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diff --git a/16713.txt b/16713.txt new file mode 100644 index 0000000..232772c --- /dev/null +++ b/16713.txt @@ -0,0 +1,22422 @@ +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] + +Language: English + +Character set encoding: ASCII + +*** 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 + + + + + +[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. + +The most common units used were: + + the Penny, abbreviated: d. (from the Roman penny, denarius) + the Shilling, abbreviated: s. + the Pound, abbreviated: L + +There was 12 Pennies to a Shilling and 20 Shillings to a Pound, so there +was 240 Pennies in a Pound. + +To further complicate things, there were many coins which were various +fractional values of Pennies, Shillings or Pounds. + + Farthing 1/4d. + + Half-penny 1/2d. + + Penny 1d. + + Three-penny 3d. + + Sixpence (or tanner) 6d. + + Shilling (or bob) 1s. + + Florin or two shilling piece 2s. + + Half-crown (or half-dollar) 2s. 6d. + + Double-florin 4s. + + Crown (or dollar) 5s. + + Half-Sovereign 10s. + + Sovereign (or Pound) L1 or 20s. + +This is by no means a comprehensive list, but it should be adequate to +solve the puzzles in this book. + +Exponents are represented in this text by ^, e.g. '3 squared' is 3^2. + +Numbers with fractional components (other than 1/4, 1/2 and 3/4) have a + +symbol separating the whole number component from the fraction. It makes +the fraction look odd, but yeilds correct solutions no matter how it is +interpreted. E.G., 4 and eleven twenty-thirds is 4+11/23, not 411/23 or +4-11/23. + +] + + + + +AMUSEMENTS IN MATHEMATICS + + by + +HENRY ERNEST DUDENEY + + + In Mathematicks he was greater + Than Tycho Brahe or Erra Pater: + For he, by geometrick scale, + Could take the size of pots of ale; + Resolve, by sines and tangents, straight, + If bread or butter wanted weight; + And wisely tell what hour o' th' day + The clock does strike by algebra. + + BUTLER'S _Hudibras_. + + +1917 + + + + +PREFACE + + +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. + +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"--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. + +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--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. + +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 x and y 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--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." + +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. + +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. + +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." + +I have to express my thanks in particular to the proprietors of _The +Strand Magazine_, _Cassell's Magazine_, _The Queen_, _Tit-Bits_, and +_The Weekly Dispatch_ for their courtesy in allowing me to reprint some +of the puzzles that have appeared in their pages. + +THE AUTHORS' CLUB _March_ 25, 1917 + + + + +CONTENTS + + + PREFACE v + ARITHMETICAL AND ALGEBRAICAL PROBLEMS 1 + Money Puzzles 1 + Age and Kinship Puzzles 6 + Clock Puzzles 9 + Locomotion and Speed Puzzles 11 + Digital Puzzles 13 + Various Arithmetical and Algebraical Problems 17 + GEOMETRICAL PROBLEMS 27 + Dissection Puzzles 27 + Greek Cross Puzzles 28 + Various Dissection Puzzles 35 + Patchwork Puzzles 46 + Various Geometrical Puzzles 49 + POINTS AND LINES PROBLEMS 56 + MOVING COUNTER PROBLEMS 58 + UNICURSAL AND ROUTE PROBLEMS 68 + COMBINATION AND GROUP PROBLEMS 76 + CHESSBOARD PROBLEMS 85 + The Chessboard 85 + Statical Chess Puzzles 88 + The Guarded Chessboard 95 + Dynamical Chess Puzzles 96 + Various Chess Puzzles 105 + MEASURING, WEIGHING, AND PACKING PUZZLES 109 + CROSSING RIVER PROBLEMS 112 + PROBLEMS CONCERNING GAMES 114 + PUZZLE GAMES 117 + MAGIC SQUARE PROBLEMS 119 + Subtracting, Multiplying, and Dividing Magics 124 + Magic Squares of Primes 125 + MAZES AND HOW TO THREAD THEM 127 + THE PARADOX PARTY 137 + UNCLASSIFIED PROBLEMS 142 + SOLUTIONS 148 + INDEX 253 + + + + + + +AMUSEMENTS IN MATHEMATICS. + + + + +ARITHMETICAL AND ALGEBRAICAL PROBLEMS. + + "And what was he? + Forsooth, a great arithmetician." + _Othello_, I. i. + + +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--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. + +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. + + + + + + +MONEY PUZZLES. + + "Put not your trust in money, but put your money in trust." + + OLIVER WENDELL HOLMES. + + +1.--A POST-OFFICE PERPLEXITY. + +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? + + +2.--YOUTHFUL PRECOCITY. + +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. + +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." + +Now, how long will it take the reader to say correctly just how much +Fred paid for his rare and refreshing fruit? + + +3.--AT A CATTLE MARKET. + +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, and then you'll have six times as +many animals as I've got here." + +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. + + +4.--THE BEANFEAST PUZZLE. + +A number of men went out together on a bean-feast. There were four +parties invited--namely, 25 cobblers, 20 tailors, 18 hatters, and 12 +glovers. They spent altogether L6, 13s. 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. + + +5.--A QUEER COINCIDENCE. + +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--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--that when they had +finished play each of the seven men had exactly the same amount--two +shillings and eightpence--in his pocket. The puzzle is to find out how +much money each man had with him before he sat down to play. + + +6.--A CHARITABLE BEQUEST. + +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. + + +7.--THE WIDOW'S LEGACY. + +A gentleman who recently died left the sum of L8,000 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? + + +8.--INDISCRIMINATE CHARITY. + +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? + + +9.--THE TWO AEROPLANES. + +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 L600 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? + + +10.--BUYING PRESENTS. + +"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? + + +11.--THE CYCLISTS' FEAST. + + 'Twas last Bank Holiday, so I've been told, + Some cyclists rode abroad in glorious weather. + Resting at noon within a tavern old, + They all agreed to have a feast together. + "Put it all in one bill, mine host," they said, + "For every man an equal share will pay." + The bill was promptly on the table laid, + And four pounds was the reckoning that day. + But, sad to state, when they prepared to square, + 'Twas found that two had sneaked outside and fled. + So, for two shillings more than his due share + Each honest man who had remained was bled. + They settled later with those rogues, no doubt. + How many were they when they first set out? + + +12.--A QUEER THING IN MONEY. + +It will be found that L66, 6s. 6d. 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? + + +13.--A NEW MONEY PUZZLE. + +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 L98,765, 4s. 31/2d. 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--pounds, shillings, pence, +and farthings--and the nought may not be used. It requires just a little +judgment and thought. + + +14.--SQUARE MONEY. + +"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. + + +15.--POCKET MONEY. + +What is the largest sum of money--all in current silver coins and no +four-shilling piece--that I could have in my pocket without being able +to give change for a half-sovereign? + +16.--THE MILLIONAIRE'S PERPLEXITY. + +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--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. + +17.--THE PUZZLING MONEY-BOXES. + +Four brothers--named John, William, Charles, and Thomas--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 2s. more in his box than at +present, if William had had 2s. 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. + +Now, when I add that all four boxes together contained 45s., 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. + + +18.--THE MARKET WOMEN. + +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--2s. 21/2d. 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. + + +19.--THE NEW YEAR'S EVE SUPPERS. + +The proprietor of a small London cafe 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. + + +20.--BEEF AND SAUSAGES. + +"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?" + +"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." + +"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." + + +21.--A DEAL IN APPLES. + +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? + + +22.--A DEAL IN EGGS. + +A man went recently into a dairyman's shop to buy eggs. He wanted them +of various qualities. 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. + + +23.--THE CHRISTMAS-BOXES. + +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 L1, 10s. 1d. 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. + + +24.--A SHOPPING PERPLEXITY. + +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--both different? + + +25.--CHINESE MONEY. + +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. + +[Illustration] + +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. + + +26.--THE JUNIOR CLERK'S PUZZLE. + +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--that is, commencing at the rate of L50 a +year, payable half-yearly. Moggs had a yearly rise of L10, and Snoggs +was offered the same, only he asked, for reasons that do not concern our +puzzle, that he might take his rise at L2, 10s. half-yearly, to which +his employer (not, perhaps, unnaturally!) had no objection. + +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 L268, 15s. How much had each saved? The +question of interest can be ignored. + + +27.--GIVING CHANGE. + +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. + + +28.--DEFECTIVE OBSERVATION. + +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. 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. + + +29.--THE BROKEN COINS. + +A man had three coins--a sovereign, a shilling, and a penny--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--that is, that the sovereign was worth a +pound, the shilling worth a shilling, and the penny worth a penny--what +proportion of each coin has been lost if the value of the three +remaining fragments is exactly one pound? + + +30.--TWO QUESTIONS IN PROBABILITIES. + +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. + + +31.--DOMESTIC ECONOMY. + +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 L190 +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--will you believe it?--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?" + +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--something absurdly +in excess of the correct answer! + + +32.--THE EXCURSION TICKET PUZZLE. + +When the big flaming placards were exhibited at the little provincial +railway station, announcing that the Great ---- 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. + +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." + +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. + + +33.--PUZZLE IN REVERSALS. + +Most people know that if you take any sum of money in pounds, shillings, +and pence, in which the number of pounds (less than L12) 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 L12, 18s. 11d. But if we omit the condition, "less than L12," and +allow nought to represent shillings or pence--(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 L14, +15s. 3d. it may be written L3, 16s. 2d., which is the same as L3, 15s. +14d. + + +34.--THE GROCER AND DRAPER. + +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 +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? + +35.--JUDKINS'S CATTLE. + +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? + +36.--BUYING APPLES. + +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--I'll take _one_!" + +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. + +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?" + +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. + +An old woman had apples of three sizes for sale--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? + + +37.--BUYING CHESTNUTS. + +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. + +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? + + +38.--THE BICYCLE THIEF. + +Here is a little tangle that is perpetually cropping up in various +guises. A cyclist bought a bicycle for L15 and gave in payment a cheque +for L25. The seller went to a neighbouring shopkeeper and got him to +change the cheque for him, and the cyclist, having received his L10 +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 L25 +from a friend, as the cyclist forgot to leave his address, and could not +be found. Now, as the bicycle cost the salesman L11, how much money did +he lose altogether? + + +39.--THE COSTERMONGER'S PUZZLE. + +"How much did yer pay for them oranges, Bill?" + +"I ain't a-goin' to tell yer, Jim. But I beat the old cove down +fourpence a hundred." + +"What good did that do yer?" + +"Well, it meant five more oranges on every ten shillin's-worth." + +Now, what price did Bill actually pay for the oranges? There is only one +rate that will fit in with his statements. + + + + +AGE AND KINSHIP PUZZLES. + + "The days of our years are threescore years and ten." + + --_Psalm_ xc. 10. + +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 +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. + + +40.--MAMMA'S AGE. + +Tommy: "How old are you, mamma?" + +Mamma: "Let me think, Tommy. Well, our three ages add up to exactly +seventy years." + +Tommy: "That's a lot, isn't it? And how old are you, papa?" + +Papa: "Just six times as old as you, my son." + +Tommy: "Shall I ever be half as old as you, papa?" + +Papa: "Yes, Tommy; and when that happens our three ages will add up to +exactly twice as much as to-day." + +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--" + +Mamma: "Supposing, Tommy, we talk about bed. Come along, darling. You'll +have a headache." + +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? + + +41.--THEIR AGES. + +"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." + + +42.--THE FAMILY AGES. + +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? + + +43.--MRS. TIMPKINS'S AGE. + +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?" + +Angelina: "Then how old was Mrs. Timpkins on the wedding day?" + +Can you answer Angelina's question? + + +44--A CENSUS PUZZLE. + +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! + + +45.--MOTHER AND DAUGHTER. + +"Mother, I wish you would give me a bicycle," said a girl of twelve the +other day. + +"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." + +Now, the mother's age is forty-five years. When may the young lady +expect to receive her present? + + +46.--MARY AND MARMADUKE. + +Marmaduke: "Do you know, dear, that in seven years' time our combined +ages will be sixty-three years?" + +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." + +Now, what are the ages of Mary and Marmaduke? + + +47--ROVER'S AGE. + +"Now, then, Tommy, how old is Rover?" Mildred's young man asked her +brother. + +"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." + +Can you tell Rover's age? + + +48.--CONCERNING TOMMY'S AGE. + +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--I forget the +year--my only sister, Ann, happened to be just one-quarter the age 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?" + +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. + + +49.--NEXT-DOOR NEIGHBOURS. + +There were two families living next door to one another at Tooting +Bec--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? + + +50.--THE BAG OF NUTS. + +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 171/2 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. + + +51.--HOW OLD WAS MARY? + +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? + +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. + + +52.--QUEER RELATIONSHIPS. + +"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." + +"The man married his deceased wife's sister under the recent Act?" put +in the Lawyer. + +"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." + +"He is married to her and she is not married to him!" said the Doctor. + +"Quite so. And the child is the legitimate son of his father, but the +illegitimate son of his mother." + +"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. + +"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." + +"And this reminds me," went on the Parson, "of a man in my parish who +married the sister of his widow. This man--" + +"Stop a moment, sir," said the Professor. "Married the sister of his +widow? Do you marry dead men in your parish?" + +"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." + +The Lawyer was the first to get at the heart of the mystery. What was +his solution? + + +53.--HEARD ON THE TUBE RAILWAY. + +First Lady: "And was he related to you, dear?" + +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." + +First Lady: "Oh, indeed!" (But you could see that she was not much +wiser.) + +How was the gentleman related to the Second Lady? + + +54.--A FAMILY PARTY. + +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? + + +55.--A MIXED PEDIGREE. + +Joseph Bloggs: "I can't follow it, my dear boy. It makes me dizzy!" + +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--" + +But Mr. Bloggs refused to hear any more. Can the reader show how this +extraordinary triple relationship might have come about? + + +56.--WILSON'S POSER. + +"Speaking of perplexities--" said Mr. Wilson, throwing down a magazine +on the table in the commercial room of the Railway Hotel. + +"Who was speaking of perplexities?" inquired Mr. Stubbs. + +"Well, then, reading about them, if you want to be exact--it just +occurred to me that perhaps you three men may be interested in a little +matter connected with myself." + +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. + +"What is the difficulty?" asked Mr. Packhurst. + +"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--reckless, I should call him--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--" + +"Here, stop, Wilson! What was your nephew doing there? You said your +uncle," interrupted Mr. Stubbs. + +"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--" + +"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." + +"What I said is quite right. Parker called out to my uncle to do +something or other, when my nephew--" + +"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?" + +"Certainly. I thought I made that clear. Where was I? Well, my nephew +shouted back to Parker--" + +"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?" + +"Of course. I said at the start that it only carried two." + +"Then what in the name of aerostation do you mean by saying that there +were three persons on board?" shouted Mr. Stubbs. + +"Who said there were three?" + +"You have told us that Parker, your uncle, and your nephew went up on +this blessed flying-machine." + +"That's right." + +"And the thing would only carry two!" + +"Right again." + +"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." + +"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." + +"He is no relation to me whatever." + +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. + +"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." + +"What!" exclaimed all three together. + +"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." + +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. + + + + +CLOCK PUZZLES. + + "Look at the clock!" + + _Ingoldsby Legends_. + + +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. + +"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 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." + +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. + + +57.--WHAT WAS THE TIME? + +"I say, Rackbrane, what is the time?" an acquaintance asked our friend +the professor the other day. The answer was certainly curious. + +"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." + +What was the time of day when the professor spoke? + + +58.--A TIME PUZZLE. + +How many minutes is it until six o'clock if fifty minutes ago it was +four times as many minutes past three o'clock? + + +59.--A PUZZLING WATCH. + +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? + + +60.--THE WAPSHAW'S WHARF MYSTERY. + +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. + +What was the exact time at which the watchman's watch stopped? The watch +is, of course, assumed to have been an accurate one. + + +61.--CHANGING PLACES. + +[Illustration] + +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? + + +62.--THE CLUB CLOCK. + +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 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? + +[Illustration] + + +63.--THE STOP-WATCH. + +[Illustration] + +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? + + +64.--THE THREE CLOCKS. + +On Friday, April 1, 1898, three new clocks were all set going precisely +at the same time--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? + + +65.--THE RAILWAY STATION CLOCK. + +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? + + +66.--THE VILLAGE SIMPLETON. + +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:-- + +"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." + +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. + + + + +LOCOMOTION AND SPEED PUZZLES. + +"The race is not to the swift."--_Ecclesiastes_ ix. II. + + +67.--AVERAGE SPEED. + +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. + + +68.--THE TWO TRAINS. + +I put this little question to a stationmaster, and his correct answer +was so prompt that I am convinced there is no necessity to seek talented +railway officials in America or elsewhere. + +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? + + +69.--THE THREE VILLAGES. + +I set out the other day to ride in a motor-car from Acrefield to +Butterford, but by mistake I took the road going _via_ 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. + + +70.--DRAWING HER PENSION. + +"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?" + +[Illustration] + + +71.--SIR EDWYN DE TUDOR. + +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. + + +72.--THE HYDROPLANE QUESTION. + +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 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--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. + + +73.--DONKEY RIDING. + +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:--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? + + +74.--THE BASKET OF POTATOES. + +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--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. + + +75.--THE PASSENGER'S FARE. + +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 L3. 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? + + + + +DIGITAL PUZZLES. + + "Nine worthies were they called." + DRYDEN: _The Flower and the Leaf._ + +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--the first that occurs to me. + +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 squared + n)/2) if its digital root be 2, 4, 5, 7, or 8. + + +76.