, meaning that they only ran 6 ft. each, and consequently the
+race was unfinished? If not, I think the following must be the
+solution, supposing the gardener to be the 'he' and the cook the
+'she.'" Though his solution was wrong even in the case he supposed,
+yet he was the only person who suspected the question of sex.</p>
+
+<hr style="width: 30%;" />
+<p><a name="X_429_PLACING_HALFPENNIESa" id="X_429_PLACING_HALFPENNIESa"></a><a href="#X_429_PLACING_HALFPENNIES"><b>429.&mdash;PLACING HALFPENNIES.&mdash;<i>solution</i></b></a></p>
+
+<p>Thirteen coins may be placed as shown.</p>
+
+<div class="figcenter" style="width: 400px;">
+<img src="images/a429.png" width="400" height="248" alt="" title="" />
+</div>
+
+<hr style="width: 30%;" />
+<p><a name="X_430_FIND_THE_MANS_WIFEa" id="X_430_FIND_THE_MANS_WIFEa"></a><a href="#X_430_FIND_THE_MANS_WIFE"><b>430.&mdash;FIND THE MAN'S WIFE.&mdash;<i>solution</i></b></a></p>
+
+<p>There is no guessing required in this puzzle. It is all a question of
+elimination. If we can pair off any five of the ladies with their
+respective husbands, other than husband No. 10, then the remaining
+lady must be No. 10's wife.</p>
+
+<p><span class='pagenum'>Pg 252<a name="Page_252" id="Page_252"></a></span></p>
+
+<p>I will show how this may be done. No. 8 is seen carrying a lady's
+parasol in the same hand with his walking-stick. But every lady is
+provided with a parasol, except No. 3; therefore No. 3 may be safely
+said to be the wife of No. 8. Then No. 12 is holding a bicycle, and
+the dress-guard and make disclose the fact that it is a lady's
+bicycle. The only lady in a cycling skirt is No. 5; therefore we
+conclude that No. 5 is No. 12's wife. Next, the man No. 6 has a dog,
+and lady No. 11 is seen carrying a dog chain. So we may safely pair
+No. 6 with No. 11. Then we see that man No. 2 is paying a newsboy for
+a paper. But we do not pay for newspapers in this way before receiving
+them, and the gentleman has apparently not taken one from the boy. But
+lady No. 9 is seen reading a paper. The inference is obvious&mdash;that she
+has sent the boy to her husband for a penny. We therefore pair No. 2
+with No. 9. We have now disposed of all the ladies except Nos. 1 and
+7, and of all the men except Nos. 4 and 10. On looking at No. 4 we
+find that he is carrying a coat over his arm, and that the buttons are
+on the left side;&mdash;not on the right, as a man wears them. So it is a
+lady's coat. But the coat clearly does not belong to No. 1, as she is
+seen to be wearing a coat already, while No. 7 lady is very lightly
+clad. We therefore pair No. 7 lady with man No. 4. Now the only lady
+left is No. 1, and we are consequently forced to the conclusion that
+she is the wife of No. 10. This is therefore the correct answer.</p>
+
+
+
+<hr style="width: 65%;" /><p><span class='pagenum'>Pg 253<a name="Page_253" id="Page_253"></a></span></p>
+<h2><a name="INDEX" id="INDEX"></a><a href="#CONTENTS">INDEX.</a></h2>
+
+
+<p>
+<span style="margin-left: 2em;">Abbot's Puzzle, The, <a href="#Page_20">20</a>, <a href="#Page_161">161</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Window, The, <a href="#Page_87">87</a>, <a href="#Page_213">213</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Academic Courtesies, <a href="#Page_18">18</a>, <a href="#Page_160">160</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Acrostic Puzzle, An, <a href="#Page_84">84</a>, <a href="#Page_210">210</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Adam and Eve and the Apples, <a href="#Page_18">18</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Aeroplanes, The Two, <a href="#Page_2">2</a>, <a href="#Page_148">148</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Age and Kinship Puzzles, <a href="#Page_6">6</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Concerning Tommy's, <a href="#Page_7">7</a>, <a href="#Page_153">153</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Mamma's, <a href="#Page_7">7</a>, <a href="#Page_152">152</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Mrs. Timpkins's, <a href="#Page_7">7</a>, <a href="#Page_152">152</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Rover's, <a href="#Page_7">7</a>, <a href="#Page_152">152</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Ages, The Family, <a href="#Page_7">7</a>, <a href="#Page_152">152</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Their, <a href="#Page_7">7</a>, <a href="#Page_152">152</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Alcuin, Abbot, <a href="#Page_20">20</a>, <a href="#Page_112">112</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Almonds, The Nine, <a href="#Page_64">64</a>, <a href="#Page_195">195</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Amazons, The, <a href="#Page_94">94</a>, <a href="#Page_221">221</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Andrews, W.S., <a href="#Page_125">125</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Apples, A Deal in, <a href="#Page_3">3</a>, <a href="#Page_149">149</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Buying, <a href="#Page_6">6</a>, <a href="#Page_151">151</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Ten, <a href="#Page_64">64</a>, <a href="#Page_195">195</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Approximations in Dissection, <a href="#Page_28">28</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Arithmetical and Algebraical Problems, <a href="#Page_1">1</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Various, <a href="#Page_17">17</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Arthur's Knights, King, <a href="#Page_77">77</a>, <a href="#Page_203">203</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Artillerymen's Dilemma, <a href="#Page_26">26</a>, <a href="#Page_167">167</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Asparagus, Bundles of, <a href="#Page_140">140</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Aspects all due South, <a href="#Page_137">137</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Associated Magic Squares, <a href="#Page_120">120</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Axiom, A Puzzling, <a href="#Page_138">138</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Bachet de M&eacute;ziriac, <a href="#Page_90">90</a>, <a href="#Page_109">109</a>, <a href="#Page_112">112</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Bachet's Square, <a href="#Page_90">90</a>, <a href="#Page_216">216</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Ball Problem, The, <a href="#Page_51">51</a>, <a href="#Page_183">183</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Ball, W.W. Rouse, <a href="#Page_109">109</a>, <a href="#Page_204">204</a>, <a href="#Page_248">248</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Balls, The Glass, <a href="#Page_78">78</a>, <a href="#Page_204">204</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Banker's Puzzle, The, <a href="#Page_25">25</a>, <a href="#Page_165">165</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Bank Holiday Puzzle, A, <a href="#Page_73">73</a>, <a href="#Page_201">201</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Banner Puzzle, The, <a href="#Page_46">46</a>, <a href="#Page_179">179</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; St. George's, <a href="#Page_50">50</a>, <a href="#Page_182">182</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Barrel Puzzle, The, <a href="#Page_109">109</a>, <a href="#Page_235">235</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Barrels of Balsam, The, <a href="#Page_82">82</a>, <a href="#Page_208">208</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Beanfeast Puzzle, A, <a href="#Page_2">2</a>, <a href="#Page_148">148</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Beef and Sausages, <a href="#Page_3">3</a>, <a href="#Page_149">149</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Beer, The Barrel of, <a href="#Page_13">13</a>, <a href="#Page_155">155</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Bell-ropes, Stealing the, <a href="#Page_49">49</a>, <a href="#Page_181">181</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Bells, The Peal of, <a href="#Page_78">78</a>, <a href="#Page_204">204</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Bergholt, E., <a href="#Page_116">116</a>, <a href="#Page_119">119</a>, <a href="#Page_125">125</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Betsy Ross Puzzle, The, <a href="#Page_40">40</a>, <a href="#Page_176">176</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Bicycle Thief, The, <a href="#Page_6">6</a>, <a href="#Page_152">152</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Bishops&mdash;Guarded, <a href="#Page_88">88</a>, <a href="#Page_214">214</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; in Convocation, <a href="#Page_89">89</a>, <a href="#Page_215">215</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzle, A New, <a href="#Page_98">98</a>, <a href="#Page_225">225</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Unguarded, <a href="#Page_88">88</a>, <a href="#Page_214">214</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Board, The Chess-, <a href="#Page_85">85</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; in Compartments, The, <a href="#Page_102">102</a>, <a href="#Page_228">228</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Setting the, <a href="#Page_105">105</a>, <a href="#Page_231">231</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Boards with Odd Number of Squares, <a href="#Page_86">86</a>, <a href="#Page_212">212</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Boat, Three Men in a, <a href="#Page_78">78</a>, <a href="#Page_204">204</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Bookworm, The Industrious, <a href="#Page_143">143</a>, <a href="#Page_248">248</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Boothby, Guy, <a href="#Page_154">154</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Box, The Cardboard, <a href="#Page_49">49</a>, <a href="#Page_181">181</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Paper, <a href="#Page_40">40</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Boys and Girls, <a href="#Page_67">67</a>, <a href="#Page_197">197</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Bridges, The Monk and the, <a href="#Page_75">75</a>, <a href="#Page_202">202</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Brigands, The Five, <a href="#Page_25">25</a>, <a href="#Page_164">164</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Brocade, The Squares of, <a href="#Page_47">47</a>, <a href="#Page_180">180</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Bun Puzzle, The, <a href="#Page_35">35</a>, <a href="#Page_170">170</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Busschop, Paul, <a href="#Page_172">172</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Buttons and String Method, <a href="#Page_230">230</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Cab Numbers, The, <a href="#Page_15">15</a>, <a href="#Page_157">157</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Calendar Puzzle, A, <a href="#Page_142">142</a>, <a href="#Page_247">247</a>.</span><br />