--THE BARREL OF BEER. + +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 bought them, +without manipulating in any way the contents. + +[Illustration: + + ( 15 Gals ) + + (31 Gals) (19 Gals) + + (20 Gals) (16 Gals) (18 Gals) + +] + + +77.--DIGITS AND SQUARES. + +[Illustration: + + +---+---+---+ + | 1 | 9 | 2 | + +---+---+---+ + | 3 | 8 | 4 | + +---+---+---+ + | 5 | 7 | 6 | + +---+---+---+ + +] + +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? + + +78.--ODD AND EVEN DIGITS. + +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. + + +79.--THE LOCKERS PUZZLE. + +[Illustration: + + A B C + ================== ================== ================== + | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | + | | | | | | | | | | | | | | | | | | | | | | | | + | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | + | | | | | | + | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | + | | | | | | | | | | | | | | | | | | | | | | | | + | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | + | | | | | | + ================== ================== ================== + | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | + | | | | | | | | | | | | | | | | | | | | | | | | + | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | + ------------------ ------------------ ------------------ + +] + +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. + +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. + + +80.--THE THREE GROUPS. + +There appeared in "Nouvelles Annales de Mathematiques" 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 x +483 = 5,796. I now also propose to include the cases where there are +one, four, and four digits, such as 4 x 1,738 = 6,952. Can you find all +the possible solutions in both cases? + + +81.--THE NINE COUNTERS. + +[Illustration: + + (1)(5)(8) (7)(9) + (2)(3) (4)(6) + +] + +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. + + + + +82.--THE TEN COUNTERS. + +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. + + +83.--DIGITAL MULTIPLICATION. + +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, 7 x 658 and 14 x 329 contain all the digits +once, and the product in each case is the same--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? + + +84.--THE PIERROT'S PUZZLE. + +[Illustration] + +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. + + +85.--THE CAB NUMBERS. + +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 _highest possible_) +containing also all the nine digits? The nought is not allowed anywhere. + + +86.--QUEER MULTIPLICATION. + +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 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. + + +87.--THE NUMBER-CHECKS PUZZLE. + +[Illustration] + +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--and you +have to be on the alert to seize it when it does so happen. + +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--8 9 0 7--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. + + +88.--DIGITAL DIVISION. + +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. + + +89.--ADDING THE DIGITS. + +If I write the sum of money, L987, 5s. 41/2d., 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. + + +90.--THE CENTURY PUZZLE. + +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+5742/638. 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? + + +91.--MORE MIXED FRACTIONS. + +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. + + +92.--DIGITAL SQUARE NUMBERS. + +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--(I) the smallest possible, and (II) the largest +possible? + + +93.--THE MYSTIC ELEVEN. + +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. + + +94.--THE DIGITAL CENTURY. + +1 2 3 4 5 6 7 8 9 = 100. + +It is required to place arithmetical signs between the nine figures so +that they shall equal 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 x) will thus count as +two strokes, the sign of subtraction (-) as one stroke, the sign of +division (/) as three, and so on. + + +95.--THE FOUR SEVENS. + +[Illustration] + +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+9/9, but the four 7's call for rather more +ingenuity. Can you discover the little trick? + + +96.--THE DICE NUMBERS. + +[Illustration] + +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? + + + + +VARIOUS ARITHMETICAL AND ALGEBRAICAL PROBLEMS. + + "Variety's the very spice of life, + That gives it all its flavour." + + COWPER: _The Task._ + + + + +97.--THE SPOT ON THE TABLE. + +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. + +[Illustration] + +"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?" + +The boy was overheard to tell a friend, "It 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. + + +98.--ACADEMIC COURTESIES. + +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. + + +99.--THE THIRTY-THREE PEARLS. + +[Illustration] + +"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 L100 more +than the preceding one, right up to the big pearl. From the other end +the pearls increase in value by L150 up to the large pearl. The whole +string is worth L65,000. What is the value of that large pearl?" + +"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." + +"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." + +"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." + +"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." + +"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!" + +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. + + +100.--THE LABOURER'S PUZZLE. + +Professor Rackbrane, during one of his rambles, chanced to come upon a +man digging a deep hole. + +"Good morning," he said. "How deep is that hole?" + +"Guess," replied the labourer. "My height is exactly five feet ten +inches." + +"How much deeper are you going?" said the professor. + +"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." + +Rackbrane now asks if you could tell how deep that hole would be when +finished. + + +101.--THE TRUSSES OF HAY. + +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? + + +102.--MR. GUBBINS IN A FOG. + +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 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. + +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? + + +103.--PAINTING THE LAMP-POSTS. + +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? + + +104.--CATCHING THE THIEF. + +"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?" + +"Yes, sir." + +"And you swear that he takes eight steps to your five?" + +"That is so." + +"Then I ask you, constable, as an intelligent man, to explain how you +ever caught him, if that is the case?" + +"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." + +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? + + +105.--THE PARISH COUNCIL ELECTION. + +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. + + +106.--THE MUDDLETOWN ELECTION. + +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? + + +107.--THE SUFFRAGISTS' MEETING. + +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? + + +108.--THE LEAP-YEAR LADIES. + +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. + +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? + + +109.--THE GREAT SCRAMBLE. + +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. + +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 gathered up one-third, David got two-sevenths, while +Charlie and Edgar divided equally what was left of that stock. + +[Illustration] + +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? + + +110.--THE ABBOT'S PUZZLE. + +The first English puzzlist whose name has come down to us was a +Yorkshireman--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?" + +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? + + +111.--REAPING THE CORN. + +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? + + +112.--A PUZZLING LEGACY. + +A man left a hundred acres of land to be divided among his three +sons--Alfred, Benjamin, and Charles--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? + +113.--THE TORN NUMBER. + +[Illustration] + +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 0 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. + + +114.--CURIOUS NUMBERS. + +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--the smallest possible +numbers. What are they? + + +115.--A PRINTER'S ERROR. + +In a certain article a printer had to set up the figures 5^4x2^3, 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 5^4x2^3 +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? + + +116.--THE CONVERTED MISER. + +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. + +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? + +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. + + +117.--A FENCE PROBLEM. + +[Illustration] + +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? + +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 (161/2 ft.): that is to say, there are fourteen rails +to the pole, lineal measure. Now, what must be the size of the field? + + +118.--CIRCLING THE SQUARES. + +[Illustration] + +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--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. + +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. + + +119.--RACKBRANE'S LITTLE LOSS. + +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 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? + + +120.--THE FARMER AND HIS SHEEP. + +[Illustration] + +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." + +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. + + +121.--HEADS OR TAILS. + +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--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? + + +122.--THE SEE-SAW PUZZLE. + +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. + +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. + +Now, what was that boy's weight, if a brick weighs equal to a +three-quarter brick and three-quarters of a pound? + + +123.--A LEGAL DIFFICULTY. + +"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--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?" + + +124.--A QUESTION OF DEFINITION. + +"My property is exactly a mile square," said one landowner to another. + +"Curiously enough, mine is a square mile," was the reply. + +"Then there is no difference?" + +Is this last statement correct? + + +125.--THE MINERS' HOLIDAY. + +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? + + +126.--SIMPLE MULTIPLICATION. + +If we number six cards 1, 2, 4, 5, 7, and 8, and arrange them on the +table in this order:-- + + 1 4 2 8 5 7 + +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? + + +127.--SIMPLE DIVISION. + +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. + + +128.--A PROBLEM IN SQUARES. + +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. + + + + +129.--THE BATTLE OF HASTINGS. + +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. + +"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!' 'Godemite!'" + +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. + +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? + +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 x 4 squared + 1 = 31 squared, +and 62 x 8 squared + 1 = 63 squared. 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. + + +130.--THE SCULPTOR'S PROBLEM. + +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 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. + +[Illustration] + + +131.--THE SPANISH MISER. + +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. + +[Illustration] + +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? + + +132.--THE NINE TREASURE BOXES. + +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--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. + +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 increased 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. + +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? + + +133.--THE FIVE BRIGANDS. + +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? + +There are a good many equally correct answers to this question. Here is +one of them: + + A 6 x 12 = 72 + B 12 x 3 = 36 + C 17 x 1 = 17 + D 120 x 1/2 = 60 + E 45 x 1/3 = 15 + ___ ___ + 200 200 + +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--only doubloons in every case. + +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? + + +134.--THE BANKER'S PUZZLE. + +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. + + +135.--THE STONEMASON'S PROBLEM. + +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 + 8 = 9 is a square, 1 + +8 + 27 = 36 is a square, 1 + 8 + 27 + 64 = 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. + +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 _declined to take the single block_ because it contained a +flaw. What was the smallest possible number of blocks of stone that the +mason had to supply? + + +136.--THE SULTAN'S ARMY. + +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--81 and 49, or 121 +and 9. Of course, all the men must be used on every occasion. + + +137.--A STUDY IN THRIFT. + +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--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. + +Now, half the sum of any number and its square is always a triangular +number. Thus half of 2 + 2 squared = 3; half of 3 + 3 squared = 6; half of 4 + +4 squared = 10; half of 5 + 5 squared= 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 squared, 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 squared, 3 squared, 6 squared, 10 squared, 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--that is, the squares of 1, +3, 6, 10, 15, 21, 28, etc., any of which numbers we have seen will form +a triangle. + +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. + +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 L36 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. + +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? + +What you have to do is to find five numbers, the smallest possible, +higher than 36, that can be displayed in the four ways--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. + + +138.--THE ARTILLERYMEN'S DILEMMA. + +[Illustration: [Pyramid of cannon-balls]] + + + MMMMMMMr + MM MM: + M 0 rWZX + M : MWM + aX ,BM + M 0M M + aMMMM2MW 02 MMWMMr + ZM. M@M 8MM 7 XM2 + MS2 M.MMMWMMMM MM + M MX iMM M7W + 8 . M r W M@ Z;M + M 0r ; M M M W + 22 W M @ M M M.M2WMMMMZ + ;MM@X:7MMMB; MMM ZM M:MM0;8: ,MS + Ma 8 MMMMMMMi rM 2MMMMMM MB + M 7 XM, ,: BMM: r7S .,MM MM MB + M i ,M , 2 ; aMMMMMMMMM XM; MZM + M . M 7 M . Z M M M8 + M M S M .0 M 8MM aMi: + MMMM7M ,7 .iM X M @ aZ M M 8 ,@MMMMBMMMa + SMW 7M,XZ@MM M 8M M .M MMMM@X MMr + Ma MMMMMMMMM@ M .WM M @WM7WMM .WX MZS + M 8M :MMMWMMMM 8X MMMBMMM7 7aM 2MM + r, 8r ZM2 Mr2 aMM; Mai :MS :iM ZiM @MX + M M . M Wr.MMMaBMMMB M M MZ. ,M MMZ + Mr M M B0 Z 2S iM S XM 7 WMM + MM @.M M M W M. M M 0;M2M;MMMM: + WW8aMM M S@ M M M : MaMMMMMM + MM0W;MZM: M i M M MM MMMZMBZa0ar + B20rMMM Si i BW MMM02 7MM0 2MMM + MMMMMMMM . M SM@aiMM20BWM + XM 0ZMMM:MMMMW; r. + 0WMBM2XrB: . + + +"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 _must_ 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 _pile_ one ball into a pyramid!" Is it +really possible to obey both orders? + + +139.--THE DUTCHMEN'S WIVES. + +I wonder how many of my readers are acquainted with the puzzle of the +"Dutchmen's Wives"--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 _Ladies' Diary_ 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. + +Three Dutchmen, named Hendrick, Elas, and Cornelius, and their wives, +Gurtruen, Katruen, 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 Katruen, +and Elas eleven more than Gurtruen. Now, what was the name of each man's +wife? + +[Illustration] + + +140.--FIND ADA'S SURNAME. + +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 8s. 51/4d. more than her daughter. Mrs. Robinson spent 6s. +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--one of the girls. Annie bought 16 yards more than Mary and spent +L3, 0s. 8d. more than Emily. The Christian name of the other girl was +Ada. Now, what was her surname? + + +141.--SATURDAY MARKETING. + +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 1s., Mary spent 2s., Jane spent 3s., and Kate spent +4s. 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? + + + + +GEOMETRICAL PROBLEMS. + + "God geometrizes continually." + + PLATO. + +"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 aesthetic +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. + + + +DISSECTION PUZZLES. + + + "Take him and cut him out in little stars." + + _Romeo and Juliet_, iii. 2. + +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 +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. + +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 _Weekly Dispatch_, 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. + +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--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--or six pieces +in all. It is, therefore, not a solution in three pieces. + +[Illustration: Fig. 1] + +[Illustration: Fig. 2] + +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--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. + +[Illustration: Fig. 3] + +[Illustration: Fig. 4] + +Before presenting a varied series of cutting-out puzzles, some very easy +and others difficult, I propose to consider one family alone--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. + + +GREEK CROSS PUZZLES. + + "To fret thy soul with crosses." + + SPENSER. + + "But, for my part, it was Greek to me." + + _Julius Caesar_, i. 2. + +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 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 _boun_ +of this kind. The cross and ball, so frequently found on Egyptian +figures, is a circle and the _tau_ cross. The circle signified the +eternal preserver of the world, and the T, named from the Greek letter +_tau_, is the monogram of Thoth, the Egyptian Mercury, meaning wisdom. +This _tau_ 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--the male and +female spirit of everything that was everlasting. + +[Illustration: Fig. 5.] + +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 "--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 a +to b. 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. + +[Illustration: Fig. 6] + +[Illustration: Fig. 7] + +[Illustration: Fig. 8] + +[Illustration: Fig. 9] + +[Illustration: Fig. 10] + +[Illustration: Fig. 11] + +It will be seen that every one of these puzzles has its reverse +puzzle--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. + +Cut a square into five pieces that will form two separate Greek crosses +of _different sizes_. 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. + +[Illustration: FIG. 12.] + +[Illustration: FIG. 13.] + +Cut a square into five pieces that will form two separate Greek crosses +of exactly the _same size_. 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. + +[Illustration: FIG. 14.] + +[Illustration: FIG. 15.] + +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 c to d is exactly +the same as the distance a to b 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 a b is parallel to c d. If +you place the point c exactly over a the lines will indicate the +solution (Figs. 10 and 11). If you place c 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 c 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 c need not necessarily be +placed within the dotted square. It may be placed, for example, at point +e to give a solution in four pieces. Here the joins at a and f may +be as slender as you like. Yet if you once get over the edge at a or +f 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. + +[Illustration: FIG. 16.] + +[Illustration: FIG. 17.] + +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 a to b in Fig. 16. This must necessarily be so, and +I will presently try to make the point quite clear. + +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--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. + +[Illustration: FIG. 18.] + +[Illustration: FIG. 19.] + +[Illustration: FIG. 20.] + +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, 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. + +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. + +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--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. + +[Illustration: FIG. 21.] + +[Illustration: FIG. 22.] + +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--not from one diagram--in order to produce a perfect result. + +[Illustration: FIG. 23.] + +[Illustration: FIG. 24.] + +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. + +[Illustration: FIG. 25.] + +[Illustration: FIG. 26.] + +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. + +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 triangle, 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. + +[Illustration: FIG. 27.] + +[Illustration: FIG. 28.] + +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. + +[Illustration: FIG. 29.] + +[Illustration: FIG. 30.] + +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 _ab_ equal to the side _cd_ the directions of +the cuts are very evident. From what we have just been considering, you +will at once see why _bc_ 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. + +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. + +[Illustration: FIG. 31.] + +[Illustration: FIG. 32.] + +[Illustration: FIG. 33.] + +[Illustration: FIG. 34.] + +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 a bad knot in the position shown in Fig. 33--a +knot that we must not attempt to cut through--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. + +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. + +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. + +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. + +I give another puzzle in Fig. 36. The dotted lines are added merely to +show the correct proportions of the figure--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. + +[Illustration: FIG. 35.] + +[Illustration: FIG. 36.] + +The solution to the first of the two puzzles last given--to cut a +rectangle of the shape of a half-square into three pieces that will form +a Greek cross--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. + +[Illustration: FIG. 37.] + +[Illustration: FIG. 38.] + +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--a square number--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. + +[Illustration: FIG. 39] + +[Illustration: FIG. 40] + +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. + +[Illustration: FIG. 41] + +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. + +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. + + +142.--THE SILK PATCHWORK. + +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. + + +143.--TWO CROSSES FROM ONE. + +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. + + +144.--THE CROSS AND THE TRIANGLE. + +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"). + + +145.--THE FOLDED CROSS. + +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. + + + + +VARIOUS DISSECTION PUZZLES. + + +We will now consider a small miscellaneous selection of cutting-out +puzzles, varying in degrees of difficulty. + + +146.--AN EASY DISSECTION PUZZLE. + +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. + + +147.