+<br />
+<span style="margin-left: 2em;"><i>Canterbury Puzzles, The</i>, <a href="#Page_14">14</a>, <a href="#Page_28">28</a>, <a href="#Page_58">58</a>, <a href="#Page_117">117</a>, <a href="#Page_121">121</a>, <a href="#Page_195">195</a>, <a href="#Page_202">202</a>, <a href="#Page_205">205</a>, <a href="#Page_206">206</a>, <a href="#Page_212">212</a>, <a href="#Page_213">213</a>, <a href="#Page_217">217</a>, <a href="#Page_233">233</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Card Frame Puzzle, The, <a href="#Page_114">114</a>, <a href="#Page_238">238</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Magic Squares, <a href="#Page_123">123</a>, <a href="#Page_244">244</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Players, A Puzzle for, <a href="#Page_78">78</a>, <a href="#Page_203">203</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzle, The "T," <a href="#Page_115">115</a>, <a href="#Page_239">239</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Triangles, <a href="#Page_115">115</a>, <a href="#Page_239">239</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cards, The Cross of, <a href="#Page_115">115</a>, <a href="#Page_238">238</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cardan, <a href="#Page_142">142</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Carroll, Lewis, <a href="#Page_43">43</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Castle Treasure, Stealing the, <a href="#Page_113">113</a>, <a href="#Page_237">237</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cats, the Wizard's, <a href="#Page_42">42</a>, <a href="#Page_178">178</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cattle, Judkins's, <a href="#Page_6">6</a>, <a href="#Page_151">151</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Market, At a, <a href="#Page_1">1</a>, <a href="#Page_148">148</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Census Puzzle, A, <a href="#Page_7">7</a>, <a href="#Page_152">152</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Century Puzzle, The, <a href="#Page_16">16</a>, <a href="#Page_158">158</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Digital, <a href="#Page_16">16</a>, <a href="#Page_159">159</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Chain Puzzle, A, <a href="#Page_144">144</a>, <a href="#Page_249">249</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Antiquary's, <a href="#Page_83">83</a>, <a href="#Page_209">209</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Cardboard, <a href="#Page_40">40</a>, <a href="#Page_176">176</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Change, Giving, <a href="#Page_4">4</a>, <a href="#Page_150">150</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Ways of giving, <a href="#Page_151">151</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Changing Places, <a href="#Page_10">10</a>, <a href="#Page_154">154</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Channel Island, <a href="#Page_138">138</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Charitable Bequest, A, <a href="#Page_2">2</a>, 148</span><br />
+<br />
+<span style="margin-left: 2em;">Charity, Indiscriminate, <a href="#Page_2">2</a>, <a href="#Page_148">148</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Checkmate, <a href="#Page_107">107</a>, <a href="#Page_233">233</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cheesemonger, The Eccentric, <a href="#Page_66">66</a>, <a href="#Page_196">196</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Chequered Board Divisions, <a href="#Page_85">85</a>, <a href="#Page_210">210</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cherries and Plums, <a href="#Page_56">56</a>, <a href="#Page_189">189</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Chess Puzzles, Dynamical, <a href="#Page_96">96</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Statical, <a href="#Page_88">88</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Various, <a href="#Page_105">105</a>.</span><br /><span class='pagenum'>Pg 254<a name="Page_254" id="Page_254"></a></span>
+<span style="margin-left: 3em;">&mdash;&mdash; Queer, <a href="#Page_107">107</a>, <a href="#Page_233">233</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Chessboard, The, <a href="#Page_85">85</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Fallacy, A, <a href="#Page_141">141</a>, <a href="#Page_247">247</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Guarded, <a href="#Page_95">95</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Non-attacking Arrangements, <a href="#Page_96">96</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Problems, <a href="#Page_84">84</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Sentence, The, <a href="#Page_87">87</a>, <a href="#Page_214">214</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Solitaire, <a href="#Page_108">108</a>, <a href="#Page_234">234</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Chinese, <a href="#Page_87">87</a>, <a href="#Page_213">213</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Crowded, <a href="#Page_91">91</a>, <a href="#Page_217">217</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Chestnuts, Buying, <a href="#Page_6">6</a>, <a href="#Page_152">152</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Chinese Money, <a href="#Page_4">4</a>, <a href="#Page_150">150</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzle, Ancient, <a href="#Page_107">107</a>, <a href="#Page_233">233</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; <i>The Fashionable</i>, <a href="#Page_43">43</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Christmas Boxes, The, <a href="#Page_4">4</a>, <a href="#Page_150">150</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Present, Mrs. Smiley's, <a href="#Page_46">46</a>, <a href="#Page_179">179</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Pudding, The, <a href="#Page_43">43</a>, <a href="#Page_178">178</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cigar Puzzle, The, <a href="#Page_119">119</a>, <a href="#Page_242">242</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Circle, The Dissected, <a href="#Page_69">69</a>, <a href="#Page_197">197</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cisterns, How to Make, <a href="#Page_54">54</a>, <a href="#Page_188">188</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Civil Service "Howler," <a href="#Page_154">154</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Clare, John, <a href="#Page_58">58</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Clock Formul&aelig;, <a href="#Page_154">154</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzles, <a href="#Page_9">9</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Club, <a href="#Page_10">10</a>, <a href="#Page_154">154</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Railway Station, <a href="#Page_11">11</a>, <a href="#Page_155">155</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Clocks, The Three, <a href="#Page_11">11</a>, <a href="#Page_154">154</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Clothes Line Puzzle, The, <a href="#Page_50">50</a>, <a href="#Page_182">182</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Coast, Round the, <a href="#Page_63">63</a>, <a href="#Page_195">195</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Coincidence, A Queer, <a href="#Page_2">2</a>, <a href="#Page_148">148</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Coins, The Broken, <a href="#Page_5">5</a>, <a href="#Page_150">150</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Ten, <a href="#Page_57">57</a>, <a href="#Page_190">190</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Two Ancient, <a href="#Page_140">140</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Combination and Group Problems, <a href="#Page_76">76</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Compasses Puzzle, The, <a href="#Page_53">53</a>, <a href="#Page_186">186</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Composite Magic Squares, <a href="#Page_127">127</a>, <a href="#Page_246">246</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cone Puzzle, The, <a href="#Page_55">55</a>, <a href="#Page_188">188</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Corn, Reaping the, <a href="#Page_20">20</a>, <a href="#Page_161">161</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cornfields, Farmer Lawrence's, <a href="#Page_101">101</a>, <a href="#Page_227">227</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Costermonger's Puzzle, The, <a href="#Page_6">6</a>, <a href="#Page_152">152</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Counter Problems, Moving, <a href="#Page_58">58</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzle, A New, <a href="#Page_98">98</a>, <a href="#Page_225">225</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Solitaire, <a href="#Page_107">107</a>, <a href="#Page_234">234</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Counters, The Coloured, <a href="#Page_91">91</a>, <a href="#Page_217">217</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Forty-nine, <a href="#Page_92">92</a>, <a href="#Page_217">217</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Nine, <a href="#Page_14">14</a>, <a href="#Page_156">156</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Ten, <a href="#Page_15">15</a>, <a href="#Page_156">156</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Crescent Puzzle, The, <a href="#Page_52">52</a>, <a href="#Page_184">184</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Crescents of Byzantium, The Five, <a href="#Page_92">92</a>, <a href="#Page_219">219</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cricket Match, The Village, <a href="#Page_116">116</a>, <a href="#Page_239">239</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Slow, <a href="#Page_116">116</a>, <a href="#Page_239">239</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cross and Triangle, <a href="#Page_35">35</a>, <a href="#Page_169">169</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; of Cards, <a href="#Page_115">115</a>, <a href="#Page_238">238</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Folded, <a href="#Page_35">35</a>, <a href="#Page_169">169</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Southern, <a href="#Page_93">93</a>, <a href="#Page_220">220</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Crosses, Counter,&nbsp; <a href="#Page_81">81</a>, <a href="#Page_207">207</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; from One, Two, <a href="#Page_35">35</a>, <a href="#Page_168">168</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; Three, <a href="#Page_169">169</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Crossing River Problems, <a href="#Page_112">112</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Crusader, The, <a href="#Page_106">106</a>, <a href="#Page_232">232</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cubes, Sums of, <a href="#Page_165">165</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cushion Covers, The, <a href="#Page_46">46</a>, <a href="#Page_179">179</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cutting-out Puzzle, A, <a href="#Page_37">37</a>, <a href="#Page_172">172</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Cyclists' Feast, The, <a href="#Page_2">2</a>, <a href="#Page_148">148</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Dairyman, The Honest, <a href="#Page_110">110</a>, <a href="#Page_235">235</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Definition, A Question of, <a href="#Page_23">23</a>, <a href="#Page_163">163</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">De Fonteney, <a href="#Page_112">112</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Deified Puzzle, The, <a href="#Page_74">74</a>, <a href="#Page_202">202</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Delannoy, <a href="#Page_112">112</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">De Morgan, A., <a href="#Page_27">27</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">De Tudor, Sir Edwyn, <a href="#Page_12">12</a>, <a href="#Page_155">155</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Diabolique Magic Squares, <a href="#Page_120">120</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Diamond Puzzle, The, <a href="#Page_74">74</a>, <a href="#Page_202">202</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Dice, A Trick with, <a href="#Page_116">116</a>, <a href="#Page_239">239</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Game, The Montenegrin, <a href="#Page_119">119</a>, <a href="#Page_242">242</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Numbers, The, <a href="#Page_17">17</a>, <a href="#Page_160">160</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Die, Painting the, <a href="#Page_84">84</a>, <a href="#Page_210">210</a>,</span><br />
+<br />
+<span style="margin-left: 2em;">Digital Analysis, <a href="#Page_157">157</a>, <a href="#Page_158">158</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Division, <a href="#Page_16">16</a>, <a href="#Page_158">158</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Multiplication, <a href="#Page_15">15</a>, <a href="#Page_156">156</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzles, <a href="#Page_13">13</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Digits, Adding the, <a href="#Page_16">16</a>, <a href="#Page_158">158</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; and Squares, <a href="#Page_14">14</a>, <a href="#Page_155">155</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Odd and Even, <a href="#Page_14">14</a>, <a href="#Page_156">156</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Dilemma, An Amazing, <a href="#Page_106">106</a>, <a href="#Page_233">233</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Diophantine Problem, <a href="#Page_164">164</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Dissection Puzzle, An Easy, <a href="#Page_35">35</a>, <a href="#Page_170">170</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzles, <a href="#Page_27">27</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; Various, <a href="#Page_35">35</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Dividing Magic Squares, <a href="#Page_124">124</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Division, Digital, <a href="#Page_16">16</a>, <a href="#Page_158">158</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Simple, <a href="#Page_23">23</a>, <a href="#Page_163">163</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Doctor's Query, The, <a href="#Page_109">109</a>, <a href="#Page_235">235</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Dogs Puzzle, The Five, <a href="#Page_92">92</a>, <a href="#Page_218">218</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Domestic Economy, <a href="#Page_5">5</a>, <a href="#Page_151">151</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Domino Frame Puzzle, The, <a href="#Page_114">114</a>, <a href="#Page_238">238</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Dominoes in Progression, <a href="#Page_114">114</a>, <a href="#Page_237">237</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Eighteen, <a href="#Page_123">123</a>, <a href="#Page_245">245</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Fifteen, <a href="#Page_83">83</a>, <a href="#Page_209">209</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Five, <a href="#Page_114">114</a>, <a href="#Page_238">238</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Donkey Riding, <a href="#Page_13">13</a>, <a href="#Page_155">155</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Dormitory Puzzle, A, <a href="#Page_81">81</a>, <a href="#Page_208">208</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Dovetailed Block, The, <a href="#Page_145">145</a>, <a href="#Page_249">249</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Drayton's <i>Polyolbion</i>, <a href="#Page_58">58</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Dungeon Puzzle, A, <a href="#Page_97">97</a>, <a href="#Page_224">224</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Dungeons, The Siberian, <a href="#Page_123">123</a>, <a href="#Page_244">244</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Spanish, <a href="#Page_122">122</a>, <a href="#Page_244">244</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Dutchmen's Wives, The, <a href="#Page_26">26</a>, <a href="#Page_167">167</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Dynamical Chess Puzzles, <a href="#Page_96">96</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Earth's Girdle, The, <a href="#Page_139">139</a>.</span><br />
+<br />
+<span style="margin-left: 2em;"><i>Educational Times Reprints</i>, <a href="#Page_204">204</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Eggs, A Deal in, <a href="#Page_3">3</a>, <a href="#Page_149">149</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Obtaining the, <a href="#Page_140">140</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Election, The Muddletown, <a href="#Page_19">19</a>, <a href="#Page_161">161</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Parish Council, <a href="#Page_19">19</a>, <a href="#Page_161">161</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Eleven, The Mystic, <a href="#Page_16">16</a>, <a href="#Page_159">159</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Elopements, The Four, <a href="#Page_113">113</a>, <a href="#Page_237">237</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Elrick, E., <a href="#Page_231">231</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Engines, The Eight, <a href="#Page_61">61</a>, <a href="#Page_194">194</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Episcopal Visitation, An, <a href="#Page_98">98</a>, <a href="#Page_225">225</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Estate, Farmer Wurzel's, <a href="#Page_51">51</a>, <a href="#Page_184">184</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Estates, The Yorkshire, <a href="#Page_51">51</a>, <a href="#Page_183">183</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Euclid, <a href="#Page_31">31</a>, <a href="#Page_138">138</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Euler, L., <a href="#Page_165">165</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Exchange Puzzle, The, <a href="#Page_66">66</a>, <a href="#Page_196">196</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Fallacy, A Chessboard, <a href="#Page_141">141</a>, <a href="#Page_247">247</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Family Party, A, <a href="#Page_8">8</a>, <a href="#Page_153">153</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Fare, The Passenger's, <a href="#Page_13">13</a>, <a href="#Page_155">155</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Farmer and his Sheep, The, <a href="#Page_22">22</a>, <a href="#Page_163">163</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Fence Problem, A, <a href="#Page_21">21</a>, <a href="#Page_162">162</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Fences, The Landowner's, <a href="#Page_42">42</a>, <a href="#Page_178">178</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Fermat, <a href="#Page_164">164</a>, <a href="#Page_168">168</a>.</span><br />