--AN EASY SQUARE PUZZLE. + +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. + + +148.--THE BUN PUZZLE. + +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." + + +149.--THE CHOCOLATE SQUARES. + +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. + + +150.--DISSECTING A MITRE. + +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 four pieces, based +on what is known as the "step principle," but it is a fallacy. + +[Illustration] + +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. + +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 _not_ 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 x 63. The +steps must be 101/2 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 +101/2 in. the side 63 in. Hence our final rectangle must be 72 in. x 731/2 +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+5/7 +(instead of 63), then the step method would apply. For the steps would +then be 10+2/7 in. in height and 12 in. in breadth. Note that 61+5/7 x +84 = the square of 72. At present no solution has been found in four +pieces, and I do not believe one possible. + + +151.--THE JOINER'S PROBLEM. + +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. + +[Illustration] + +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 of +material. How should he go to work? How many pieces would you require? + + +152.--ANOTHER JOINER'S PROBLEM. + +[Illustration] + +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. + + +153--A CUTTING-OUT PUZZLE. + +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. + +[Illustration] + + +154.--MRS. HOBSON'S HEARTHRUG. + +[Illustration] + +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 x 27 (it does not matter +whether we say inches or yards), and each piece cut away measured 12 and +6 on the outside. + + +155.--THE PENTAGON AND SQUARE. + +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 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. + +[Illustration] + +Having formed your pentagon, the puzzle is to cut it into the fewest +possible pieces that will fit together and form a perfect square. + +[Illustration] + + +156.--THE DISSECTED TRIANGLE. + +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--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? + +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--all +equilateral. + + +157.--THE TABLE-TOP AND STOOLS. + +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. + +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 descendant of the +original boy who had most reason to be interested in the matter. + +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. + +[Illustration] + +Now what was the method the boy proposed? + +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. + + +158.--THE GREAT MONAD. + +[Illustration] + +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--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. + +I will introduce the Monad in its elementary form. Here are three easy +questions respecting this great symbol:-- + +(I.) Which has the greater area, the inner circle containing the Yin and +the Yan, or the outer ring? + +(II.) Divide the Yin and the Yan into four pieces of the same size and +shape by one cut. + +(III.) Divide the Yin and the Yan into four pieces of the same size, but +different shape, by one straight cut. + + +159.--THE SQUARE OF VENEER. + +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 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. + +[Illustration] + + +160.--THE TWO HORSESHOES. + +[Illustration] + +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 _Victory_. 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. + +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. + +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. + + +161.--THE BETSY ROSS PUZZLE. + +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. + +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. + + +162.--THE CARDBOARD CHAIN. + +[Illustration] + +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. + + +163.--THE PAPER BOX. + +It may be interesting to introduce here, though it is not strictly a +puzzle, an ingenious method for making a paper box. + +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. + +This box will be found to serve excellently for the production of vortex +rings. These rings, 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. + +[Illustration: + + .-----------.-----------.-----------.-----------. + | . . .|||||||. .|||||||. . . | + | . .|||. .|||. . . | + | . . . . | + | . . . . . . . . | + | . . . . . . . . | + . . . . . + ||. . . . . /|\ . .|| + ||||. . . . . \|/ . .|||| + ||||||. . . .|||||| + ||||. . . . . . . .|||| + ||. . . . . . . .|| + . . . . . + ||. . . . . . . .|| + ||||. . . . . . . .|||| + ||||||. . . .|||||| + ||||. . . . . . . .|||| + ||. . . . . . . .|| + . . . . . + | . . . . . . . . | + | . . . . . . . . | + | . . . . | + | . .|||. .|||. . . | + | . . .|||||||. .|||||||. . . | + .-----------.-----------.-----------.-----------. + +] + +[Illustration] + + +164.--THE POTATO PUZZLE. + +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? + +[Illustration: + + -------- + / \ 1/ \ + / \ 2 \/ 3 / \ + / \ /\ / \ + / \ 4 \/ 5\/ 6 / \ + | \ /\ /\ / | + \ 7\/ 8\/ 9\/10 / + \ /\ /\ /\ / + \/11\/12\/13\/ + \ /\ /\ / + \/14\/15\/ + \ /\ / + \/16\/ + ----- + +] + +The illustration shows how to make sixteen pieces. This can, of course, +be easily beaten. + + +165.--THE SEVEN PIGS. + +[Illustration] + + +------------------------------+ + | | + | P | + | | + | P | + | P | + | P | + | P | + | P | + | 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. + +[Illustration] + +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--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 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? + + +166.--THE LANDOWNER'S FENCES. + +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. + + +167.--THE WIZARD'S CATS. + +[Illustration] + +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 +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. + + +168.--THE CHRISTMAS PUDDING. + +[Illustration] + +"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. + +"'Problem: To find the contents,' I suppose," said the Eton boy. + +"No; the proof of that is in the eating. I will read you the +conditions." + +"'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.'" + +"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. + +"It is just a puzzle--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. + + +169.--A TANGRAM PARADOX. + +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. + +[Illustration] + +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. + +A few years ago a little book came into my possession, from the library +of the late Lewis Carroll, entitled _The Fashionable Chinese Puzzle_. 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. + +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--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. + +As I have referred to the author of _Alice in Wonderland_, I give also +my designs of the March 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 _The Strand Magazine_ for November, 1908. + +[Illustration: 2] + +[Illustration: 3] + +[Illustration: 4] + +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:--"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." + +[Illustration: 5] + +"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 _ch'i ch'iao t'u_; 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 _t'an_, 'to extend'; or +_t'ang_, 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." + +[Illustration: 6] + +[Illustration: 7] + +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:--"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+1/8 in. square. My great uncle, +----, 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." + +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 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. + +[Illustration: 8] + +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. + +[Illustration: 9] + +[Illustration: 10] + +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. + +[Illustration: 11] + +[Illustration: 12] + +Finally, I give an example from the many curious paradoxes that one +happens upon in manipulating Tangrams. I show designs of 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? + +[Illustration: 13] + +[Illustration: 14] + +[Illustration: 15] + +[Illustration: 16] + + + +PATCHWORK PUZZLES. + +"Of shreds and patches."--_Hamlet_, iii. 4. + + +170.--THE CUSHION COVERS. + +[Illustration] + +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? + + +171.--THE BANNER PUZZLE. + +[Illustration] + +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. + + +172.--MRS. SMILEY'S CHRISTMAS PRESENT. + +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. + +[Illustration] + + +173.--MRS. PERKINS'S QUILT. + +[Illustration] + +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. + + +174.--THE SQUARES OF BROCADE. + +[Illustration] + +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--both of the same design, a delicate +chequered pattern. + +"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 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." + +[Illustration] + +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? + + +175--ANOTHER PATCHWORK PUZZLE. + +[Illustration] + +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--one 12 x 12 and the +other 5 x 5. She proposes to join them together and make one square +patchwork quilt, 13 x 13, but, of course, she will not cut any of the +material--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? + + +176.--LINOLEUM CUTTING. + +[Illustration] + +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 x 10, so +that the pattern shall properly match, and so that the larger piece +shall have as small a portion as possible cut from it. + + +177.--ANOTHER LINOLEUM PUZZLE. + +[Illustration] + +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. + + + + +VARIOUS GEOMETRICAL PUZZLES. + + "So various are the tastes of men." + MARK AKENSIDE. + + +178.--THE CARDBOARD BOX. + +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? + + +179.--STEALING THE BELL-ROPES. + +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? + + +180.--THE FOUR SONS. + +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. + +[Illustration] + + +181.--THE THREE RAILWAY STATIONS. + +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. + +"How far have you to drive to your place from the railway station?" +asked the stranger. + +"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." + +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? + + +182.--THE GARDEN PUZZLE. + +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, "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? + + +183.--DRAWING A SPIRAL. + +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? + +[Illustration] + + +184.--HOW TO DRAW AN OVAL. + +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. + + +185.--ST. GEORGE'S BANNER. + +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. + +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? + +[Illustration] + + +186.--THE CLOTHES LINE PUZZLE. + +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? + + +187.--THE MILKMAID PUZZLE. + +[Illustration] + +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." + +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. + + "Where are you going to, my pretty maid?" + "Down to the river, sir," she said. + "I'll _not_ choose your dairy, my pretty maid." + "Nobody axed you, sir," she said. + +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. + +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? + + +188.--THE BALL PROBLEM. + +[Illustration] + +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. + +"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?" + +The boy at once gave the correct answer, and then put this little +question to the mason:-- + +"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?" + +The stonemason could not give an answer. Could you have replied +correctly to the mason's and the boy's questions? + + +189.--THE YORKSHIRE ESTATES. + +[Illustration] + +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. + +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? + + +190.--FARMER WURZEL'S ESTATE. + +[Illustration] + +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. + +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 four intervening triangular +fields. The puzzle is to discover what was then the whole area of his +estate. + + +191.--THE CRESCENT PUZZLE. + +[Illustration] + +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? + + +192.--THE PUZZLE WALL. + +[Illustration] + +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? + + +193.--THE SHEEPFOLD. + +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:--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? + + +194.--THE GARDEN WALLS. + +[Illustration] + +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. + +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. + + +195.--LADY BELINDA'S GARDEN. + +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 +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--that is, it need not be a graduated measure. + +[Illustration] + + +196.--THE TETHERED GOAT. + +[Illustration] + +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. + + +197.--THE COMPASSES PUZZLE. + +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:-- + +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--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. + + +198.--THE EIGHT STICKS. + +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. + + + +199.--PAPA'S PUZZLE. + +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 +_Mathematical Collections_. 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 found by trial clippings. A curious and pretty +point is involved in this setting of the puzzle. Can the reader discover +it? + +[Illustration] + + +200.--A KITE-FLYING PUZZLE. + +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? + +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--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. + + +201.--HOW TO MAKE CISTERNS. + +[Illustration] + +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 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? + + +202.--THE CONE PUZZLE. + +[Illustration] + +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? + + +203.--CONCERNING WHEELS. + +[Illustration] + +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. + +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. + + +204.--A NEW MATCH PUZZLE. + +[Illustration] + +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. + + +205.--THE SIX SHEEP-PENS. + +[Illustration] + +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. + + + + +POINTS AND LINES PROBLEMS. + + +"Line upon line, line upon line; here a little and there a +little."--_Isa_. xxviii. 10. + +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--published in 1821--_Rational Amusement for Winter Evenings_, by +John Jackson. The author gives ten examples of "Trees planted in Rows." + +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. + +[Illustration] + + +206.--THE KING AND THE CASTLES. + +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. + +[Illustration] + + +207.--CHERRIES AND PLUMS. + +[Illustration] + +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 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. + + +208.--A PLANTATION PUZZLE. + +[Illustration] + +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? + + +209.--THE TWENTY-ONE TREES. + +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? + + +210.--THE TEN COINS. + +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. + +[Illustration] + + +211.--THE TWELVE MINCE-PIES. + +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 _seven_ straight rows with four in every row. Which four +would you remove, and where would you replace them? + +[Illustration] + + +212.--THE BURMESE PLANTATION. + +[Illustration] + +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. + +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. + +Of course there may not be more than four trees in any row. + + +213.--TURKS AND RUSSIANS. + +This puzzle is on the lines of the Afridi problem published by me in +_Tit-Bits_ some years ago. + +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. + + + + +MOVING COUNTER PROBLEMS. + + + "I cannot do't without counters." + + _Winter's Tale_, iv. 3. + +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 _The Canterbury Puzzles_), from +which "Noughts and Crosses" seems to be derived, is still more ancient. + +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):-- + + "The nine-men's morris is filled up with mud; + And the quaint mazes in the wanton green, + For lack of tread, are undistinguishable." + +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:--"Oft we track his haunts .... By nine-peg-morris nicked +upon the green." It is also mentioned by Drayton in his "Polyolbion." + +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--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. + +The game has been played with different 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 _Sports and Pastimes_, describes the +game, and it agrees with the way I played it as a boy:--"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." + +[Illustration] + + +214.--THE SIX FROGS. + +[Illustration] + +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. + + +215.--THE GRASSHOPPER PUZZLE. + +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? + +[Illustration] + + +216.--THE EDUCATED FROGS. + +[Illustration] + +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--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 +successive 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? + + +217.--THE TWICKENHAM PUZZLE. + +[Illustration: + + + ( I ) ((N)) + + ( M ) ((A)) + + ( H ) ((T)) + + ( E ) ((W)) + + ( C ) ((K)) + ( ) + + +] + +In the illustration we have eleven discs in a circle. On five of the +discs we place white counters with black letters--as shown--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. + + +218.--THE VICTORIA CROSS PUZZLE. + +[Illustration: + + +---------------------+ + | \... A .../ | + | (I) |.......| (V) | + |\_____|_______|_____/| + |......| |------| + |.. R .| |. I ..| + |......| |......| + | _____|_______|_____ | + |/ |.......| \| + | (O) |.. T ..| (C) | + | /.........\ | + +---------------------+ + +] + +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." + +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. + +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. + +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. + + +219.--THE LETTER BLOCK PUZZLE. + +[Illustration: + + + +-----+-----+-----+\ + | | | | | + | G | E | F | | + | | | | | + +-----+-----+-----+\| + | | | | | + | H | C | B | | + | | | | | + +-----+-----+-----+\| + | |\____| | | + | D || | A | | + | || | | | + +-----+-----+-----+ | + \_________________\| + +] + +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-- + + A B C + D E F + G H + +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. + + +220.--A LODGING-HOUSE DIFFICULTY. + +[Illustration] + +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. + + +221.--THE EIGHT ENGINES. + +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? + +[Illustration] + + +222.--A RAILWAY PUZZLE. + +[Illustration] + +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-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? + + +223.--A RAILWAY MUDDLE. + +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. + +[Illustration] + + +224.--THE MOTOR-GARAGE PUZZLE. + +[Illustration] + +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. + +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--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? + +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. + + +225.--THE TEN PRISONERS. + +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 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? + +[Illustration] + + +226.--ROUND THE COAST. + +[Illustration] + +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--1, 7--2, 8--3, +7--4, 8--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--"Glasg"--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? + + +227.--CENTRAL SOLITAIRE. + +[Illustration] + +This ancient puzzle was a great favourite with our grandmothers, and +most of us, I imagine, have on occasions come across a "Solitaire" +board--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. + +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 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. + +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. + + +228.--THE TEN APPLES. + +[Illustration] + +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--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. + + +229.--THE NINE ALMONDS. + +"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." + +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. + +"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--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." + +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 should be. Remember to remove +those you jump over. Any number of jumps in succession with the same +almond count as one move. + +[Illustration] + + +230.--THE TWELVE PENNIES. + +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. + +[Illustration] + + +231.--PLATES AND COINS. + +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. + +[Illustration] + + +232.--CATCHING THE MICE. + +[Illustration] + +"Play fair!" said the mice. "You know the rules of the game." + +"Yes, I know the rules," said the cat. "I've got to go round and round +the circle, in the 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." + +"Hurry up, then!" shouted the mice. + +"Give a fellow time to think," said the cat. "I don't know which of you +to start at. I must figure it out." + +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? + +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? + +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. + + +233.--THE ECCENTRIC CHEESEMONGER. + +[Illustration] + +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. + + +234.--THE EXCHANGE PUZZLE. + + +Here is a rather entertaining little puzzle with moving counters. You +only need twelve counters--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:-- + + A B C D + E F G H + I J K L + +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? + +[Illustration] + +It cannot be done in fewer moves. The puzzle is really much easier than +it looks, if properly attacked. + + +235.--TORPEDO PRACTICE. + +[Illustration] + +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--provided the enemy will allow you to arrange his fleet for +your convenience and promise to lie still and do nothing! + + +236.--THE HAT PUZZLE. + +Ten hats were hung on pegs as shown in the illustration--five silk hats +and five felt "bowlers," alternately silk and felt. The two pegs at the +end of the row were empty. + +[Illustration] + +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. + +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. + +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. + + +237.