+<br /><span class='pagenum'>Pg 255<a name="Page_255" id="Page_255"></a></span>
+<span style="margin-left: 2em;">Find the Man's Wife, <a href="#Page_147">147</a>, <a href="#Page_251">251</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Fly on the Octahedron, The, <a href="#Page_70">70</a>, <a href="#Page_198">198</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Fog, Mr. Gubbins in a, <a href="#Page_18">18</a>, <a href="#Page_161">161</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Football Players, The, <a href="#Page_116">116</a>, <a href="#Page_240">240</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Fraction, A Puzzling, <a href="#Page_138">138</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Fractions, More Mixed, <a href="#Page_16">16</a>, <a href="#Page_159">159</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Frame Puzzle, The Card, <a href="#Page_114">114</a>, <a href="#Page_238">238</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; The Domino, <a href="#Page_114">114</a>, <a href="#Page_238">238</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Frankenstein, E.N., <a href="#Page_232">232</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Fr&eacute;nicle, B., <a href="#Page_119">119</a>, <a href="#Page_168">168</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Frogs, The Educated, <a href="#Page_59">59</a>, 194-</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Four, <a href="#Page_103">103</a>, <a href="#Page_229">229</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Six, <a href="#Page_59">59</a>, <a href="#Page_193">193</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Frost, A.H., <a href="#Page_120">120</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Games, Puzzle, <a href="#Page_117">117</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Problems concerning, <a href="#Page_114">114</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Garden, Lady Belinda's, <a href="#Page_52">52</a>, <a href="#Page_186">186</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzle, The, <a href="#Page_49">49</a>, <a href="#Page_182">182</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Gardener and the Cook, The, <a href="#Page_146">146</a>, <a href="#Page_251">251</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Geometrical Problems, <a href="#Page_27">27</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzles, Various, <a href="#Page_49">49</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">George and the Dragon, St., <a href="#Page_101">101</a>, <a href="#Page_227">227</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Getting Upstairs, Such a, <a href="#Page_143">143</a>, <a href="#Page_248">248</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Girdle, the Earth's, <a href="#Page_139">139</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Goat, The Tethered, <a href="#Page_53">53</a>, <a href="#Page_186">186</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Grand Lama's Problem, The, <a href="#Page_86">86</a>, <a href="#Page_212">212</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Grasshopper Puzzle, The, <a href="#Page_59">59</a>, <a href="#Page_193">193</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Greek Cross Puzzles, <a href="#Page_28">28</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; Three from One, <a href="#Page_169">169</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Greyhound Puzzle, The, <a href="#Page_101">101</a>, <a href="#Page_227">227</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Grocer and Draper, The, <a href="#Page_5">5</a>, <a href="#Page_151">151</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Gros, L., <a href="#Page_248">248</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Group Problems, Combination and, <a href="#Page_76">76</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Groups, The Three, <a href="#Page_14">14</a>, <a href="#Page_156">156</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Guarini, <a href="#Page_229">229</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Hairdresser's Puzzle, The, <a href="#Page_137">137</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Halfpennies, Placing, <a href="#Page_147">147</a>, <a href="#Page_251">251</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Hampton Court Maze solved, <a href="#Page_133">133</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Hannah's Puzzle, <a href="#Page_75">75</a>, <a href="#Page_202">202</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Hastings, The Battle of, <a href="#Page_23">23</a>, <a href="#Page_164">164</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Hatfield Maze solved, <a href="#Page_136">136</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Hat Puzzle, The, <a href="#Page_67">67</a>, <a href="#Page_196">196</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Hat-peg Puzzle, The, <a href="#Page_93">93</a>, <a href="#Page_221">221</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Hats, The Wrong, <a href="#Page_78">78</a>, <a href="#Page_203">203</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Hay, The Trusses of, <a href="#Page_18">18</a>, <a href="#Page_161">161</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Heads or Tails, <a href="#Page_22">22</a>, <a href="#Page_163">163</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Hearthrug, Mrs. Hobson's, <a href="#Page_37">37</a>, <a href="#Page_172">172</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Helmholtz, Von, <a href="#Page_41">41</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Honey, The Barrels of, <a href="#Page_111">111</a>, <a href="#Page_236">236</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Honeycomb Puzzle, The, <a href="#Page_75">75</a>, <a href="#Page_202">202</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Horse Race Puzzle, The, <a href="#Page_117">117</a>, <a href="#Page_240">240</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Horseshoes, The Two, <a href="#Page_40">40</a>, <a href="#Page_175">175</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Houdin, <a href="#Page_68">68</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Hydroplane Question, The, <a href="#Page_12">12</a>, <a href="#Page_155">155</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Hymn-board Poser, The, <a href="#Page_145">145</a>, <a href="#Page_250">250</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Icosahedron Puzzle, The, <a href="#Page_70">70</a>, <a href="#Page_198">198</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Jack and the Beanstalk, <a href="#Page_145">145</a>, <a href="#Page_249">249</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Jackson, John, <a href="#Page_56">56</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Jaenisch, C.F. de, <a href="#Page_92">92</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Jampots, Arranging the, <a href="#Page_68">68</a>, <a href="#Page_197">197</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Jealous Husbands, Five, <a href="#Page_113">113</a>, <a href="#Page_236">236</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Joiner's Problem, The, <a href="#Page_36">36</a>, <a href="#Page_171">171</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; Another, <a href="#Page_37">37</a>, <a href="#Page_171">171</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Jolly Gaol-Birds, Eight, <a href="#Page_122">122</a>, <a href="#Page_243">243</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; Nine, <a href="#Page_122">122</a>, <a href="#Page_243">243</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Journey, The Queen's, <a href="#Page_100">100</a>, <a href="#Page_227">227</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Rook's, <a href="#Page_96">96</a>, <a href="#Page_224">224</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Junior Clerks' Puzzle, The, <a href="#Page_4">4</a>, <a href="#Page_150">150</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Juvenile Puzzle, A, <a href="#Page_68">68</a>, <a href="#Page_197">197</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Kangaroos, The Four, <a href="#Page_102">102</a>, <a href="#Page_228">228</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Kelvin, Lord, <a href="#Page_41">41</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Kennel Puzzle, The, <a href="#Page_105">105</a>, <a href="#Page_231">231</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">King and the Castles, The, <a href="#Page_56">56</a>, <a href="#Page_189">189</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Forsaken, <a href="#Page_106">106</a>, <a href="#Page_232">232</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Kite-flying Puzzle, A, <a href="#Page_54">54</a>, <a href="#Page_187">187</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Knight-guards, The, <a href="#Page_95">95</a>, <a href="#Page_222">222</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Knights, King Arthur's, <a href="#Page_77">77</a>, <a href="#Page_203">203</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Tour, Magic, <a href="#Page_127">127</a>, <a href="#Page_247">247</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; The Cubic, <a href="#Page_103">103</a>, <a href="#Page_229">229</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; The Four, <a href="#Page_103">103</a>, <a href="#Page_229">229</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Labosne, A., <a href="#Page_25">25</a>, <a href="#Page_90">90</a>, <a href="#Page_216">216</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Labourer's Puzzle, The, <a href="#Page_18">18</a>, <a href="#Page_160">160</a>.</span><br />