--BOYS AND GIRLS. + +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. + +The puzzle is to remove two children who are occupying adjoining chairs +and place them in two empty chairs, _making them first change sides_; +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, 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. + +[Illustration] + + +238.--ARRANGING THE JAMPOTS. + +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. + +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--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. + +[Illustration] + + + + +UNICURSAL AND ROUTE PROBLEMS. + + "I see them on their winding way." + REGINALD HEBER. + +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. + + +239.--A JUVENILE PUZZLE. + +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." + +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 in a good deal +of the figure with one continuous stroke, but it will always appear as +if four strokes are necessary. + +[Illustration] + +Another form of the puzzle is to draw the diagram on a slate and then +rub it out in three rubs. + + +240.--THE UNION JACK. + +[Illustration] + +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 _how much_ 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. + + +241.--THE DISSECTED CIRCLE. + +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. + +[Illustration] + + +242.--THE TUBE INSPECTOR'S PUZZLE. + +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? + +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? + +[Illustration] + + +243.--VISITING THE TOWNS. + +[Illustration] + +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. + + +244.--THE FIFTEEN TURNINGS. + +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? + +[Illustration] + + +245.--THE FLY ON THE OCTAHEDRON. + +"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?" + +"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?" + +[Illustration] + +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--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. + + +246.--THE ICOSAHEDRON PUZZLE. + +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. + +Now, a Platonic body does not mean a 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--that is, have traversed every one of +the edges? + +[Illustration] + + +247.--INSPECTING A MINE. + +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. + +[Illustration] + + +248.--THE CYCLISTS' TOUR. + +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. + +[Illustration] + + +249.--THE SAILOR'S PUZZLE. + +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 +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--thus, for example, A, I, O, L, G, etc.--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--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? + +[Illustration] + + +250.--THE GRAND TOUR. + +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. + +I give a rough map of a country--it is not necessary to say what +particular country--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. 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. + +[Illustration] + + +251.--WATER, GAS, AND ELECTRICITY. + +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. + +[Illustration] + + +252.--A PUZZLE FOR MOTORISTS. + +[Illustration] + +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. + + +253.--A BANK HOLIDAY PUZZLE. + +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 destination, always +travelling either due south or due east. The puzzle is to discover which +town is their destination. + +[Illustration] + +Of course, if you find that there are more than 1,365 different routes +to a town it cannot be the right one. + + +254.--THE MOTOR-CAR TOUR. + +[Illustration] + +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. + + +255.--THE LEVEL PUZZLE. + +[Illustration] + +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--that is to say, if you come to a letter you must use it. + + +256.--THE DIAMOND PUZZLE. + +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? + +[Illustration] + + +257.--THE DEIFIED PUZZLE. + +In how many different ways may the word DEIFIED be read in this +arrangement under the same conditions as in the last puzzle, with the +addition that you can use any letters twice in the same reading? + +[Illustration] + + +258.--THE VOTERS' PUZZLE. + +[Illustration] + +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. + + +259.--HANNAH'S PUZZLE. + +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:-- + + H H H H H H + H A A A A H + H A N N A H + H A N N A H + H A A A A H + H H H H H H + +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? + + +260.--THE HONEYCOMB PUZZLE. + +[Illustration] + +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. + + +261.--THE MONK AND THE BRIDGES. + +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 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. + +[Illustration] + + + + +COMBINATION AND GROUP PROBLEMS. + + "A combination and a form indeed." + _Hamlet_, iii. 4. + +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 Probleme des Menages, may be interesting. If +n married ladies are seated at a round table in any determined order, +in how many different ways may their n husbands be placed so that +every man is between two ladies but never next to his own wife? + +This difficult problem was first solved by Laisant, and the method shown +in the following table is due to Moreau:-- + + 4 0 2 + 5 3 13 + 6 13 80 + 7 83 579 + 8 592 4738 + 9 4821 43387 + 10 43979 439792 + +The first column shows the number of married couples. The numbers in the +second column are obtained in this way: 5 x 3 + 0 - 2 = 13; 6 x 13 + 3 + +2 = 83; 7 x 83 + 13 - 2 = 592; 8 x 592 + 83 + 2 = 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 - 0 = 13; 83 - 3 = 80; +592 - 13 = 579; 4821 - 83 = 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. + +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. + + +262.--THOSE FIFTEEN SHEEP. + +A certain cyclopaedia 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. + +[Illustration] + +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 cyclopaedia. What was it? Can you place those fifteen sheep? + + +263.--KING ARTHUR'S KNIGHTS. + +King Arthur sat at the Round Table on three successive evenings with his +knights--Beleobus, Caradoc, Driam, Eric, Floll, and Galahad--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? + + +264.--THE CITY LUNCHEONS. + +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-- + + (A B) (C D) (E F) (G H) (I J) (K L). + +Then give any pairing you like for the next day, say-- + + (A C) (B D) (E G) (F H) (I K) (J L), + +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. + + +265.--A PUZZLE FOR CARD-PLAYERS. + +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. + + +266.--A TENNIS TOURNAMENT. + +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. + + +267.--THE WRONG HATS. + +"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?" + +"The first thing," said Mr. Waterson, "is to see in how many different +ways the eight hats could be taken." + +"That is quite easy," Mr. Stubbs explained. "Multiply together the +numbers, 1, 2, 3, 4, 5, 6, 7, and 8. Let me see--half a minute--yes; +there are 40,320 different ways." + +"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. + +"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." + +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? + + +268.--THE PEAL OF BELLS. + +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:-- + + 1 2 3 + 2 1 3 + 2 3 1 + 3 2 1 + 3 1 2 + 1 3 2 + +How are we to give him a correct solution for his four bells? + + +269.--THREE MEN IN A BOAT. + +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. + +"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." + +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: + + 1 2 3 4 5 + First Day: (ABC) (DEF) (GHI) (JKL) (MNO). + +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. + + +270.--THE GLASS BALLS. + +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 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? + +[Illustration] + + +271.--FIFTEEN LETTER PUZZLE. + + ALE FOE HOD BGN + CAB HEN JOG KFM + HAG GEM MOB BFH + FAN KIN JEK DFL + JAM HIM GCL LJH + AID JIB FCJ NJD + OAK FIG HCK MLN + BED OIL MCD BLK + ICE CON DGK + +The above is the solution of a puzzle I gave in _Tit-bits_ 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. + +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. + +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 groups in accordance with the conditions, and +show as many good English words as possible. + + +272.--THE NINE SCHOOLBOYS. + +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 _side by side_ with any other boy more than +once. How would you arrange them? + +If we represent them by the first nine letters of the alphabet, they +might be grouped on the first day as follows:-- + + A B C + D E F + G H I + +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. + + +273.--THE ROUND TABLE. + +Seat the same n persons at a round table on + + (n - 1)(n - 2) + -------------- + 2 + +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. + + +274.--THE MOUSE-TRAP PUZZLE. + +[Illustration + + + 6 20 2 19 + 13 21 + 7 5 + 3 18 + 17 8 + 15 11 + 14 16 + 1 9 + 10 4 12 + +] + +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. + + +275.--THE SIXTEEN SHEEP. + +[Illustration: + + +========================+ + || | | | || + || 0 | 0 | 0 | 0 || + +-----+-----+-----+------+ + || | | | || + || 0 | 0 | 0 | 0 || + +========================+ + || || | || || + || 0 || 0 | 0 || 0 || + +-----+=====+=====+------+ + || | || | || + || 0 | 0 || 0 | 0 || + +========================+ + +] + +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. + + +276.--THE EIGHT VILLAS. + +In one of the outlying suburbs of London a man had a square plot of +ground on which he 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. + + +[Illustration: + + /\ /\ /\ + |2 | |5 | |2 | + + /\ /\ + |5 | |5 | + + /\ /\ /\ + |2 | |5 | |2 | + + +---+---+---+ +---+---+---+ +---+---+---+ + | 1 | 6 | 2 | | 2 | 6 | 1 | | 1 | 6 | 2 | + +---+---+---+ +---+---+---+ +---+---+---+ + | 6 | | 6 | | 6 | | 6 | | 4 | | 4 | + +---+---+---+ +---+---+---+ +---+---+---+ + | 2 | 6 | 1 | | 1 | 6 | 2 | | 4 | 2 | 3 | + +---+---+---+ +---+---+---+ +---+---+---+ + A B C + +] + + +277.--COUNTER CROSSES. + +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? + +[Illustration: + + (1) (2) + + (2) (4) (5) (1) (6) (7) + + (3) (4) (9) (5) (6) (3) + + (7) (8) + + A (8) B (9) + +] + + +278.--A DORMITORY PUZZLE. + +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 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. + +[Illustration + + +---+---+---+ + | | | | + | | | | + | | | | + +---+---+---+ + | |\|/| | + | |-*-| | + | |/|\| | + +---+---+---+ + | | | | + | | | | + | | | | + +---+---+---+ + +] + +279.--THE BARRELS OF BALSAM. + +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--such, for example, as this:-- + + 1 2 5 7 8 + 3 4 6 9 10 + +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? + + +280.--BUILDING THE TETRAHEDRON. + +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? + +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 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? + +[Illustration] + + +281.--PAINTING A PYRAMID. + +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--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." + +[Illustration] + +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. + + +282.--THE ANTIQUARY'S CHAIN. + +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. + +[Illustration] + + +283.--THE FIFTEEN DOMINOES. + +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 that +have a five or a six on them and limit ourselves to the fifteen that +remain, where the double-four is the highest. + +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--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. + + +384.--THE CROSS TARGET. + + +-+-+ + |*|*| + +-+-+ + |*|*| + +-+-+-+-+-+-+ + | | | |*| | | + +-+-+-+-+-+-+ + | | |*| |*| | + +-+-+-+-+-+-+ + | |*| + +-+-+ + | | | + +-+-+ + +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. + + +285.--THE FOUR POSTAGE STAMPS. + + +---+----+----+----+ + | 1 | 2 | 3 | 4 | + +---+----+----+----+ + | 5 | 6 | 7 | 8 | + +---+----+----+----+ + | 9 | 10 | 11 | 12 | + +---+----+----+----+ + +"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--three by four--and a friend asks you to oblige him with four +stamps, all joined together--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? + + +286.--PAINTING THE DIE. + +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. + + +287.--AN ACROSTIC PUZZLE. + +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--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? + + + + +CHESSBOARD PROBLEMS. + + "You and I will goe to the chesse." + + GREENE'S _Groatsworth of Wit._ + + +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 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 _vice versa_. Occasionally a man will be found strong on +both subjects, but he is the exception to the rule. + +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. + +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 formulae 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. + + + +THE CHESSBOARD. + + "Good company's a chessboard." + BYRON'S _Don Juan_, xiii. 89. + +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. + + +288.--CHEQUERED BOARD DIVISIONS. + +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. + +[Illustration: + + +---+---*---+---+ +---+---+---*---+ +---+---+---*---+ + | | H | | | | | H | | | | H | + +---+---*---+---+ +---+---*===*---+ +---*===*---*---+ + | | H | | | | H | | | H H H | + +---+---*---+---+ +---+---*---+---+ +---*---*---*---+ + | | H | | | | H | | | H H H | + +---+---*---+---+ +---*===*---+---+ +---*---*===*---+ + | | H | | | H | | | | H | | | + +---+---*---+---+ +---*---+---+---+ +---*---+---+---+ + + +---+---+---+---+---+---+ + | | | | | | | + +---+---+---+---+---+---+ + | | | | | | | + +---+---+---+---+---+---+ + | | | | | | | + +---+---+---+---+---+---+ + | | | | | | | + +---+---+---+---+---+---+ + | | | | | | | + +---+---+---+---+---+---+ + | | | | | | | + +---+---+---+---+---+---+ + + +---+---*---+---+ +---+---+---*---+ +---+---+---*---+ + | | H | | | | | H | | | | H | + +---*===*---+---+ +---*===*===*---+ +---+---*===*---+ + | H | | | | H | | | | | H | | + +---*===*===*---+ +---*===*===*---+ +---+---*---+---+ + | | | H | | | | H | | | H | | + +---+---*===*---+ +---*===*===*---+ +---*===*---+---+ + | | H | | | H | | | | H | | | + +---+---*---+---+ +---*---+---+---+ +---*---+---+---+ + +] + +It is obvious that a board of four squares can only be so divided in one +way--by a straight cut down the centre--because we shall not count +reversals and reflections as different. In the case of a board of +sixteen squares--four by four--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. + + +289.--LIONS AND CROWNS. + +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 communicated 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. + +[Illustration: + + +-+-+-+-+-+-+ + | | | | | | | + +-+-+-+-+-+-+ + | |L|L|L| | | + +-+-+-+-+-+-+ + | | |C|C| | | + +-+-+-+-+-+-+ + | | |C|C| | | + +-+-+-+-+-+-+ + |L| | | | | | + +-+-+-+-+-+-+ + | | | | | | | + +-+-+-+-+-+-+ + +] + + +290.--BOARDS WITH AN ODD NUMBER OF SQUARES. + +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. + +[Illustration: + + +---*---+---+ + | H | | + +---*===*---+ + | HHHHH | + +---*===*---+ + | | H | + +---+---*---+ + +Fig 1] + +[Illustration: + + +---+---+---+---+---+ + | | | | | | + *===*===*===*---+---+ + | | | H | | + +---+---*===*---+---+ + | | HHHHH | | + +---+---*===*---+---+ + | | H | | | + +---+---*===*===*===* + | H | | | | + +---*---+---+---+---+ + +Fig 2] + + +291.--THE GRAND LAMA'S PROBLEM. + +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--possibly +of chagrin. + +[Illustration: + + +---+---+---+---+---+---+---+---+ + | * | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | * | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | * | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | * | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + +] + +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: "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? + + +292.--THE ABBOT'S WINDOW. + +[Illustration] + +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. + +"Father John," said the Abbot, "dost thou know that I came into this +wicked world on a Christmas Even?" + +The monk nodded assent. + +"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!" + +The Abbot pointed to the large dormitory window, of which I give a +sketch. The monk looked, and was perplexed. + +"Dost thou not see that the sixty-four lights add up an even number +vertically and horizontally, but that all the _diagonal_ lines, except +fourteen are of a number that is odd? Why is this?" + +"Of a truth, my Lord Abbot, it is of the very nature of things, and +cannot be changed." + +"Nay, but it _shall_ 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." + +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? + + +293.--THE CHINESE CHESSBOARD. + +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. + + +294.--THE CHESSBOARD SENTENCE. + +[Illustration] + +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. + +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? + + + + +STATICAL CHESS PUZZLES. + + "They also serve who only stand and wait." + MILTON. + + +295.--THE EIGHT ROOKS. + + +[Illustration: + + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | R | R | R | R | R | R | R | R | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + +FIG. 1.] + +[Illustration: + + +---+---+---+---+---+---+---+---+ + | R | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | R | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | R | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | R | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | R | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | R | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | R | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | R | + +---+---+---+---+---+---+---+---+ + +FIG. 2.] + +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. + + +296.--THE FOUR LIONS. + +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. + +[Illustration + + +---+---+---+---+ + | L | | | | + +---+---+---+---+ + | | L | | | + +---+---+---+---+ + | | | L | | + +---+---+---+---+ + | | | | L | + +---+---+---+---+ + +] + + +297.--BISHOPS--UNGUARDED. + +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. + + +298.--BISHOPS--GUARDED. + +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? + + +299.--BISHOPS IN CONVOCATION. + +[Illustration: + + +---+---+---+---+---+---+---+---+ + | B | B | B | B | B | B | B | B | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | B | B | B | B | B | B | | + +---+---+---+---+---+---+---+---+ + +] + +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. + + +300.--THE EIGHT QUEENS. + +[Illustration: + + +---+---+---+---+---+---+---+---+ + | | | | ..Q | | | | + +---+---+---+...+---+---+---+---+ + | | ..Q.. | | | | | + +---+...+---+---+---+---+---+---+ + | Q.. | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | Q | | + +---+---+---+---+---+---+---+---+ + | | Q | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | ..Q | + +---+---+---+---+---+---+...+---+ + | | | | | ..Q.. | | + +---+---+---+---+...+---+---+---+ + | | | | Q.. | | | | + +---+---+---+---+---+---+---+---+ + +] + +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. + +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? + + +301.--THE EIGHT STARS. + +[Illustration: + + +---+---+---+---+---+---+---+---+ + |///| | | | | | |///| + +---+---+---+---+---+---+---+---+ + | |///| | | | |///| * | + +---+---+---+---+---+---+---+---+ + | | |///| | |///| | | + +---+---+---+---+---+---+---+---+ + | | | |///|///| | | | + +---+---+---+---+---+---+---+---+ + | | | |///|///| | | | + +---+---+---+---+---+---+---+---+ + | | |///| | |///| | | + +---+---+---+---+---+---+---+---+ + | |///| | | | |///| | + +---+---+---+---+---+---+---+---+ + |///| | | | | | |///| + +---+---+---+---+---+---+---+---+ + +] + +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. + + +302.--A PROBLEM IN MOSAICS. + +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. + +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. + +[Illustration: + + +---+---+---+---+---+---+---+---+ + | V | R | Y | G | O | P | W | B | + +---+---+---+---+---+---+---+---+ + | W | B | O | P | Y | G | V | R | + +---+---*===*---+---*===*---+---+ + | G | P H W H V | B H R H Y | O | + +---+---*===*---+---*===*---+---+ + | R | Y | B | O | G | V | P | W | + +---+---+---+---+---+---+---+---+ + | B | G | R | Y | P | W | O | V | + +---+---+---+---+---+---+---+---+ + | O | V | P | W | R | Y | B | G | + +---+---+---+---+---+---+---+---+ + | P | W | G | B | V | O | R | Y | + +---+---+---+---+---+---+---+---+ + |///| O | V | R | W | B | G |///| + +---+---+---+---+---+---+---+---+ + +] + +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? + + +303.--UNDER THE VEIL. + +[Illustration: + + +---+---+---+---+---+---+---+---+ + | | | V | E | I | L | | | + +---+---+---+---+---+---+---+---+ + | | | I | L | V | E | | | + +---+---+---+---+---+---+---+---+ + | I | V | | | | | L | E | + +---+---+---+---+---+---+---+---+ + | L | E | | | | | I | V | + +---+---+---+---+---+---+---+---+ + | V | I | | | | | E | L | + +---+---+---+---+---+---+---+---+ + | E | L | | | | | V | I | + +---+---+---+---+---+---+---+---+ + | | | E | V | L | I | | | + +---+---+---+---+---+---+---+---+ + | | | L | I | E | V | | | + +---+---+---+---+---+---+---+---+ + +] + +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. + +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. + +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. + + +304.--BACHET'S SQUARE. + +One of the oldest card puzzles is by Claude Caspar Bachet de Meziriac, +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. + +Note "row of four cards," so that the only diagonals we have here to +consider are the two long ones. + + +305.--THE THIRTY-SIX LETTER-BLOCKS. + +[Illustration] + +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. + + +306.--THE CROWDED CHESSBOARD. + +[Illustration] + +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--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. + + +307.--THE COLOURED COUNTERS. + +[Illustration] + +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? + + +308.--THE GENTLE ART OF STAMP-LICKING. + +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 x 4), +and are provided with plenty of stamps of the values 1d., 2d., 3d., 4d., +and 5d., 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 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--but no matter. + + +309.--THE FORTY-NINE COUNTERS. + +[Illustration] + +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. + + +310.--THE THREE SHEEP. + +[Illustration] + +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--in the maximum number of different ways. + + +311.