+<br />
+<span style="margin-left: 2em;"><i>Ladies' Diary</i>, <a href="#Page_26">26</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Lagrange, J.L., <a href="#Page_9">9</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Laisant, C.A., <a href="#Page_76">76</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Lamp-posts, Painting the, <a href="#Page_19">19</a>, <a href="#Page_161">161</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Leap Year, <a href="#Page_155">155</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; Ladies, The, <a href="#Page_19">19</a>, <a href="#Page_161">161</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Legacy, A Puzzling, <a href="#Page_20">20</a>, <a href="#Page_161">161</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Legal Difficulty, A, <a href="#Page_23">23</a>, <a href="#Page_163">163</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Le Plongeon, Dr., <a href="#Page_29">29</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Letter Block Puzzle, The, <a href="#Page_60">60</a>, <a href="#Page_194">194</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Blocks, The Thirty-six, <a href="#Page_91">91</a>, <a href="#Page_216">216</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzle, The Fifteen, <a href="#Page_79">79</a>, <a href="#Page_205">205</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Level Puzzle, The, <a href="#Page_74">74</a>, <a href="#Page_202">202</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Linoleum Cutting, <a href="#Page_48">48</a>, <a href="#Page_181">181</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzle, Another, <a href="#Page_49">49</a>, <a href="#Page_181">181</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Lion and the Man, The, <a href="#Page_97">97</a>, <a href="#Page_224">224</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Hunting, <a href="#Page_94">94</a>, <a href="#Page_222">222</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Lions and Crowns, <a href="#Page_85">85</a>, <a href="#Page_212">212</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Four, <a href="#Page_88">88</a>, <a href="#Page_214">214</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Lockers Puzzle, The, <a href="#Page_14">14</a>, <a href="#Page_156">156</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Locomotion and Speed Puzzles, <a href="#Page_11">11</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Lodging-house Difficulty, A, <a href="#Page_61">61</a>, <a href="#Page_194">194</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">London and Wise, <a href="#Page_131">131</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Loyd, Sam, <a href="#Page_8">8</a>, <a href="#Page_43">43</a>, <a href="#Page_44">44</a>, <a href="#Page_98">98</a>, <a href="#Page_144">144</a>, <a href="#Page_232">232</a>, <a href="#Page_233">233</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Lucas, Edouard, <a href="#Page_16">16</a>, <a href="#Page_76">76</a>, <a href="#Page_112">112</a>, <a href="#Page_121">121</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Luncheons, The City, <a href="#Page_77">77</a>, <a href="#Page_203">203</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">MacMahon, Major, <a href="#Page_109">109</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Magic Knight's Tour, <a href="#Page_127">127</a>, <a href="#Page_247">247</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Square Problems, <a href="#Page_119">119</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; Card, <a href="#Page_123">123</a>, <a href="#Page_244">244</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; of Composites, <a href="#Page_127">127</a>, <a href="#Page_246">246</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; of Primes, <a href="#Page_125">125</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; of Two Degrees, <a href="#Page_125">125</a>, <a href="#Page_245">245</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; Two New, <a href="#Page_125">125</a>, <a href="#Page_245">245</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Strips, <a href="#Page_121">121</a>, <a href="#Page_243">243</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Magics, Subtracting, Multiplying, and Dividing, <a href="#Page_124">124</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Maiden, The Languishing, <a href="#Page_97">97</a>, <a href="#Page_224">224</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Mandarin's Puzzle, The, <a href="#Page_103">103</a>, <a href="#Page_230">230</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; "T" Puzzle, The, <a href="#Page_126">126</a>, <a href="#Page_246">246</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Marketing, Saturday, <a href="#Page_27">27</a>, <a href="#Page_168">168</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Market Women, The, <a href="#Page_3">3</a>, <a href="#Page_149">149</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Mary and Marmaduke, <a href="#Page_7">7</a>, <a href="#Page_152">152</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Mary, How Old was, <a href="#Page_8">8</a>, <a href="#Page_153">153</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Massacre of Innocents, <a href="#Page_139">139</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Match Mystery, A, <a href="#Page_118">118</a>, <a href="#Page_241">241</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzle, A New, <a href="#Page_55">55</a>, <a href="#Page_188">188</a>.</span><br />
+<br /><span class='pagenum'>Pg 256<a name="Page_256" id="Page_256"></a></span>
+<span style="margin-left: 2em;">Mates, Thirty-six, <a href="#Page_106">106</a>, <a href="#Page_233">233</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Mazes and how to thread Them, <a href="#Page_127">127</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Measuring, Weighing, and Packing Puzzles, <a href="#Page_109">109</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzle, New, <a href="#Page_110">110</a>, <a href="#Page_235">235</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Meeting, The Suffragists', <a href="#Page_19">19</a>, <a href="#Page_161">161</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Mellor, W.M.F., <a href="#Page_242">242</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">M&eacute;nages, Probl&ecirc;me de, <a href="#Page_76">76</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Mersenne, M., <a href="#Page_168">168</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Mice, Catching the, <a href="#Page_65">65</a>, <a href="#Page_196">196</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Milkmaid Puzzle, The, <a href="#Page_50">50</a>, <a href="#Page_183">183</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Millionaire's Perplexity, The, <a href="#Page_3">3</a>, <a href="#Page_149">149</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Mince Pies, The Twelve, <a href="#Page_57">57</a>, <a href="#Page_191">191</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Mine, Inspecting a, <a href="#Page_71">71</a>, <a href="#Page_199">199</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Miners' Holiday, The, <a href="#Page_23">23</a>, <a href="#Page_163">163</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Miser, The Converted, <a href="#Page_21">21</a>, <a href="#Page_162">162</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Mitre, Dissecting a, <a href="#Page_35">35</a>, <a href="#Page_170">170</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Monad, The Great, <a href="#Page_39">39</a>, <a href="#Page_174">174</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Money, A Queer Thing in, <a href="#Page_2">2</a>, <a href="#Page_148">148</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Boxes, The Puzzling, <a href="#Page_3">3</a>, <a href="#Page_149">149</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash;, Pocket, <a href="#Page_3">3</a>, <a href="#Page_149">149</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzles, <a href="#Page_1">1</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzle, A New, <a href="#Page_2">2</a>, <a href="#Page_148">148</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash;, Square, <a href="#Page_3">3</a>, <a href="#Page_149">149</a>.</span><br />
+<br />
+<span style="margin-left: 2em;"><i>Monist, The</i>, <a href="#Page_125">125</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Monk and the Bridges, The, <a href="#Page_75">75</a>, <a href="#Page_202">202</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Monstrosity, The, <a href="#Page_108">108</a>, <a href="#Page_234">234</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Montenegrin Dice Game, The, <a href="#Page_119">119</a>, <a href="#Page_242">242</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Moreau, <a href="#Page_76">76</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Morris, Nine Men's, <a href="#Page_58">58</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Mosaics, A Problem in, <a href="#Page_90">90</a>, <a href="#Page_215">215</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Mother and Daughter, <a href="#Page_7">7</a>, <a href="#Page_152">152</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Motor-car Race, The, <a href="#Page_117">117</a>, <a href="#Page_240">240</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Tour, The, <a href="#Page_74">74</a>, <a href="#Page_201">201</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Garage Puzzle, The, <a href="#Page_62">62</a>, <a href="#Page_195">195</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Motorists, A Puzzle for, <a href="#Page_73">73</a>, <a href="#Page_201">201</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Mouse-trap Puzzle, The, <a href="#Page_80">80</a>, <a href="#Page_206">206</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Moving Counter Problems, <a href="#Page_58">58</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Multiplication, Digital, <a href="#Page_15">15</a>, <a href="#Page_156">156</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Queer, <a href="#Page_15">15</a>, <a href="#Page_157">157</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Simple, <a href="#Page_23">23</a>, <a href="#Page_163">163</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Multiplying Magic Squares, <a href="#Page_124">124</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Muncey, J.N., <a href="#Page_125">125</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Murray, Sir James, <a href="#Page_44">44</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Napoleon, <a href="#Page_43">43</a>, <a href="#Page_44">44</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Nasik Magic Squares, <a href="#Page_120">120</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Neighbours, Next-Door, <a href="#Page_8">8</a>, <a href="#Page_153">153</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Newton, Sir Isaac, <a href="#Page_56">56</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Nine Men's Morris, <a href="#Page_58">58</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Notation, Scales of, <a href="#Page_149">149</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Noughts and Crosses, <a href="#Page_58">58</a>, <a href="#Page_117">117</a>.</span><br />
+<br />
+<span style="margin-left: 2em;"><i>Nouvelles Annales de Math&eacute;matiques</i>, <a href="#Page_14">14</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Number Checks Puzzle, The, <a href="#Page_16">16</a>, <a href="#Page_158">158</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Numbers, Curious, <a href="#Page_20">20</a>, <a href="#Page_162">162</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Nuts, The Bag of, <a href="#Page_8">8</a>, <a href="#Page_153">153</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Observation, Defective, <a href="#Page_4">4</a>, <a href="#Page_150">150</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Octahedron, The Fly on the, <a href="#Page_70">70</a>, <a href="#Page_198">198</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Oval, How to draw an, <a href="#Page_50">50</a>, <a href="#Page_182">182</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Ovid's Game, <a href="#Page_58">58</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Packing in Russia, Gold, <a href="#Page_111">111</a>, <a href="#Page_236">236</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzles, Measuring, Weighing, and, <a href="#Page_109">109</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzle, A, <a href="#Page_111">111</a>, <a href="#Page_236">236</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pandiagonal Magic Squares, <a href="#Page_120">120</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Papa's Puzzle, <a href="#Page_53">53</a>, <a href="#Page_187">187</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pappus, <a href="#Page_53">53</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Paradox Party, The, <a href="#Page_137">137</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Party, A Family, <a href="#Page_8">8</a>, <a href="#Page_153">153</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Patchwork Puzzles, <a href="#Page_46">46</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzle, Another, <a href="#Page_48">48</a>, <a href="#Page_180">180</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Silk, <a href="#Page_34">34</a>, <a href="#Page_168">168</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Patience, <i>Strand</i>, <a href="#Page_116">116</a>, <a href="#Page_239">239</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pawns, A Puzzle with, <a href="#Page_94">94</a>, <a href="#Page_222">222</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Immovable, <a href="#Page_106">106</a>, <a href="#Page_233">233</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Six, <a href="#Page_107">107</a>, <a href="#Page_233">233</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Two, <a href="#Page_105">105</a>, <a href="#Page_231">231</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pearls, The Thirty-three, <a href="#Page_18">18</a>, <a href="#Page_160">160</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pebble Game, The, <a href="#Page_117">117</a>, <a href="#Page_240">240</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pedigree, A Mixed, <a href="#Page_8">8</a>, <a href="#Page_153">153</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pellian Equation, <a href="#Page_164">164</a>, <a href="#Page_167">167</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pennies, The Five, <a href="#Page_143">143</a>, <a href="#Page_248">248</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Twelve, <a href="#Page_65">65</a>, <a href="#Page_195">195</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pension, Drawing her, <a href="#Page_12">12</a>, <a href="#Page_155">155</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pentagon and Square, The, <a href="#Page_37">37</a>, <a href="#Page_172">172</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Drawing a, <a href="#Page_37">37</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pfeffermann, M., <a href="#Page_125">125</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pheasant-Shooting, <a href="#Page_146">146</a>, <a href="#Page_251">251</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Philadelphia Maze solved, <a href="#Page_137">137</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pierrot's Puzzle, The, <a href="#Page_15">15</a>, <a href="#Page_156">156</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pigs, The Seven, <a href="#Page_41">41</a>, <a href="#Page_177">177</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Planck, C., <a href="#Page_220">220</a>, <a href="#Page_246">246</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Plane Paradox, <a href="#Page_138">138</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Plantation Puzzle, A, <a href="#Page_57">57</a>, <a href="#Page_189">189</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Burmese, <a href="#Page_58">58</a>, <a href="#Page_191">191</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Plates and Coins, <a href="#Page_65">65</a>, <a href="#Page_195">195</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Plums, The Baskets of, <a href="#Page_126">126</a>, <a href="#Page_245">245</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Poe, E.A., <a href="#Page_249">249</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Points and Lines Problems, <a href="#Page_56">56</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Postage Stamps, The Four, <a href="#Page_84">84</a>, <a href="#Page_210">210</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Post-Office Perplexity, A, <a href="#Page_1">1</a>, <a href="#Page_148">148</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Potato Puzzle, The, <a href="#Page_41">41</a>, <a href="#Page_177">177</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Potatoes, The Basket of, <a href="#Page_13">13</a>, <a href="#Page_155">155</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Precocious Baby, The, <a href="#Page_139">139</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Presents, Buying, <a href="#Page_2">2</a>, <a href="#Page_148">148</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Prime Magic Squares, <a href="#Page_125">125</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Printer's Error, A, <a href="#Page_20">20</a>, <a href="#Page_162">162</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Prisoners, Exercise for, <a href="#Page_104">104</a>, <a href="#Page_230">230</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Ten, <a href="#Page_62">62</a>, <a href="#Page_195">195</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Probabilities, Two Questions in, <a href="#Page_5">5</a>, <a href="#Page_150">150</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Problems concerning Games, <a href="#Page_114">114</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Puss in the Corner, <a href="#Page_118">118</a>, <a href="#Page_240">240</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Puzzle Games, <a href="#Page_117">117</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pyramid, Painting a, <a href="#Page_83">83</a>, <a href="#Page_208">208</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pyramids, Square and Triangular, <a href="#Page_167">167</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Pythagoras, <a href="#Page_31">31</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">"Queen, The," <a href="#Page_120">120</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Queens and Bishop Puzzle, <a href="#Page_93">93</a>, <a href="#Page_219">219</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Eight, <a href="#Page_89">89</a>, <a href="#Page_215">215</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Queen's Journey, The, <a href="#Page_100">100</a>, <a href="#Page_227">227</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Tour, The, <a href="#Page_98">98</a>, <a href="#Page_225">225</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Quilt, Mrs. Perkins's, <a href="#Page_47">47</a>, <a href="#Page_180">180</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Race Puzzle, The Horse-, <a href="#Page_117">117</a>, <a href="#Page_240">240</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Motor-car, <a href="#Page_117">117</a>, <a href="#Page_240">240</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Rackbrane's Little Loss, <a href="#Page_21">21</a>, <a href="#Page_163">163</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Railway Muddle, A, <a href="#Page_62">62</a>, <a href="#Page_194">194</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzle, A, <a href="#Page_61">61</a>, <a href="#Page_194">194</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Stations, The Three, <a href="#Page_49">49</a>, <a href="#Page_182">182</a>.</span><br />