--THE FIVE DOGS PUZZLE. + +In 1863, C.F. de Jaenisch first discussed the "Five Queens Puzzle"--to +place five queens on the chessboard so that every square shall be +attacked or occupied--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. + +[Illustration] + +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--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 _in a straight row_, so that every kennel shall always be in +line with at least one dog. Reversals and reflections are here counted +as different. + + +312.--THE FIVE CRESCENTS OF BYZANTIUM. + +When Philip of Macedon, the father of Alexander the Great, found himself +confronted with great difficulties in the siege of Byzantium, he 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:-- + +[Illustration] + +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. + +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. + + +313.--QUEENS AND BISHOP PUZZLE. + +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. + +[Illustration] + + +314.--THE SOUTHERN CROSS. + +[Illustration] + +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. + +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. + + +315.--THE HAT-PEG PUZZLE. + +Here is a five-queen puzzle that I gave in a fanciful dress in 1897. As +the queens were 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 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. + +[Illustration] + + +316.--THE AMAZONS. + +[Illustration] + +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. + + +317.--A PUZZLE WITH PAWNS. + +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. + +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--at the centres of the squares. See +dotted lines in the case of No. 300, "The Eight Queens." + + +318.--LION-HUNTING. + +[Illustration] + +My friend Captain Potham Hall, the renowned hunter of big game, says +there is nothing more exhilarating than a brush with a herd--a pack--a +team--a flock--a swarm (it has taken me a full quarter of an hour to +recall the right word, but I have it at last)--a _pride_ 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. + +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 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. + +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. + + +319.--THE KNIGHT-GUARDS. + +[Illustration] + +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--without success. It really consists in moving one +square like a rook, and then another square like a bishop--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. + +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. + + +THE GUARDED CHESSBOARD. + +On an ordinary chessboard, 8 by 8, every square can be guarded--that is, +either occupied or attacked--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. + +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. 295, "The Eight Rooks.") I have not enumerated the ways +in which every rook shall be protected by another rook. + +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. 297 and 298, +"Bishops unguarded" and "Bishops guarded.") + +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. 319, "The Knight-Guards.") + +Dealing with the queen on n squared boards generally, where n is less +than 8, the following results will be of interest:-- + + +1 queen guards 2 squared board in 1 fundamental way. + +1 queen guards 3 squared board in 1 fundamental way. + +2 queens guard 4 squared board in 3 fundamental ways (protected). + +3 queens guard 4 squared board in 2 fundamental ways (not protected). + +3 queens guard 5 squared board in 37 fundamental ways (protected). + +3 queens guard 5 squared board in 2 fundamental ways (not protected). + +3 queens guard 6 squared board in 1 fundamental way (protected). + +4 queens guard 6 squared board in 17 fundamental ways (not protected). + +4 queens guard 7 squared board in 5 fundamental ways (protected). + +4 queens guard 7 squared board in 1 fundamental way (not protected). + + +NON-ATTACKING CHESSBOARD ARRANGEMENTS. + +We know that n queens may always be placed on a square board of n squared +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:-- + +Where n = 4 there is 1 fundamental solution and 2 in all. + +Where n = 5 there are 2 fundamental solutions and 10 in all. + +Where n = 6 there is 1 fundamental solution and 4 in all. + +Where n = 7 there are 6 fundamental solutions and 40 in all. + +Where n = 8 there are 12 fundamental solutions and 92 in all. + +Where n = 9 there are 46 fundamental solutions. + +Where n = 10 there are 92 fundamental solutions. + +Where n = 11 there are 341 fundamental solutions. + +Obviously n rooks may be placed without attack on an n squared 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. 296, "The Four Lions.") + +We can place 2n-2 bishops on an n squared board in 2^{n} ways. (See No. 299, +"Bishops in Convocation.") 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^{1/2(n-1)} such arrangements, each giving 4 by reversals and +reflections, and 2^{n-3} - 2^{1/2(n-3)} giving 8. Where n is even there +are 2^{1/2(n-2)}, each giving 4 by reversals and reflections, and 2^{n-3} +- 2^{1/2(n-4)}, each giving 8. + +We can place 1/2(n squared+1) knights on an n squared board without attack, when n +is odd, in 1 fundamental way; and 1/2n squared knights on an n squared 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. + + +THE TWO PIECES PROBLEM. + +On a board of n squared squares, two queens, two rooks, two bishops, or two +knights can always be placed, irrespective of attack or not, in 1/2(n^{4} +- n squared) ways. The following formulae will show in how many of these ways +the two pieces may be placed with attack and without:-- + + With Attack. Without Attack. + + 2 Queens 5n cubed - 6n squared + n 3n^{4} - 10n cubed + 9n squared - 2n + ------------------- ------------------------------ + 3 6 + + 2 Rooks n cubed - n squared n^{4} - 2n cubed + n squared + ---------------------- + 2 + + 2 Bishops 4n cubed - 6n squared + 2n 3n^{4} - 4n cubed + 3n squared - 2n + -------------------- ----------------------------- + 6 6 + + 2 Knights 4n squared - 12n + 8 n^{4} - 9n squared + 24n + -------------------- + 2 + +(See No. 318, " Lion Hunting.") + + + + +DYNAMICAL CHESS PUZZLES. + + "Push on--keep moving." + THOS. MORTON: _Cure for the Heartache_. + + +320.--THE ROOK'S TOUR. + +[Illustration: + + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | R | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + +] + + +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. + + +321.--THE ROOK'S JOURNEY. + +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 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--that is to say, +you must make a turn after every move. + +[Illustration: + + +---+---+---+---+---+---+---+---+ + | | | | | | | | R | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + +---+---+---+---+---+---+---+---+ + | | 21| | 10| | | | | + +---+---+---+---+---+---+---+---+ + +] + + +322.--THE LANGUISHING MAIDEN. + +[Illustration: + + --+-----+-----+-----+-----+-----+-----+-----+ + | | | | | | | | + | Kt | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + | | | | | | | | | + | | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + | | | | | | | | | + | | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + | | | | | | | | | + | | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + | | | | | | | | | + | | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + | | | | | | | | | + | | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + | | | | | | | | | + | M | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + | | | | | | | | | + | | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + +] + +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. + + +323.--A DUNGEON PUZZLE. + +[Illustration: + + +-----+-----+-----+-----+-----+-----+-----+-----+ + | | | | | | | | | + | ............. ....... ............. | + | . | | . | . | . | . | | . | + +--.--+-- --+--.--+--.--+--.--+--.--+-- --+--.--+ + | . | | . | . | . | . | | . | + | ....... ....... ....... ....... | + | | . | | | | | . | | + +-- --+--.--+-- --+-- --+-- --+-- --+--.--+-- --+ + | | . | | | | | . | | + | ....... ....... ....... ....... | + | . | | . | . | . | . | | . | + +--.--+-- --+--.--+--.--+--.--+--.--+-- --+--.--+ + | . | | . | . | . | . | | . | + | ............. ....... . ....... | + | | | | | | . | . | | + +-- --+-- --+-- --+-- --+-- --+--.--+--.--+-- --+ + | | | | | | . | . | | + | ............. ....... . ....... | + | . | | . | . | . | . | | . | + +--.--+-- --+--.--+--.--+--.--+--.--+-- --+--.--+ + | . | | . | . | . | . | | . | + | ....... ....... ....... ....... | + | | . | | | | | . | | + +-- --+--.--+-- --+-- --+-- --+-- --+--.--+-- --+ + | | . | | | | | . | | + | ....... ....... ....... ....... | + | . | | . | . | . | . | | . | + +--.--+-- --+--.--+--.--+--.--+--.--+-- --+--.--+ + | . | | . | . | . | . | | . | + | ............. . 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. + + +324.--THE LION AND THE MAN. + +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 no coward, and was as anxious to find the lion as the lion +undoubtedly was to find him. + +[Illustration: + + +-----+-----+-----+-----+-----+-----+-----+-----+ + | | | | | | | | | + | L | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + | | | | | | | | | + | | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + | | | | | | | | | + | | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + | | | | | | | | | + | | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + | | | | | | | | | + | | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + | | | | | | | | | + | | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + | | | | | | | | | + | | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + | | | | | | | | | + | C | + | | | | | | | | | + +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+ + +] + +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. + + +325.--AN EPISCOPAL VISITATION. + +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. + + +326.--A NEW COUNTER PUZZLE. + +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. + +[Illustration: + + 4 8 + / \ / \ + 2 6 10 + \ / \ / + 3 7 + / \ / \ + 1 5 9 + +] + + +327.--A NEW BISHOP'S PUZZLE. + +[Illustration: + + +---+---+---+---+ + | b | b | b | b | + +---+---+---+---+ + | | | | | + +---+---+---+---+ + | | | | | + +---+---+---+---+ + | B | B | B | B | + +---+---+---+---+ + +] + +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--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. + +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. + + +328.--THE QUEEN'S TOUR. + +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 in his _Chess Strategy_. But +the solution shown below is the one he gave in _American Chess-Nuts_ in +1868. I have recorded at least six different solutions in the minimum +number of moves--fourteen--but this one is the best of all, for reasons +I will explain. + +[Illustration: + + +---+---+---+---+---+---+---+---+ + | | | | | | | | | + | ............................. | + | . | | | | | | | . | + +-.-+---+---+---+---+---+---+-.-+ + | . | | | | | | | . | + | . | ..........................| + | . | .| | | | | | . | + +-.-+---.---+---+---+---+---+..-+ + | . | |. | | | | . . | + | . | ................. | .| . | + | . | .| .| | |. | . | . | + +-.-+---.---.---+---.---+.--+-.-+ + | . | |. |. | .| . | . | + | . | . | . | . | . | .| . | . | + | . | ..| .| .|. | . |.. | . | + +-.-+-.-.---.---.---+.--.-.-+-.-+ + | . | . |. |. .|. . .| . | . | + | . | . | . | . | ..| . | . | . | + | . | . | .|. .| ..|. | . | . | + +-.-+-.-+---.---..--.---+-.-+-.-+ + | . | . | .|. .. .|. | . | . | + | . | . | . | ..| . | . | . | . | + | . | . |. | ..|. .| .| . | . | + +-.-+-.-.---+.--.---.---.-.-+-.-+ + | . | ..| . .|. |. |.. | . | + | . | . | .| . | . | . | . | . | + | . |.. | . |. | .| .| ..| . | + +-.-.-.-+.--.---+---.---.-.-.-.-+ + | ..| . . .| | |. |.. |.. | + | . | ..| ............. | . | . | + | | . | | | | | | | + +---+---+---+---+---+---+---+---+ + +] + +If you will look at the lettered square you will understand that there +are only ten really differently placed squares on a chessboard--those +enclosed by a dark line--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--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:-- + +[Illustration: + + +---+---+---+---*---+---+---+---+ + | A | B | C | G " G | C | B | A | + *===*---+---+---*---+---+---+---+ + | B " D | E | H " H | E | D | B | + +---*===*---+---*---+---+---+---+ + | C | E " F | I " I | F | E | C | + +---+---*===*---*---+---+---+---+ + | G | H | I " J " J | I | H | G | + +---+---+---*===*---+---+---+---+ + | G | H | I | J | J | I | H | G | + +---+---+---+---+---+---+---+---+ + | C | E | F | I | I | F | E | C | + +---+---+---+---+---+---+---+---+ + | B | D | E | H | H | E | D | B | + +---+---+---+---+---+---+---+---+ + | A | B | C | G | G | C | B | A | + +---+---+---+---+---+---+---+---+ + +] + +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. + + +329.--THE STAR PUZZLE. + +[Illustration: + + +---+---+---+---+---+---+---+---+ + | * | * | * | * | * | * | * | * | + +---+---+---+---+---+---+---+---+ + | * | * | * | * | * | * | * | * | + +---+---+---+---+---+---+---+---+ + | * | * | * | * | * | * | * | * | + +---+---+---+---+---+---+---+---+ + | * | * | ¤ | * | * | * | * | * | + +---+---+---+---+---+---+---+---+ + | * | * | * | ¤ | * | * | * | * | + +---+---+---+---+---+---+---+---+ + | * | * | * | * | * | * | * | * | + +---+---+---+---+---+---+---+---+ + | * | * | * | * | * | * | * | * | + +---+---+---+---+---+---+---+---+ + | * | * | * | * | * | * | * | * | + +---+---+---+---+---+---+---+---+ + +] + +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. + +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--such as from the upper white star direct to a +corner star. + + +330.--THE YACHT RACE. + +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! + +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. + +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. The +point of a lead pencil and a good nautical eye are all the outfit that +we require. + +[Illustration] + +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. + + +331.--THE SCIENTIFIC SKATER. + +[Illustration] + +It will be seen that this skater has marked on the ice sixty-four points +or stars, and he proposes to start _from his present position_ 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. + +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. + + +332.--THE FORTY-NINE STARS. + +[Illustration] + +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. + + +333.--THE QUEEN'S JOURNEY. + +[Illustration] + +Place the queen on her own square, as shown in the illustration, and +then try to discover the 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. + + +334.--ST. GEORGE AND THE DRAGON. + +[Illustration] + +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. + + +335.--FARMER LAWRENCE'S CORNFIELDS. + +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--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!" + +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. + +[Illustration] + + +336.--THE GREYHOUND PUZZLE. + +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 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. + +[Illustration] + + +337.--THE FOUR KANGAROOS. + +[Illustration] + +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 _do_ 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. + +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. + + +338.--THE BOARD IN COMPARTMENTS. + +[Illustration] + +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. + +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 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. + +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. + + +339.--THE FOUR KNIGHTS' TOURS. + +[Illustration] + +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. + + +340.--THE CUBIC KNIGHT'S TOUR. + +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. + + +341.--THE FOUR FROGS. + +[Illustration] + +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. + + +342.--THE MANDARIN'S PUZZLE. + +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 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. + +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? + +[Illustration] + +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--21--14--18--22, etc. A +rough diagram should be made on a larger scale for practice, and +numbered counters or pieces of cardboard used. + + +343.--EXERCISE FOR PRISONERS. + +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. + +[Illustration] + +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:-- + + 8 3 12 1 + 11 14 9 6 + 4 7 2 13 + 15 10 5 + +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. + +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. + +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. + + +344.--THE KENNEL PUZZLE. + +[Illustration] + +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? + + +345.--THE TWO PAWNS. + +[Illustration] + +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. + + + + +VARIOUS CHESS PUZZLES. + + "Chesse-play is a good and wittie exercise of + the minde for some kinde of men." + Burton's _Anatomy of Melancholy_. + +346.--SETTING THE BOARD. + +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. + + +347.--COUNTING THE RECTANGLES. + +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? + + +348.--THE ROOKERY. + +[Illustration] + +The White rooks cannot move outside the little square in which they are +enclosed except on the final move, in giving checkmate. The puzzle 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. + + +349.--STALEMATE. + +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? + + +350.--THE FORSAKEN KING. + +[Illustration] + +Set up the position shown in the diagram. Then the condition of the +puzzle is--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. + + +351.--THE CRUSADER. + +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 _force_ +mate in three moves. + + +352.--IMMOVABLE PAWNS. + +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. + +[Illustration] + + +353.--THIRTY-SIX MATES. + +[Illustration] + +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. + + +354.--AN AMAZING DILEMMA. + +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 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. + +[Illustration] + + +355.--CHECKMATE! + +[Illustration] + +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. + + +356.--QUEER CHESS. + +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. + + +357.--ANCIENT CHINESE PUZZLE. + +[Illustration] + +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. + + +358.--THE SIX PAWNS. + +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. + + +359.--COUNTER SOLITAIRE. + +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 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. + +[Illustration] + +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. + + +360.--CHESSBOARD SOLITAIRE. + +[Illustration] + +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. + + +361.--THE MONSTROSITY. + +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:-- + +[Illustration] + +My fellow-passenger suddenly turned his head and caught the look of +bewilderment on my face. + +"Do you play chess?" he asked. + +"Yes, a little. What is that? A problem?" + +"Problem? No; a game." + +"Impossible!" I exclaimed rather rudely. "The position is a perfect +monstrosity!" + +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." + +"It is a correspondence game." he exclaimed. "That is my friend's last +move, and I am considering my reply." + +"But you really must excuse me; the position seems utterly impossible. +How on earth, for example--" + +"Ah!" he broke in smilingly. "I see; you are a beginner; you play to +win." + +"Of course you wouldn't play to lose or draw!" + +He laughed aloud. + +"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?" + +I inwardly congratulated myself that I never had. + +"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--" + +"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--"Insane, but +quite harmless. He is in my charge." + +After that I let the poor fellow run on in his wild way until both got +out at the next station. + +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. + + + + +MEASURING, WEIGHING, AND PACKING PUZZLES. + + "Measure still for measure." + _Measure for Measure_, v. 1. + + +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 Meziriac reprinted this and other +of Tartaglia's puzzles in his _Problemes plaisans et delectables_ +(1612). It is the general opinion that puzzles of this class can only be +solved by trial, but I think formulae can be constructed for the solution +generally of certain related cases. It is a practically unexplored field +for investigation. + +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 _Mathematical Recreations_ (5th edition). + +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. + + +362.--THE WASSAIL BOWL. + +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. + + +363.--THE DOCTOR'S QUERY. + +"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?" + + +364.--THE BARREL PUZZLE. + +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 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. + +[Illustration] + + +365.--NEW MEASURING PUZZLE. + +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. + + +366.--THE HONEST DAIRYMAN. + +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--or two parts of milk and one of +water--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. + + +367.--WINE AND WATER. + +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. + +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? + + +368.--THE KEG OF WINE. + +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 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? + + +369.--MIXING THE TEA. + +"Mrs. Spooner called this morning," said the honest grocer to his +assistant. "She wants twenty pounds of tea at 2s. 41/2d. per lb. Of +course we have a good 2s. 6d. tea, a slightly inferior at 2s. 3d., and a +cheap Indian at 1s. 9d., but she is very particular always about her +prices." + +"What do you propose to do?" asked the innocent assistant. + +"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." + +How was the poor fellow to mix the three teas? Could you have shown him +how to do it? + + +370.--A PACKING PUZZLE. + +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+9/10 inches long, 22+4/5 inches +wide, and 14 inches deep. Now, what is the greatest number of the +balls that he could pack into that box? + + +371.--GOLD PACKING IN RUSSIA. + +The editor of the _Times_ 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 _Daily Mail_ 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. + +Russian officials were engaged in packing 800 gold slabs, each measuring +121/2 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. + + +372.--THE BARRELS OF HONEY. + +[Illustration] + +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. + +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--full, half-full, or empty. Can you show how they succeeded +in making a correct division of the property? + + + + +CROSSING RIVER PROBLEMS + + "My boat is on the shore." + BYRON. + + +This is another mediaeval 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. + + +[Illustration] + +373.