+<br />
+<span style="margin-left: 2em;"><i>Rational Amusement for Winter Evenings</i>, <a href="#Page_56">56</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Rectangles, Counting the, <a href="#Page_105">105</a>, <a href="#Page_232">232</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Reiss, M., <a href="#Page_58">58</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Relationships, Queer, <a href="#Page_8">8</a>, <a href="#Page_153">153</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Reversals, A Puzzle in, <a href="#Page_5">5</a>, <a href="#Page_151">151</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">River Axe, Crossing the, <a href="#Page_112">112</a>, <a href="#Page_236">236</a>.</span><br />
+<br /><span class='pagenum'>Pg 257<a name="Page_257" id="Page_257"></a></span>
+<span style="margin-left: 2em;">River Problems, Crossing, <a href="#Page_112">112</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Rookery, The, <a href="#Page_105">105</a>, <a href="#Page_232">232</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Rook's Journey, The, <a href="#Page_96">96</a>, <a href="#Page_224">224</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Tour, The, <a href="#Page_96">96</a>, <a href="#Page_223">223</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Rooks, The Eight, <a href="#Page_88">88</a>, <a href="#Page_214">214</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Two, <a href="#Page_117">117</a>, <a href="#Page_240">240</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Round Table, The, <a href="#Page_80">80</a>, <a href="#Page_205">205</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Route Problems, Unicursal and, <a href="#Page_68">68</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Ruby Brooch, The, <a href="#Page_144">144</a>, <a href="#Page_249">249</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Sabbath Puzzle, The, <a href="#Page_144">144</a>, <a href="#Page_249">249</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Sailor's Puzzle, The, <a href="#Page_71">71</a>, <a href="#Page_199">199</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Sayles, H.A., <a href="#Page_125">125</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Schoolboys, The Nine, <a href="#Page_80">80</a>, <a href="#Page_205">205</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Schoolgirls, The Fifteen, <a href="#Page_80">80</a>, <a href="#Page_204">204</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Scramble, The Great, <a href="#Page_19">19</a>, <a href="#Page_161">161</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Sculptor's Problem, The, <a href="#Page_23">23</a>, <a href="#Page_164">164</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Second Day of Week, <a href="#Page_139">139</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">See-Saw Puzzle, The, <a href="#Page_22">22</a>, <a href="#Page_163">163</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Semi-Nasik Magic Squares, <a href="#Page_120">120</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Senior and Junior, <a href="#Page_140">140</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Sevens, The Four, <a href="#Page_17">17</a>, <a href="#Page_160">160</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Sharp's Puzzle, <a href="#Page_230">230</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Sheepfold, The, <a href="#Page_52">52</a>, <a href="#Page_184">184</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Sheep Pens, The Six, <a href="#Page_55">55</a>, <a href="#Page_189">189</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Sixteen, <a href="#Page_80">80</a>, <a href="#Page_206">206</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Three, <a href="#Page_92">92</a>, <a href="#Page_217">217</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Those Fifteen, <a href="#Page_77">77</a>, <a href="#Page_203">203</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Shopping Perplexity, A, <a href="#Page_4">4</a>, <a href="#Page_150">150</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Shuldham, C.D., <a href="#Page_125">125</a>, <a href="#Page_126">126</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Siberian Dungeons, The, <a href="#Page_123">123</a>, <a href="#Page_244">244</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Simpleton, The Village, <a href="#Page_11">11</a>, <a href="#Page_155">155</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Skater, The Scientific, <a href="#Page_100">100</a>, <a href="#Page_226">226</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Skeat, Professor, <a href="#Page_127">127</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Solitaire, Central, <a href="#Page_63">63</a>, <a href="#Page_195">195</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Chessboard, <a href="#Page_108">108</a>, <a href="#Page_234">234</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Counter, <a href="#Page_107">107</a>, <a href="#Page_234">234</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Sons, The Four, <a href="#Page_49">49</a>, <a href="#Page_181">181</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Spanish Dungeons, The, <a href="#Page_122">122</a>, <a href="#Page_244">244</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Miser, The, <a href="#Page_24">24</a>, <a href="#Page_164">164</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Speed and Locomotion Puzzles, <a href="#Page_11">11</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Average, <a href="#Page_11">11</a>, <a href="#Page_155">155</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Spiral, Drawing a, <a href="#Page_50">50</a>, <a href="#Page_182">182</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Spot on the Table, The, <a href="#Page_17">17</a>, <a href="#Page_160">160</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Square Numbers, Check for, <a href="#Page_13">13</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; Digital, <a href="#Page_16">16</a>, <a href="#Page_159">159</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; of Veneer, The, <a href="#Page_39">39</a>, <a href="#Page_175">175</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Puzzle, An Easy, <a href="#Page_35">35</a>, <a href="#Page_170">170</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Squares, A Problem in, <a href="#Page_23">23</a>, <a href="#Page_163">163</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Circling the, <a href="#Page_21">21</a>, <a href="#Page_162">162</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Difference of Two, <a href="#Page_167">167</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Magic, <a href="#Page_119">119</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Sum of Two, <a href="#Page_165">165</a>, <a href="#Page_175">175</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Chocolate, <a href="#Page_35">35</a>, <a href="#Page_170">170</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Stalemate, <a href="#Page_106">106</a>, <a href="#Page_232">232</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Stamp-licking, The Gentle Art of, <a href="#Page_91">91</a>, <a href="#Page_217">217</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Star Puzzle, The, <a href="#Page_99">99</a>, <a href="#Page_226">226</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Stars, The Eight, <a href="#Page_89">89</a>, <a href="#Page_215">215</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Forty-nine, <a href="#Page_100">100</a>, <a href="#Page_226">226</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Statical Chess Puzzles, <a href="#Page_88">88</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Sticks, The Eight, <a href="#Page_53">53</a>, <a href="#Page_186">186</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Stonemason's Problem, The, <a href="#Page_25">25</a>, <a href="#Page_165">165</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Stop-watch, The, <a href="#Page_11">11</a>, <a href="#Page_154">154</a>.</span><br />
+<br />
+<span style="margin-left: 2em;"><i>Strand Magazine, The</i>, <a href="#Page_44">44</a>, <a href="#Page_116">116</a>, <a href="#Page_220">220</a>.</span><br />
+<br />
+<span style="margin-left: 2em;"><i>Strand</i> Patience, <a href="#Page_116">116</a>, <a href="#Page_239">239</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Stream, Crossing the, <a href="#Page_112">112</a>, <a href="#Page_236">236</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Strutt, Joseph, <a href="#Page_59">59</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Subtracting Magic Squares, <a href="#Page_124">124</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Sultan's Army, The, <a href="#Page_25">25</a>, <a href="#Page_165">165</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Suppers, The New Year's Eve, <a href="#Page_3">3</a>, <a href="#Page_149">149</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Surname, Find Ada's, <a href="#Page_27">27</a>, <a href="#Page_168">168</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Swastika, The, <a href="#Page_29">29</a>, <a href="#Page_31">31</a>, <a href="#Page_169">169</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">"T" Card Puzzle, The, <a href="#Page_115">115</a>, <a href="#Page_239">239</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Table, The Round, <a href="#Page_80">80</a>, <a href="#Page_205">205</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Table-top and Stools, The, <a href="#Page_38">38</a>, <a href="#Page_173">173</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Tangram Paradox, A, <a href="#Page_43">43</a>, <a href="#Page_178">178</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Target, The Cross, <a href="#Page_84">84</a>, <a href="#Page_210">210</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Tarry, <a href="#Page_112">112</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Tartaglia, <a href="#Page_25">25</a>, <a href="#Page_109">109</a>, <a href="#Page_112">112</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Tea, Mixing the, <a href="#Page_111">111</a>, <a href="#Page_235">235</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Telegraph Posts, The, <a href="#Page_139">139</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Tennis Tournament, A, <a href="#Page_78">78</a>, <a href="#Page_203">203</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Tetrahedron, Building the, <a href="#Page_82">82</a>, <a href="#Page_208">208</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Thief, Catching the, <a href="#Page_19">19</a>, <a href="#Page_161">161</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Thrift, A Study in, <a href="#Page_25">25</a>, <a href="#Page_166">166</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Thompson, W.H., <a href="#Page_232">232</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Ticket Puzzle, The Excursion, <a href="#Page_5">5</a>, <a href="#Page_151">151</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Time Puzzle, A, <a href="#Page_10">10</a>, <a href="#Page_153">153</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; What was the, <a href="#Page_10">10</a>, <a href="#Page_153">153</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Tiring Irons, The, <a href="#Page_142">142</a>, <a href="#Page_247">247</a>.</span><br />
+<br />
+<span style="margin-left: 2em;"><i>Tit-Bits</i>, <a href="#Page_58">58</a>, <a href="#Page_79">79</a>, <a href="#Page_124">124</a>, <a href="#Page_251">251</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Torn Number, The, <a href="#Page_20">20</a>, <a href="#Page_162">162</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Torpedo Practice, <a href="#Page_67">67</a>, <a href="#Page_196">196</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Tour, The Cyclists', <a href="#Page_71">71</a>, <a href="#Page_199">199</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Grand, <a href="#Page_72">72</a>, <a href="#Page_200">200</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Queen's, <a href="#Page_98">98</a>, <a href="#Page_225">225</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Rook's, <a href="#Page_96">96</a>, <a href="#Page_223">223</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Towns, Visiting the, <a href="#Page_70">70</a>, <a href="#Page_198">198</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Trains, The Two, <a href="#Page_11">11</a>, <a href="#Page_155">155</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Treasure Boxes, The Nine, <a href="#Page_24">24</a>, <a href="#Page_164">164</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Trees, The Twenty-one, <a href="#Page_57">57</a>, <a href="#Page_190">190</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Tr&eacute;maux, M., <a href="#Page_133">133</a>, <a href="#Page_135">135</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Triangle, The Dissected, <a href="#Page_38">38</a>, <a href="#Page_173">173</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Triangular Numbers, <a href="#Page_13">13</a>, <a href="#Page_25">25</a>, <a href="#Page_166">166</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; &mdash;&mdash; Check for, <a href="#Page_13">13</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Troublesome Eight, The, <a href="#Page_121">121</a>, <a href="#Page_242">242</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Tube Inspector's Puzzle, The, <a href="#Page_69">69</a>, <a href="#Page_198">198</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; Railway, Heard on the, <a href="#Page_8">8</a>, <a href="#Page_153">153</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Turks and Russians, <a href="#Page_58">58</a>, <a href="#Page_191">191</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Turnings, The Fifteen, <a href="#Page_70">70</a>, <a href="#Page_198">198</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Twickenham Puzzle, The, <a href="#Page_60">60</a>, <a href="#Page_194">194</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Two Pieces Problem, The, <a href="#Page_96">96</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Unclassified Puzzles, <a href="#Page_142">142</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Unicursal and Route Problems, <a href="#Page_68">68</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Union Jack, The, <a href="#Page_50">50</a>, <a href="#Page_69">69</a>, <a href="#Page_197">197</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Vandermonde, A., <a href="#Page_58">58</a>, <a href="#Page_103">103</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Veil, Under the, <a href="#Page_90">90</a>, <a href="#Page_216">216</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Verne, Jules, <a href="#Page_249">249</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Victoria Cross Puzzle, The, <a href="#Page_60">60</a>, <a href="#Page_194">194</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Village, A Wonderful, <a href="#Page_142">142</a>, <a href="#Page_247">247</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Villages, The Three, <a href="#Page_12">12</a>, <a href="#Page_155">155</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Villas, The Eight, <a href="#Page_80">80</a>, <a href="#Page_206">206</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Vortex Rings, <a href="#Page_40">40</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Voter's Puzzle, The, <a href="#Page_75">75</a>, <a href="#Page_202">202</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Wall, The Puzzle, <a href="#Page_52">52</a>, <a href="#Page_184">184</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Wallis, J., <a href="#Page_142">142</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; (Another), <a href="#Page_220">220</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Walls, The Garden, <a href="#Page_52">52</a>, <a href="#Page_185">185</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Wapshaw's Wharf Mystery, The, <a href="#Page_10">10</a>, <a href="#Page_153">153</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">War Puzzle Game, The, <a href="#Page_118">118</a>, <a href="#Page_240">240</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Wassail Bowl, The, <a href="#Page_109">109</a>, <a href="#Page_235">235</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Watch, A Puzzling, <a href="#Page_10">10</a>, <a href="#Page_153">153</a>.</span><br />
+<br /><span class='pagenum'>Pg 258<a name="Page_258" id="Page_258"></a></span>
+<span style="margin-left: 2em;">Water, Gas, and Electricity, <a href="#Page_73">73</a>, <a href="#Page_200">200</a>.</span><br />
+<br />
+<span style="margin-left: 2em;"><i>Weekly Dispatch, The</i>, <a href="#Page_28">28</a>, <a href="#Page_124">124</a>, <a href="#Page_125">125</a>, <a href="#Page_146">146</a>, <a href="#Page_148">148</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Weighing Puzzles, Measuring, Packing, and, <a href="#Page_109">109</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Wheels, Concerning, <a href="#Page_55">55</a>, <a href="#Page_188">188</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Who was First? <a href="#Page_142">142</a>, <a href="#Page_247">247</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Whyte, W.T., <a href="#Page_147">147</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Widow's Legacy, The, <a href="#Page_2">2</a>, <a href="#Page_148">148</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Wife, Find the Man's, <a href="#Page_147">147</a>, <a href="#Page_251">251</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Wilkinson, Rev. Mr., <a href="#Page_193">193</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Wilson, Professor, <a href="#Page_29">29</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Wilson's Poser, <a href="#Page_9">9</a>, <a href="#Page_153">153</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Wine and Water, <a href="#Page_110">110</a>, <a href="#Page_235">235</a>.</span><br />
+<span style="margin-left: 3em;">&mdash;&mdash; The Keg of, <a href="#Page_110">110</a>, <a href="#Page_235">235</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Wotherspoon, G., <a href="#Page_244">244</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Yacht race, The, <a href="#Page_99">99</a>, <a href="#Page_226">226</a>.</span><br />
+<br />
+<span style="margin-left: 2em;">Youthful Precocity, <a href="#Page_1">1</a>, <a href="#Page_148">148</a>.</span><br />
+<br />
+<br />
+<span style="margin-left: 2em;">Zeno, <a href="#Page_139">139</a>.</span><br />
+</p>
+
+<hr style="width: 65%;" /><p><span class='pagenum'>Pg 259<a name="Page_259" id="Page_259"></a></span></p>
+<h2>THE END.</h2>
+
+
+
+
+
+
+
+
+<pre>
+
+
+
+
+
+End of Project Gutenberg's Amusements in Mathematics, by Henry Ernest Dudeney
+
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author	Roger Frank <rfrank@pglaf.org>	2025-10-15 04:49:32 -0700
committer	Roger Frank <rfrank@pglaf.org>	2025-10-15 04:49:32 -0700
commit	5cacbee4d5384b61ff4ab64c5ee8c2059720d141 (patch)
tree	ec3c2c14d0fede301dd1dbc4995cb66451682cf8 /16713-h