--CROSSING THE STREAM. + +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--while the dog was a +perfect fool. + + +374--CROSSING THE RIVER AXE. + +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 L800, Jasper L500 +worth, and Timothy L300 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 tricks, such as ropes, +"flying bridges," currents, swimming, or similar dodges, may be +employed. + + +375.--FIVE JEALOUS HUSBANDS. + +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. + +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. + + +376.--THE FOUR ELOPEMENTS. + +Colonel B---- 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. + +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 _terra firma_, 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. + +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--or would have been to any one except +themselves. + +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. + +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. + +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. + + +377.--STEALING THE CASTLE TREASURE. + +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. + +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? + +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. + + + + +PROBLEMS CONCERNING GAMES. + + "The little pleasure of the game." + MATTHEW PRIOR. + +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. + + +378.--DOMINOES IN PROGRESSION. + +[Illustration] + +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. + + +379.--THE FIVE DOMINOES. + +[Illustration] + +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: -- + + (1--0) (0--0) (0--2) (2--1) (1--3) + (4--0) (0--0) (0--2) (2--1) (1--0) + (2--0) (0--0) (0--1) (1--3) (3--0) + +Now, how many similar arrangements are there of five dominoes that shall +give six instead of five in the two additions? + + +380.--THE DOMINO FRAME PUZZLE. + +[Illustration] + +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. + + +381.--THE CARD FRAME PUZZLE. + +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 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. + +[Illustration] + + +382.--THE CROSS OF CARDS. + +[Illustration] + +In this case we use only nine cards--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. + + +383.--THE "T" CARD PUZZLE. + +[Illustration] + +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? + + +384.--CARD TRIANGLES. + +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 is to find out in just how many different ways this can be +done. + +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. + +[Illustration] + + +385.--"STRAND" PATIENCE. + +The idea for this came to me when considering the game of Patience that +I gave in the _Strand Magazine_ for December, 1910, which has been +reprinted in Ernest Bergholt's _Second Book of Patience Games_, under +the new name of "King Albert." + +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--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. + + +386.--A TRICK WITH DICE. + +[Illustration] + +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. + + +387.--THE VILLAGE CRICKET MATCH. + +In a cricket match, Dingley Dell v. 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--four in all. To what +extent, if any, did this manoeuvre increase Mr. Struggles's total? + + +388.--SLOW CRICKET. + +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? + + +389.--THE FOOTBALL PLAYERS. + +"It is a glorious game!" an enthusiast was heard to exclaim. "At the +close of last season, 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? + +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. + + +390.--THE HORSE-RACE PUZZLE. + +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. + +Three horses--Acorn, Bluebottle, and Capsule--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 L13, no matter which horse +comes in first? Supposing, as an example, that I betted L5 on each +horse. Then, if Acorn won, I should receive L20 (four times L5), and +have to pay L5 each for the other two horses; thereby winning L10. But +it will be found that if Bluebottle was first I should only win L5, 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." + + + +391.--THE MOTOR-CAR RACE. + +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:-- + +"There's Gogglesmith--that man in the white car!" + +"Yes, I see," was the reply; "but how many cars are running in this +race?" + +Then came this curious rejoinder:-- + +"One-third of the cars in front of Gogglesmith added to three-quarters +of those behind him will give you the answer." + +Now, can you tell how many cars were running in the race? + + + + +PUZZLE GAMES. + + + "He that is beaten may be said + To lie in honour's truckle bed." + HUDIBRAS. + +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, "_Canterbury Puzzles_") 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. + +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. + + +392.--THE PEBBLE GAME. + +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. + + +393.--THE TWO ROOKS. + +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 +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? + +[Illustration] + + +394.--PUSS IN THE CORNER. + +[Illustration] + +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. + +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? + + +395.--A WAR PUZZLE GAME. + +[Illustration] + +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. + + +396.--A MATCH MYSTERY. + +Here is a little game that is childishly simple in its conditions. But +it is well worth investigation. + +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. + +"Here are thirty matches," he said. "I 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." + +"As I can draw any number," Mr. Wilson said, "suppose I exhibit my usual +moderation and take all the 14 heap." + +"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." + +"Very well, then," said Mr. Wilson. "I will take 6 from the 14, and +leave you 8, 11, 5." + +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. + +"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." + + +397.--THE MONTENEGRIN DICE GAME. + +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. + +The puzzle is to discover which two pairs of numbers should be selected +in order to give both players an exactly even chance. + + +398.--THE CIGAR PUZZLE. + +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. + +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 41/2 +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? + + + + +MAGIC SQUARE PROBLEMS. + + "By magic numbers." + CONGREVE, _The Mourning Bride._ + +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 Frenicle published in 1693 diagrams of all the arrangements +of the fourth order--880 in number--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 +_Nature_, 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. + +As to classification, it is largely a matter of individual +taste--perhaps an aesthetic 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 +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. + +[Illustration: SIMPLE] + +[Illustration: SEMI-NASIK] + +[Illustration: ASSOCIATED] + +[Illustration: NASIK] + + +I published in _The Queen_ 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 + 4 + 11 + 5 = 34 and 12 + 6 + +13 + 3 = 34. The third example is not only Semi-Nasik but 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 + 15, 3 + 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 + 14 + 2 + 3, and 10 + 4 + 7 + 13, and 15 + 5 + 2 + 12. As a +consequence, its properties are such that if you repeat the square in +all directions you may mark off a square, 4 x 4, wherever you please, +and it will be magic. + +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 ++ 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. + + NASIK (Type I.) . . . . . 48 + SEMI-NASIK (Type II., Transpositions + of Nasik) . 48 + " (Type III., Associated) 48 + " (Type IV.) . . . 96 + " (Type V.) . . . 96 192 + ___ + " (Type VI.) . . . 96 384 + ___ + SIMPLE. (Type VI.) . . . 208 + " (Type VII.) . . . 56 + " (Type VIII.). . . 56 + " (Type IX.) . . . 56 + " (Type X.) . . . 56 224 + ___ + " (Type XI.) . . . 8 + " (Type XII.) . . . 8 16 448 + ___ ___ ___ + 880 + ___ + +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. + +An infinite variety of puzzles may be made introducing new conditions +into the magic square. In _The Canterbury Puzzles_ 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. + + +399.--THE TROUBLESOME EIGHT. + +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. + +[Illustration] + + +400.--THE MAGIC STRIPS. + +[Illustration] + +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. + +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. + +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 be turned upside down or placed on their sides--that is, all +the strips must lie in their original direction. + +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. + + +401.--EIGHT JOLLY GAOL BIRDS. + +[Illustration] + +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. + +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--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? + + +402.--NINE JOLLY GAOL BIRDS. + +[Illustration] + +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. + + +403.--THE SPANISH DUNGEON. + +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. + +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." + +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 from cell to cell in the order in which he should +call out their numbers. + +[Illustration] + +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. + + +404.--THE SIBERIAN DUNGEONS. + +[Illustration] + +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. + +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. + + +405.--CARD MAGIC SQUARES. + +[Illustration] + +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. + + +406.--THE EIGHTEEN DOMINOES. + +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 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. + +[Illustration] + + + + +SUBTRACTING, MULTIPLYING, AND DIVIDING MAGICS. + +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 +_Tit-Bits_ in 1897. The dividing magic was apparently first discussed by +me in _The Weekly Dispatch_ 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. + +[Illustration: ADDING SUBTRACTING MULTIPLYING DIVIDING] + +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 n times that of the subtracting square derived from +it, where n 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"--a term I have explained in the introductory +article to this department. + +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-- + + 1 3 5 + 4 6 8 + 7 9 11 + +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--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-- + + 1 3 9 + 2 6 18 + 4 12 36 + +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. + +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 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. + +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 n 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 nth) of +that of the former. + +[Illustration] + +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. + + +407.--TWO NEW MAGIC SQUARES. + +Construct a subtracting magic square with the first sixteen whole +numbers that shall be "associated" by _subtraction_. 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 _division_. 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. + + +408.--MAGIC SQUARES OF TWO DEGREES. + +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. + + + + + +MAGIC SQUARES OF PRIMES. + +The problem of constructing magic squares with prime numbers only was +first discussed by myself in _The Weekly Dispatch_ 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 _The Monist_ (Chicago) for October 1913:-- + + Order of Totals of Lowest Squares + Square. Series. Constants. made by-- + (Henry E. + 3rd 333 111 { Dudeney + ( (1900). + + (Ernest Bergholt + 4th 408 102 { and C. D. + ( Shuldham. + + 5th 1065 213 H. A. Sayles. + + (C. D. Shuldham + 6th 2448 408 { and J. + ( N. Muncey. + + 7th 4893 699 do. + 8th 8912 1114 do. + 9th 15129 1681 do. + 10th 24160 2416 J. N. Muncey. + 11th 36095 3355 do. + 12th 54168 4514 do. + +For further details the reader should consult the article itself, by W. +S. Andrews and H. A. Sayles. + +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. + + +409.--THE BASKETS OF PLUMS. + +[Illustration] + +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. + +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. + +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. + + +410.--THE MANDARIN'S "T" PUZZLE. + +[Illustration] + +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 beat him easily. I present a little problem that one +learned mandarin propounded to our traveller, as depicted on the last +page. + +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? + + +411.--A MAGIC SQUARE OF COMPOSITES. + +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--the smallest +possible. + + +412.--THE MAGIC KNIGHT'S TOUR. + +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." + + + + +MAZES AND HOW TO THREAD THEM. + + "In wandering mazes lost." + _Paradise Lost._ + +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. + +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 Daedalus 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." + +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. + +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 341/2 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. + +[Illustration: FIG. 1.--Maze at St. Quentin.] + +[Illustration: FIG. 2.--Maze in Chartres Cathedral.] + +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. + +[Illustration: FIG. 3.--Maze in Lucca Cathedral.] + +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 "_Tour through Part of +Normandy_" (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." + +[Illustration: FIG. 4.--Maze at Saffron Walden, Essex.] + +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 191/2 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 +Chalons-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--where the centre was frequently called "Ciel." + +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 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"--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:-- + + "The nine men's morris is filled up with mud, + And the quaint mazes in the wanton green + For lack of tread are undistinguishable." + + _A Midsummer Night's Dream_, ii. 1. + + + + "My old bones ache: here's a maze trod indeed, + Through forth-rights and meanders!" + + _The Tempest_, iii. 3. + + +[Illustration: FIG. 5.--Maze at Sneinton, Nottinghamshire.] + +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. + +[Illustration: FIG. 6.--Maze at Alkborough, Lincolnshire.] + +In the next two mazes given--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)--the paths must in +each case be understood to be on the lines, black or white, as the case +may be. + +[Illustration: FIG. 7.--Maze at Boughton Green, Nottinghamshire.] + +I give in Fig. 6 a maze that was at Alkborough, Lincolnshire, +overlooking the Humber. This was 44 feet in diameter, and the +resemblance 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. + +[Illustration: FIG. 8.--Maze at Wing, Rutlandshire.] + +[Illustration: FIG. 9.--Maze on St. Catherine's Hill, Winchester.] + +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. + +[Illustration: FIG. 10.--Maze on Ripon Common.] + +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. + +[Illustration: FIG. 11.--Maze at Theobalds, Hertfordshire.] + +In the case of the maze at Theobalds, Hertfordshire, after you have +found the entrance within the four enclosing hedges, the path is 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, _The Retired Gard'ner_, published in 1706 +(Fig. 13). Also, I add a Dutch maze (Fig. 14). + +[Illustration: FIG. 12.--Italian Maze of Sixteenth Century.] + +[Illustration: FIG. 13.--By the Designers of Hampton Court Maze.] + +[Illustration: FIG. 14.--A Dutch Maze.] + +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 +_topiarius_ 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. + +[Illustration: FIG. 15.--Maze at Hampton Court Palace.] + +[Illustration: FIG. 16.--Maze at Hatfield House, Herts.] + +[Illustration: FIG. 17.--Maze formerly at South Kensington.] + +[Illustration: FIG. 18.--A German Maze.] + +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 +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--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 acre of ground; but it was, +unfortunately, ploughed up in 1730. + +[Illustration: FIG. 19.--Maze at Pimperne, Dorset.] + +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. + +[Illustration: FIG. 20.--M. Tremaux's Method of Solution.] + +[Illustration: FIG. 21.--How to thread the Hatfield Maze.] + +Now look at the Hampton Court plan. Follow, say to the right, the path +indicated by the 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--for ever after! + +[Illustration: FIG. 22. The Philadelphia Maze, and its Solution.] + +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 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. + +[Illustration: FIG. 23.--Simplified Diagram of Fig. 22.] + +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. Tremaux, 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. + +[Illustration: FIG. 24.--Can you find the Shortest Way to Centre?] + +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. + +[Illustration: FIG. 25.--Rosamund's Bower.] + +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, 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--by blowing your +brains out. + +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. + +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. + + + + +THE PARADOX PARTY. + + "Is not life itself a paradox?" + C.L. DODGSON, _Pillow Problems_. + + +"It is a wonderful age!" said Mr. Allgood, and everybody at the table +turned towards him and assumed an attitude of expectancy. + +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. + +"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." + +"That would appeal to me," said Mrs. Allgood, "for I cannot endure a +room with a north aspect." + +"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?" + +"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!" + +"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." + +"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 hairs on their heads.' If this is a fact, I +confess I cannot see it." + +"How do the bald-headed affect the question?" asked Uncle John. + +"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." + +"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?" + +"Let me think," said Mrs. Allgood. "Yes--yes--that is correct." + +"Very well, then. As there are only nine hundred and ninety-nine +thousand nine hundred and ninety-nine _different_ ways of bearing hair, +it is clear that the millionth person must repeat one of those ways. Do +you see?" + +"Yes; I see that--at least I think I see it." + +"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." + +"But, Mr. Filkins," said little Willie Allgood, "why could not the +millionth man have, say, ten thousand hairs and a half?" + +"That is mere hair-splitting, Willie, and does not come into the +question." + +"Here is a curious paradox," said George. "If a thousand soldiers are +drawn up in battle array on a plane"--they understood him to mean +"plain"--"only one man will stand upright." + +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. + +"In the same way," he remarked, "if a billiard-table were quite +level--that is, a perfect plane--the balls ought to roll to the centre." + +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. + +"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?" + +"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." + +"But isn't Guernsey between England and France?" asked Willie. + +"Yes, certainly." + +"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." + +"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.'" + +"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?" + +"But, Mr. Filkins," asked George, "is this tumbler half full of water +equal to a similar glass half empty?" + +"Certainly, George." + +"Then it follows from the axiom that a glass full must equal a glass +empty. Is that correct?" + +"No, clearly not. I never thought of it in that light." + +"Perhaps," suggested Mr. Allgood, "the rule does not apply to liquids." + +"Just what I was thinking, Allgood. It would seem that we must make an +exception in the case of liquids." + +"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?" + +"Well, let me see! No, of course not," Mr. Filkins replied, "because two +miles square is four square miles." + +"Then," said George, "if the axiom is not true in these cases, when is +it true?" + +Mr. Filkins promised to look into the matter, and perhaps the reader +will also like to give it consideration at leisure. + +"Look here, George," said his cousin Reginald Woolley: "by what +fractional part does four-fourths exceed three-fourths?" + +"By one-fourth!" shouted everybody at once. + +"Try another one," George suggested. + +"With pleasure, when you have answered that one correctly," was +Reginald's reply. + +"Do you mean to say that it isn't one-fourth?" + +"Certainly I do." + +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. + +"Uncle John, how do you pronounce 't-o-o'?" asked Willie. + +"'Too," my boy." + +"And how do you pronounce 't-w-o'?" + +"That is also 'too.'" + +"Then how do you pronounce the second day of the week?" + +"Well, that I should pronounce 'Tuesday,' not 'Toosday.'" + +"Would you really? I should pronounce it 'Monday.'" + +"If you go on like this, Willie," said Uncle John, with mock severity, +"you will soon be without a friend in the world." + +"Can any of you write down quickly in figures 'twelve thousand twelve +hundred and twelve pounds'?" asked Mr. Allgood. + +His eldest daughter, Miss Mildred, was the only person who happened to +have a pencil at hand. + +"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, +"L13,212." + +"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?" + +"I suppose," said Mrs. Allgood, "it is a conundrum--something to do with +their poor little 'souls.'" + +But after everybody had given it up, Mildred reminded the company that +only boys were put to death. + +"Once upon a time," began George, "Achilles had a race with a +tortoise--" + +"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." + +"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?" + +"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." + +"Quite right," confirmed Mr. Allgood. "Nothing could be easier." + +"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." + +[Illustration] + +"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." + +"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." + +But Mrs. Allgood was not convinced. + +Mr. Smoothly, the curate, at the end of the table, said at this point +that he had a little question to ask. + +"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." + +"'I'll put a girdle round about the earth in forty minutes,'" muttered +George, quoting the words of Puck in _A Midsummer Night's Dream_. + +"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?" + +"In such a great length," said Mr. Allgood, "I do not suppose the +distance would be worth mentioning." + +"What do you say, George?" asked Mr. Smoothly. + +"Well, without calculating I should imagine it would be a very minute +fraction of an inch." + +Reginald and Mr. Filkins were of the same opinion. + +"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!" + +"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. + +"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--absolutely nothing!" + +"Impossible!" insisted Mildred. "And only three months old!" + +"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?" + +"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?" + +"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?" + +"Quite simple," said Mr. Allgood. "They should have described one as +'Andrew Johnson, Senior,' and the other as 'Andrew Johnson, Junior.'" + +"But I forgot to tell you that the father died first." + +"What difference can that make?" + +"Well, you see, they wanted to be absolutely exact, and that was the +difficulty." + +"But I don't see any difficulty," said Mr. Allgood, nor could anybody +else. + +"Well," explained Mr. Smoothly, "it is like this. If the father died +first, the son was then no longer 'Junior.' Is that so?" + +"To be strictly exact, yes." + +"That is just what they wanted--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?" + +"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?" + +"Perhaps you know the man to be addicted to lying," said Reginald. + +"But that would be no proof that he was not telling the truth in this +instance." + +"Perhaps," suggested Mildred, "you know that there were no coins made at +those dates. + +"On the contrary, they were made at both periods." + +"Were they silver or copper coins?" asked Willie. + +"My friend did not state, and I really cannot see, Willie, that it makes +any difference." + +"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." + +"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.'" + +"Why not?" asked Mrs. Allgood. "He _was_ George I." + +"Yes; but they would not know it until there was a George II." + +"Then there was no George II. until George III. came to the throne?" + +"That does not follow. The second George becomes 'George II.' on account +of there having been a 'George I.'" + +"Then the first George was 'George I.' on account of there having been +no king of that name before him." + +"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." + +"But there _have_ been several Georges, and therefore he was 'George I.' +There _haven't_ been several Victorias, so the two cases are not +similar." + +They gave up the attempt to convince Mrs. Allgood, but the reader will, +of course, see the point clearly. + +"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?" + +"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." + +"Then they were equal?" + +"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." + +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. + +"Mr. Filkins, can you answer this?" asked Willie. "There is a man in the +next village who eats two eggs for breakfast every morning." + +"Nothing very extraordinary in that," George broke in. "If you told us +that the two eggs ate the man it would be interesting." + +"Don't interrupt the boy, George," said his mother. + +"Well," Willie continued, "this man neither buys, borrows, barters, +begs, steals, nor finds the eggs. He doesn't keep hens, and the eggs are +not given to him. How does he get the eggs?" + +"Does he take them in exchange for something else?" asked Mildred. + +"That would be bartering them," Willie replied. + +"Perhaps some friend sends them to him," suggested Mrs. Allgood. + +"I said that they were not given to him." + +"I know," said George, with confidence. "A strange hen comes into his +place and lays them." + +"But that would be finding them, wouldn't it?" + +"Does he hire them?" asked Reginald. + +"If so, he could not return them after they were eaten, so that would be +stealing them." + +"Perhaps it is a pun on the word 'lay,'" Mr. Filkins said. "Does he lay +them on the table?" + +"He would have to get them first, wouldn't he? The question was, How +does he get them?" + +"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. + +"The man keeps ducks!" he cried, "and his servant collects the eggs +every morning." + +"But you said he doesn't keep birds!" George protested. + +"I didn't, did I, Mr. Filkins? I said he doesn't keep hens." + +"But he finds them," said Reginald. + +"No; I said his servant finds them." + +"Well, then," Mildred interposed, "his servant gives them to him." + +"You cannot give a man his own property, can you?" + +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. + + +413.--A CHESSBOARD FALLACY. + +[Illustration] + +"Here is a diagram of a chessboard," he said. "You see there are +sixty-four squares--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." + +"It seems to be like the other old chessboard fallacy, and perhaps the +explanation is the same," said Reginald--"that the pieces do not exactly +fit." + +"But they _do_ fit," said Uncle John. "Try it, and you will see." + +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? + + + + +UNCLASSIFIED PROBLEMS. + + + "A snapper up of unconsidered trifles." + _Winter's Tale_, iv. 2. + + +414.--WHO WAS FIRST? + +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. + +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? + + +415.--A WONDERFUL VILLAGE. + +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? + + +416.--A CALENDAR PUZZLE. + +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? + + +417.--THE TIRING IRONS. + +[Illustration] + +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." + +The puzzle will be seen to consist of a simple _loop_ of wire fixed in a +handle to be held in the left hand, and a certain number of _rings_ +secured by _wires_ which pass through holes in the _bar_ 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. + +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. + +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. + +Drop the first ring; drop the third; put up the first; drop the second; +and drop the first--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 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. + +[Illustration] + +[Illustration: + + + o o o o o * * + {------------- + + o o o o * o + 1{------------- o + + o o o o o + 2{------------- + o o + + o o o o * * + 3{------------- + o + + o o o o * + 4{------------- + o o + + o o * o + 5{------------- + o o o + +] + +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? + + + +418.--SUCH A GETTING UPSTAIRS. + +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. + +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. + + +419.--THE FIVE PENNIES. + +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--place them +so that every penny shall touch every other penny--and you will find it +a different matter altogether. + + +420.--THE INDUSTRIOUS BOOKWORM. + +[Illustration] + +Our friend Professor Rackbrane is seen in the illustration to be +propounding another of his 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? + + +421.--A CHAIN PUZZLE. + +[Illustration] + +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 2s. 2d. 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. + + +422.--THE SABBATH PUZZLE. + +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. + + Christians the week's _first_ day for Sabbath hold; + The Jews the _seventh_, as they did of old; + The Turks the _sixth_, as we have oft been told. + How can these three, in the same place and day, + Have each his own true Sabbath? tell, I pray. + + +423.--THE RUBY BROOCH. + +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. + +"A fine collection of rubies, madam," said the shopkeeper, to whom her +ladyship was a stranger. + +"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." + +[Illustration] + +Six months later a brother of Lady Littlewood'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. + +"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?" + +"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." + +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. + +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. + + +424.--THE DOVETAILED BLOCK. + +[Illustration] + +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. + + +425.--JACK AND THE BEANSTALK. + +[Illustration] + +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? + + +426.--THE HYMN-BOARD POSER. + +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 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. + +[Illustration] + +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. + +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 6d., and +for the painting of each plate the charges are to be: For one plate, +1s.; for two plates alike, 113/4d. each; for three plates alike, +111/2d. each, and so on, the charge being one farthing less per plate +for each similarly painted plate. Now, what should be the lowest cost? + +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. + + +427.--PHEASANT-SHOOTING. + +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:-- + +"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?" + +It seems a simple enough question, but can the reader give a correct +answer? + + +428.--THE GARDENER AND THE COOK. + +A correspondent, signing himself "Simple Simon," suggested that I should +give a special catch puzzle in the issue of _The Weekly Dispatch_ 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. + +[Illustration] + +"The illustration is a fancy sketch of my correspondent, 'Simple Simon,' +in the act of trying to solve the following innocent little 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?" + +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--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." + + +429.--PLACING HALFPENNIES. + +[Illustration] + +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? + + +430.--FIND THE MAN'S WIFE. + +[Illustration] + +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:-- + +"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--and you say this is correct." + +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, 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--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. + +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!" + +I now leave the question in the hands of my readers. Which is really +Number 10's wife? + +The illustration is of necessity considerably reduced from the large +scale on which it originally appeared in _The Weekly Dispatch_ (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. + + + + +SOLUTIONS. + + +1.--A POST-OFFICE PERPLEXITY. + +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. + + +2.--YOUTHFUL PRECOCITY. + +The price of the banana must have been one penny farthing. Thus, 960 +bananas would cost L5, and 480 sixpences would buy 2,304 bananas. + + +3.--AT A CATTLE MARKET. + +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. + + +4.--THE BEANFEAST PUZZLE. + +The cobblers spent 35s., the tailors spent also 35s., the hatters spent +42s., and the glovers spent 21s. Thus, they spent altogether L6,13s., +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. + + +5.--A QUEER COINCIDENCE. + +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 n players, the amount held by every player at +the end will be m(2^n), the last winner must have held m(n + 1) +at the start, the next m(2n + 1), the next m(4n + 1), the next +m(8n + 1), and so on to the first player, who must have held +m(2^{n - 1}n + 1). + +Thus, in this case, n = 7, and the amount held by every player at the +end was 2^7 farthings. Therefore m = 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. + + +6.--A CHARITABLE BEQUEST. + +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. + + +7.--THE WIDOW'S LEGACY. + +The widow's share of the legacy must be L205, 2s. 6d. and 10/13 of a +penny. + + +8.--INDISCRIMINATE CHARITY + +The gentleman must have had 3s. 6d. in his pocket when he set out for +home. + + +9.--THE TWO AEROPLANES. + +The man must have paid L500 and L750 for the two machines, making +together L1,250; but as he sold them for only L1,200, he lost L50 by the +transaction. + + +10.--BUYING PRESENTS. + +Jorkins had originally L19, 18s. in his pocket, and spent L9, 19s. + + +11.--THE CYCLISTS' FEAST. + +There were ten cyclists at the feast. They should have paid 8s. each; +but, owing to the departure of two persons, the remaining eight would +pay 10s. each. + + +12.--A QUEER THING IN MONEY. + +The answer is as follows: L44,444, 4s. 4d. = 28, and, reduced to pence, +10,666,612=28. + +It is a curious little coincidence that in the answer 10,666,612 the +four central figures indicate the only other answer, L66, 6s. 6d. + + + +13.--A NEW MONEY PUZZLE. + +The smallest sum of money, in pounds, shillings, pence, and farthings, +containing all the nine digits once, and once only, is L2,567, 18s. +93/4d. + + +14.--SQUARE MONEY. + +The answer is 11/2d. and 3d. Added together they make 41/2d., and +11/2d. multiplied by 3 is also 41/2d. + + +15.--POCKET MONEY. + +The largest possible sum is 15s. 9d., composed of a crown and a +half-crown (or three half-crowns), four florins, and a threepenny piece. + + +16.--THE MILLIONAIRE'S PERPLEXITY. + +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. + +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. + +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. + +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. + + +17.--THE PUZZLING MONEY BOXES. + +The correct answer to this puzzle is as follows: John put into his +money-box two double florins (8s.), William a half-sovereign and a +florin (12s.), Charles a crown (5s.), and Thomas a sovereign (20s.). +There are six coins in all, of a total value of 45s. If John had 2s. +more, William 2s. less, Charles twice as much, and Thomas half as much +as they really possessed, they would each have had exactly 10s. + + +18.--THE MARKET WOMEN. + +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. + + +19.--THE NEW YEAR'S EVE SUPPERS. + +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 L5 +together. + + +20.--BEEF AND SAUSAGES. + +The lady bought 48 lbs. of beef at 2s., and the same quantity of +sausages at 1s. 6d., thus spending L8, 8s. Had she bought 42 lbs. of +beef and 56 lbs. of sausages she would have spent L4, 4s. on each, and +have obtained 98 lbs. instead of 96 lbs.--a gain in weight of 2 lbs. + + +21.--A DEAL IN APPLES. + +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. + + +22.--A DEAL IN EGGS. + +The man must have bought ten eggs at fivepence, ten eggs at one penny, +and eighty eggs 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. + + +23.--THE CHRISTMAS-BOXES. + +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 L1, 10s. 1d. + + +24.--A SHOPPING PERPLEXITY. + +The first purchase amounted to 1s. 53/4d., the second to 1s. 111/2d., +and together they make 3s. 51/4d. Not one of these three amounts can be +paid in fewer than six current coins of the realm. + + +25.--CHINESE MONEY. + +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--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. + + +26.--THE JUNIOR CLERKS' PUZZLE. + +Although Snoggs's _reason_ for wishing to take his rise at L2, 10s. +half-yearly did not concern our puzzle, the _fact_ 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 +L350 in five years, the artful Snoggs actually obtained L362, 10s. in +the same time. The rest is simplicity itself. It is evident that if +Moggs saved L87, 10s. and Snoggs L181, 5s., 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 +L268, 15s. + + +27.--GIVING CHANGE. + +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. + + +28.--DEFECTIVE OBSERVATION. + +Of course the date on a penny is on the same side as Britannia--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. + + +29.--THE BROKEN COINS. + +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 +13/253 of the original value has been lost. And as the same fraction of +each coin has been broken away, each coin has lost 13/253 of its +original bulk. + + +30.--TWO QUESTIONS IN PROBABILITIES. + +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:-- + + (a) 5 heads 1 way + (b) 5 tails 1 way + (c) 4 heads and 1 tail 5 ways + (d) 4 tails and 1 head 5 ways + (e) 3 heads and 2 tails 10 ways + (f) 3 tails and 2 heads 10 ways + +Now, it will be seen that the only favourable cases are a, b, c, +and d--12 cases. The remaining 20 cases are unfavourable, because they +do 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. + +The amount that should be paid for a draw from the bag that contains +three sovereigns and one shilling is 15s. 3d. 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 15s., overlooking the fact that +one must draw at least a shilling--there being no blanks. + + + +31.--DOMESTIC ECONOMY. + +Without the hint that I gave, my readers would probably have been +unanimous in deciding that Mr. Perkins's income must have been L1,710. +But this is quite wrong. Mrs. Perkins says, "We have spent a third of +his yearly income in rent," etc., etc.--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 _not_ say that they have spent _each year_ this sum, +whatever it is, but that _during the two years_ that amount has been +spent. The only possible answer, according to the exact reading of her +words, is, therefore, that his income was L180 per annum. Thus the +amount spent in two years, during which his income has amounted to L360, +will be L60 in rent, etc., L90 in domestic expenses, L20 in other ways, +leaving the balance of L190 in the bank as stated. + + +32.--THE EXCURSION TICKET PUZZLE. + +Nineteen shillings and ninepence may be paid in 458,908,622 different +ways. + +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. + +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 (n + 1) squared ways where n always represents the number of +pence. If threepenny-pieces are admitted, there are + + 2n cubed + 15n squared + 33n + --------------------- + 1 ways. + 18 + +If sixpences are also used there are + + n^{4} + 22n cubed + 159n squared + 414n + 216 + --------------------------------- + 216 + +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. + +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 + + Farthing in 0 way. + Halfpenny in 1 way. + Penny in 3 ways. + Threepenny-piece in 16 ways. + Sixpence in 66 ways. + Shilling in 402 ways. + Florin in 3,818 ways. + Half-crown in 8,709 ways. + Double florin in 60,239 ways. + Crown in 166,651 ways. + Half-sovereign in 6,261,622 ways. + Sovereign in 500,291,833 ways. + +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. + + +33.--A PUZZLE IN REVERSALS. + +(i) L13. (2) L23, 19s. 11d. The words "the number of pounds exceeds that +of the pence" exclude such sums of money as L2, 16s. 2d. and all sums +under L1. + + +34.--THE GROCER AND DRAPER. + +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 _forty-seven_ 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! + + +35.--JUDKINS'S CATTLE. + +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--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. + + +36.--BUYING APPLES. + +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 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--three girls and three +boys. + + +37.--BUYING CHESTNUTS. + +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. + +"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 _of a chestnut more_,' he remarked. 'But if I give you one +chestnut more,' the shopman replied, 'you will have _five-sixths_ too +many.' Now, strange to say, they were both right. How many chestnuts +should the buyer receive for half a crown?" + +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+1/6 +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. + + +38.--THE BICYCLE THIEF. + +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. + + +39.--THE COSTERMONGER'S PUZZLE. + +Bill must have paid 8s. per hundred for his oranges--that is, 125 for +10s. At 8s. 4d. per hundred, he would only have received 120 oranges for +10s. This exactly agrees with Bill's statement. + + +40.--MAMMA'S AGE. + +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. + + +41.--THEIR AGES. + +The gentleman's age must have been 54 years and that of his wife 45 +years. + + +42.--THE FAMILY AGES. + +The ages were as follows: Billie, 31/2 years; Gertrude, 13/4 year; +Henrietta, 51/4 years; Charlie, 101/2; years; and Janet, 21 years. + + +43.--MRS. TIMPKINS'S AGE. + +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. + + +44.--A CENSUS PUZZLE. + +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 241/2 +and 31/2 were correct. + + +45.--MOTHER AND DAUGHTER. + +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. + + +46.--MARY AND MARMADUKE. + +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. + + +47.--ROVER'S AGE. + +Rover's present age is ten years and Mildred's thirty years. Five years +ago their respective 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. 44.) + + +48.--CONCERNING TOMMY'S AGE. + +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. + + +49.--NEXT-DOOR NEIGHBOURS. + +Mr. Jupp 39, Mrs. Jupp 34, Julia 14, and Joe 13; Mr. Simkin 42; Mrs. +Simkin 40; Sophy 10; and Sammy 8. + + +50.--THE BAG OF NUTS. + +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 171/2 years or half the sum of 12, 9, and 14, +their respective ages must be 6, 41/2, and 7 years. + + +51.--HOW OLD WAS MARY? + +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 271/2 and Ann 161/2. 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 (271/2) twice as old as Ann +was (133/4) when Mary was half as old (243/4) as Ann will be (491/2) when Ann +is three times as old (491/2) as Mary was (161/2) when Mary was (161/2) three +times as old as Ann (51/2)." Now, check this backwards. When Mary was +three times as old as Ann, Mary was 161/2 and Ann 51/2 (11 years younger). +Then we get 491/2 for the age Ann will be when she is three times as old +as Mary was then. When Mary was half this she was 243/4. And at that time +Ann must have been 133/4 (11 years younger). Therefore Mary is now twice +as old--271/2, and Ann 11 years younger--161/2. + + +52.--QUEER RELATIONSHIPS. + +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. + +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. + + +53.--HEARD ON THE TUBE RAILWAY. + +The gentleman was the second lady's uncle. + + +54.--A FAMILY PARTY. + +The party consisted of two little girls and a boy, their father and +mother, and their father's father and mother. + + +55.--A MIXED PEDIGREE. + +[Illustration: + + Thos. Bloggs m . . . . . + | + +------------------------+------------+ + | | | + | | | + | W. Snoggs m Kate Bloggs. | + | | | + | | | + . . m Henry Bloggs. | Joseph Bloggs m + | | | + | +--------+-------------+ | + | | | | + | | | | + Jane John Alf. Mary + Bloggs m Snoggs Snoggs m Bloggs + +] + +The letter m stands for "married." It will be seen that John Snoggs +can say to Joseph Bloggs, "You are my _father's brother-in-law_, because +my father married your sister Kate; you are my _brother's +father-in-law_, because my brother Alfred married your daughter Mary; +and you are my _father-in-law's brother_, because my wife Jane was your +brother Henry's daughter." + + +56.--WILSON'S POSER. + +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. + + +57.--WHAT WAS THE TIME? + +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. + + +58.--A TIME PUZZLE. + +Twenty-six minutes. + + +59.--A PUZZLING WATCH. + +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+5/11 +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 5/11 of a minute in +65 minutes, or 60/143 of a minute per hour. + + +60.--THE WAPSHAW'S WHARF MYSTERY. + +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 +hours by 11 we get 1 hr. 5 min. 27+3/11 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+6/11 sec. +(twice the above time); next at 3 hr. 16 min. 21+9/11 sec.; next at 4 +hr. 21 min. 49+1/11 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 _Across the World for a +Wife_, 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+1/11 sec. out in his reckoning. + + +61.--CHANGING PLACES. + +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 (n) to midnight is the sum of 12 - (n + 1) natural numbers. In +the case of the puzzle n = 3; therefore 12 - (3 + 1) = 8 and 1 + 2 + 3 ++ 4 + 5 + 6 + 7 + 8 = 36, the required answer. + +The first pair of times is 3 hr. 21+57/143 min. and 4 hr. 16+112/143 +min., and the last pair is 10 hr. 59+83/143 min. and 11 hr. 54+138/143 +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:-- + + 720b + 60a 720a + 60b min. + a hr ---------- min. and b hr. --------------- + 143 143 + +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. + +By the aid of this formula there is no difficulty in discovering the +answer to the second question: a = 8 and b = 11 will give the pair 8 hr. +58+106/143 min. and 11 hr. 44+128/143 min., the latter being the time +when the minute hand is nearest of all to the point IX--in fact, it is +only 15/143 of a minute distant. + +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 = 0 and b = 1 in the above +expressions we find the first case, and enter hr. 5+5/143 min. at the +head of the first column, and 1 hr. 0+60/143 min. at the head of the +second column. Now, by successively adding 5+5/143 min. in the first, +and 1 hr. 0+60/143 min. in the second column, we get all the _eleven_ +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 = 1 and b = 2, and then by successively +adding these two times as before you will get all the _ten_ pairs after +1 o'clock. Then there is another "jump," and you will be able to get by +addition all the _nine_ 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 + 10 + 9 ++ 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 66 pairs of times, which result agrees +with the formula in the first paragraph of this article. + +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+5/143 +min. past XII;" and this he finally gave as correct, together with the +reason for it that at that time _the time indicated is the same +whichever hand you may assume as hour hand!_ + + +62.--THE CLUB CLOCK. + +The positions of the hands shown in the illustration could only indicate +that the clock stopped at 44 min. 51+1143/1427 sec. after eleven +o'clock. The second hand would next be "exactly midway between the other +two hands" at 45 min. 52+496/1427 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+106/1427 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. + + +63.--THE STOP-WATCH. + +The time indicated on the watch was 5+5/11 min. past 9, when the second +hand would be at 27+3/11 sec. The next time the hands would be similar +distances apart would be 54+6/11 min. past 2, when the second hand would +be at 32+8/11 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. + + +64.--THE THREE CLOCKS. + +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? + +I published this little puzzle in 1898 to see 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. + + +65.--THE RAILWAY STATION CLOCK. + +The time must have been 43+7/11 min. past two o'clock. + + +66.--THE VILLAGE SIMPLETON. + +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. + + +67.--AVERAGE SPEED. + +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. + + +68.--THE TWO TRAINS. + +One train was running just twice as fast as the other. + + +69.--THE THREE VILLAGES. + +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. + + +70.--DRAWING HER PENSION. + +The distance must be 63/4 miles. + + +71.--SIR EDWYN DE TUDOR. + +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--an hour too +soon. If he rode 10 miles an hour, he would arrive at six o'clock--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--the time appointed. + + +72.--THE HYDROPLANE QUESTION. + +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. + + +73.--DONKEY RIDING. + +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. + + +74.--THE BASKET OF POTATOES. + +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--a nice little recreation after a day's work. + + +75.--THE PASSENGER'S FARE. + +Mr. Tompkins should have paid fifteen shillings as his correct share of +the motor-car fare. He only shared half the distance travelled for L3, +and therefore should pay half of thirty shillings, or fifteen shillings. + + +76.--THE BARREL OF BEER. + +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). + + +77.--DIGITS AND SQUARES. + +The top row must be one of the four following numbers: 192, 219, 273, +327. The first was the example given. + + + +78.--ODD AND EVEN DIGITS. + +As we have to exclude complex and improper fractions and recurring +decimals, the simplest solution is this: 79 + 5+1/3 and 84 + 2/6, both +equal 84+1/3. Without any use of fractions it is obviously impossible. + + +79.--THE LOCKERS PUZZLE. + +The smallest possible total is 356 = 107 + 249, and the largest sum +possible is 981 = 235 + 746, or 657+324. The middle sum may be either +720 = 134 + 586, or 702 = 134 + 568, or 407 = 138 + 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:-- + + 107 134 235 + 249 586 746 + --- --- --- + 356 720 981 + + +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. 9/9 - 7/10 - 5/11 -3/12 - 1/13 - +8/14 - 6/15 - 4/16 - 2/17 - 0/18. 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. + + +80.--THE THREE GROUPS. + +There are nine solutions to this puzzle, as follows, and no more:-- + + 12 x 483 = 5,796 27 x 198 = 5,346 + 42 x 138 = 5,796 39 x 186 = 7,254 + 18 x 297 = 5,346 48 x 159 = 7,632 + 28 x 157 = 4,396 + 4 x 1,738 = 6,952 + 4 x 1,963 = 7,852 + +The seventh answer is the one that is most likely to be overlooked by +solvers of the puzzle. + + +81.--THE NINE COUNTERS. + +In this case a certain amount of mere "trial" is unavoidable. But there +are two kinds of "trials"--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. + + +82.--THE TEN COUNTERS. + +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--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. + +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, 3,45O, etc., and it may be ascertained that 3,485 is the lowest +possible. One of the required answers is 3,485 x 2 = 6,970, and 6,970 x +1 = 6,970. + +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 x 64 = 58,560, and 732 x 80 = 58,560. + + +83.--DIGITAL MULTIPLICATION. + +The solution that gives the smallest possible sum of digits in the +common product is 23 x 174 = 58 x 69 = 4,002, and the solution that +gives the largest possible sum of digits, 9x654 =18x327=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. + + +84.--THE PIERROT'S PUZZLE. + +There are just six different solutions to this puzzle, as follows:-- + + 8 multiplied by 473 equals 3784 + 9 " 351 " 3159 + 15 " 93 " 1395 + 21 " 87 " 1287 + 27 " 81 " 2187 + 35 " 41 " 1435 + +It will be seen that in every case the two multipliers contain exactly +the same figures as the product. + + +85.--THE CAB NUMBERS. + +The highest product is, I think, obtained by multiplying 8,745,231 by +96--namely, 839,542,176. + +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. + +These cases have all been given. With five digits there are just +twenty-two solutions, as follows:-- + + 3 x 4128 = 12384 + 3 x 4281 = 12843 + 3 x 7125 = 21375 + 3 x 7251 = 21753 + 2541 x 6 = 15246 + 651 x 24 = 15624 + 678 x 42 = 28476 + 246 x 51 = 12546 + 57 x 834 = 47538 + 75 x 231 = 17325 + 624 x 78 = 48672 + 435 x 87 = 37845 + ------ + 9 x 7461 = 67149 + 72 x 936 = 67392 + ------ + 2 x 8714 = 17428 + 2 x 8741 = 17482 + 65 x 281 = 18265 + 65 x 983 = 63985 + ------ + 4973 x 8 = 39784 + 6521 x 8 = 52168 + 14 x 926 = 12964 + 86 x 251 = 21586 + +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--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. + +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. + +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. + +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 x 4,128 and 3 x 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 x 843 and 21 x 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. + +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. + + +86.--QUEER MULTIPLICATION. + +If we multiply 32547891 by 6, we get the product, 195287346. In both +cases all the nine digits are used once and once only. + + +87.--THE NUMBER CHECKS PUZZLE. + +Divide the ten checks into the following three groups: 7 1 5--4 6--3 2 8 +9 0, and the first multiplied by the second produces the third. + + +88.--DIGITAL DIVISION. + +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:-- + + 6729/13458 = 1/2 + + 5823/17469 = 1/3 + + 3942/15768 = 1/4 + + 2697/13485 = 1/5 + + 2943/17658 = 1/6 + + 2394/16758 = 1/7 + + 3187/25496 = 1/8 + + 6381/57429 = 1/9 + +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--9, +8--1, 7--2, 6--3, or 5--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--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--3, but of course the higher +root may occur either in the numerator or in the denominator; thus +2697/13485, 2769/13845, 2973/14865, 3729/18645, where, in the first two +arrangements, the roots of the numerator and denominator are +respectively 6--3, and in the last two 3--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. + +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 5832/17496 +and 5823/17469, where the 2/6 and 3/9 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--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. + + +89.--ADDING THE DIGITS. + +The smallest possible sum of money is L1, 8s. 93/4d., the digits of which +add to 25. + + +90.--THE CENTURY PUZZLE. + +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:-- + + 2148 1752 1428 1578 + 96 ----, 96 ----, 96 ----, 94 ----, + 537 438 357 263 + + + 7524 5823 5742 3546 + 91 ----, 91 ----, 91 ----, 82 ----, + 836 647 638 197 + + + 7524 5643 69258 + 81 ----, 81 ----, 3 -----. + 396 297 714 + +Now, as all the fractions necessarily represent whole numbers, it will +be convenient to deal with them in the following form: 96 + 4, 94 + 6, +91 + 9, 82 + 18, 81 + 19, and 3 + 97. + +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 ++ 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 + 6 the roots of the numerator and +denominator will be respectively 3--2, in the case of 91 + 9 and of 82 + +18 they will be 9--8, in the case of 81 + 19 they will be 9--9, and in +the case of 3 + 97 they will be 3--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. + +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 that such +combinations as 98 + 2, 92 + 8, 86 + 14, 83 + 17, 74 + 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. + +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 + 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 + 3 there is a +possible form for the digital roots of the fraction, namely, 6--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. + + +91.--MORE MIXED FRACTIONS. + +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:-- + + 9+5472/1368 = 13; + 9+6435/1287 = 14; + 12+3576/894 = 16; + 6+13258/947 = 20; + 15+9432/786 = 27; + 24+9756/813 = 36; + 27+5148/396 = 40; + 65+1892/473 = 69; + 59+3614/278 = 72; + 75+3648/192 = 94. + +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:-- + + 3952 + ---- + 746 = 15; + 3 ---- + 1 + + 5742 + ---- + 638 = 18. + 9 ---- + 1 + +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--such as 26, 35, 44, 53, etc.--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. + + +92.--DIGITAL SQUARE NUMBERS. + +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. + + +93.--THE MYSTIC ELEVEN. + +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. + + +94.--THE DIGITAL CENTURY. + +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. + +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. + + Signs. Strokes. + 1 + 2 + 3 + 4 + 5 + 6 + 7 + (8 x 9) = 100 ( 9 18) + + - (1 x 2) - 3 - 4 - 5 + (6 x 7) + (8 x 9) + = 100 (12 20) + + 1 + (2 x 3) + (4 x 5) - 6 + 7 + (8 x 9) + = 100 (11 21) + + (1 + 2 - 3 - 4)(5 - 6 - 7 - 8 - 9) = 100 ( 9 12) + + 1 + (2 x 3) + 4 + 5 + 67 + 8 + 9 =100 (8 16) + + (1 x 2) + 34 + 56 + 7 - 8 + 9 = 100 (7 13) + + 12 + 3 - 4 + 5 + 67 + 8 + 9 = 100 (6 11) + + 123 - 4 - 5 - 6 - 7 + 8 - 9 = 100 (6 7) + + 123 + 4 - 5 + 67 - 8 - 9 = 100 (4 6) + + 123 + 45 - 67 + 8 - 9 = 100 (4 6) + + 123 - 45 - 67 + 89 = 100 (3 4) + +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. + + +95.--THE FOUR SEVENS. + +The way to write four sevens with simple arithmetical signs so that they +represent 100 is as follows:-- + + 7 7 + -- x -- = 100. + .7 .7 + +Of course the fraction, 7 over decimal 7, equals 7 divided by 7/10, +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. + + +96.--THE DICE NUMBERS. + +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--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. + +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--that is, if the condition were that you could use any +four figures so long as they summed to a given amount--then we have to +remember that several combinations of four digits will, in many cases, +make the same sum. + + 10 11 12 13 14 15 16 17 18 19 20 + 1 1 2 3 5 6 8 9 11 11 12 + + 21 22 23 24 25 26 27 28 29 30 + 11 11 9 8 6 5 3 2 1 1 + +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. + +The following general solution for any number of digits will doubtless +interest readers. Let n represent number of digits, then 5 (10^n - 1) 8! +divided by (9 - n)! equals the required sum. Note that 0! equals 1. This +may be reduced to the following practical rule: Multiply together 4 x 7 +x 6 x 5 ... to (n - 1) factors; now add (n + 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 = 4 (as in the case last mentioned), 4 x +7 x 6 = 168. Therefore 16,800,000 less 1,680 gives us 16,798,320 in +another way. + + +97.--THE SPOT ON THE TABLE. + +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--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. + + +98.--ACADEMIC COURTESIES. + +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. + + +99.--THE THIRTY-THREE PEARLS. + +The value of the large central pearl must have been L3,000. The pearl at +one end (from which they increased in value by L100) was L1,400; the +pearl at the other end, L600. + + +100.--THE LABOURER'S PUZZLE. + +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 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. + + +101.--THE TRUSSES OF HAY. + +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--54 lbs., 56 lbs., 58 lbs., 59 lbs., and 62 +lbs. respectively. + + +102.--MR. GUBBINS IN A FOG. + +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. + + +103.--PAINTING THE LAMP-POSTS. + +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 + + +104.--CATCHING THE THIEF. + +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. + + +105.--THE PARISH COUNCIL ELECTION, + +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. + + +106.--THE MUDDLETOWN ELECTION. + +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. + + +107.--THE SUFFRAGISTS' MEETING. + +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. + + +108.--THE LEAP-YEAR LADIES. + +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. + + +109.--THE GREAT SCRAMBLE. + +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--that is, one-eighth of the +three-quarters that Bob and Andrew had last acquired. + + +110.--THE ABBOT'S PUZZLE. + +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. + + +111.--REAPING THE CORN. + +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. + + +112.--A PUZZLING LEGACY. + +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--that is in the proportion of +four-twelfths to three-twelfths, which is the same as four to three. +Therefore Alfred takes four-sevenths of the hundred acres and Benjamin +three-sevenths. + + +113.--THE TORN NUMBER. + +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. + +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^n - 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 = 3. The factors of 10^n +- 1 are 1¹ x 37¹ (not considering the 3 cubed), and the product of 2 x 2 = +4, the number of solutions. This always includes the special cases 98 - +01, 00 - 01, 998 - 01, 000 - 001, etc. The solutions are obtained as +follows:--Factorize 10 cubed - 1 in all possible ways, always keeping the +powers of 3 together, thus, 37 x 27, 999 x 1. Then solve the equation +37x = 27y + 1. Here x = 19 and y = 26. Therefore, 19 x 37 = 703, the +square of which gives one label, 494,209. A complementary solution +(through 27x = 37x + 1) can at once be found by 10^n - 703 = 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 - 04641 = 4879 squared, where 0238 - 4641 would not work.) The +special case 999 x 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. + + +114.--CURIOUS NUMBERS. + +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. + + +115.--A PRINTER'S ERROR. + +The answer is that 2^5 .9^2 is the same as 2592, and this is the only +possible solution to the puzzle. + + +116.--THE CONVERTED MISER. + +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. + +There are four cases to be considered--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:-- + + 313 weekdays, 52 Sundays L112,055 + 312 weekdays, 53 Sundays 19,345 + 314 weekdays, 52 Sundays No solution possible. + 313 weekdays, 53 Sundays L69,174 + +The lowest possible amount, and therefore the correct answer, is +L19,345, distributed in an ordinary year that began on a Sunday. The +last year of this kind was 1911. He would have paid L53 on every day of +the year, or L62 on every weekday, with L1 left over, as required, in +the latter event. + + +117.--A FENCE PROBLEM. + +Though this puzzle presents no great difficulty to any one possessing a +knowledge of algebra, it has perhaps rather interesting features. + +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. + +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. + + +118.--CIRCLING THE SQUARES. + +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. + +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 x 4) + 1, or 9; also, the square of 7 (49) less the +square of 3 (9) equals (7 + 3) x (7 - 3), or 40. + +Now, the number 192, referred to above, may be divided into five +different pairs of even factors: 2 x 96, 4 x 48, 6 x 32, 8 x 24, and 12 +x 16, and these divided by 2 give us, 1 x 48, 2 x 24, 3 x 16, 4 x 12, +and 6 x 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. + +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 + 6; that is, it must be a number composed of 6 plus a +multiple of 4. + + +119.--RACKBRANE'S LITTLE LOSS. + +The professor must have started the game with thirteen shillings, Mr. +Potts with four shillings, and Mrs. Potts with seven shillings. + + +120.--THE FARMER AND HIS SHEEP. + +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--that is, one-third. Any +two fractions will do if the denominator equals the sum of the two +numerators. + + +121.--HEADS OR TAILS. + +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. + + +122.--THE SEE-SAW PUZZLE. + +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. + + +123.--A LEGAL DIFFICULTY. + +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. + + +124.--A QUESTION OF DEFINITION. + +There is, of course, no difference in _area_ between a mile square and a +square mile. But there may be considerable difference in _shape_. 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. + + +125.--THE MINERS' HOLIDAY. + +Bill Harris must have spent thirteen shillings and sixpence, which would +be three shillings more than the average for the seven men--half a +guinea. + + +126.--SIMPLE MULTIPLICATION. + +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. + + +127.--SIMPLE DIVISION. + +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 x 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. + + +128.--A PROBLEM IN SQUARES. + +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 113/120, +337/120, 463/120; and with a common difference of 13 are 80929/19380, +106921/19380, and 127729/19380. 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. + + +129.--THE BATTLE OF HASTINGS. + +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. + +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"--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. + +Next to 61, the most difficult number under 100 is 97, where 97 x +6,377,352 squared + 1 = a square. + +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. + + +130.--THE SCULPTOR'S PROBLEM. + +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 8/7 ft., and of the smaller 3/7 ft., if we are to have the +answer in the smallest possible figures. Here the lineal measurement is +11/7 ft.--that is, 1+4/7 ft. What are the cubic contents of the two +cubes? First 8/7 x 3/7 x 8/7 = 512/343, and secondly 3/7 x 3/7 x 3/7 = +27/343. Add these together and the result is 539/343, which reduces to +11/7 or 1+4/7 ft. We thus see that the answers in cubic feet and lineal +feet are precisely the same. + +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, a, b, c, where a is the largest and c the smallest. + +Then ab + c squared = denominator, and a squared - c squared, b squared - c squared, and a squared - b squared will be +the three numerators. Thus, using the generators 3, 2, 1, we get 8/7, +3/7, 5/7 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 6n + 1, or +composed of such primes. Thus you can have 13, 19, etc., as +denominators, but not 25, 55, 187, etc. + +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: 99999999/99990001 and +19999/99990001. + + +131.--THE SPANISH MISER. + +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. + + +132.--THE NINE TREASURE BOXES. + +Here is the answer that fulfils the conditions:-- + + A = 4 B = 3,364 C = 6,724 + D = 2,116 E = 5,476 F = 8,836 + G = 9,409 H = 12,769 I = 16,129 + + +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. + + +133.--THE FIVE BRIGANDS. + +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 distributed 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:-- + + Table I. Table II. + A = 6. A = 6. + B = n. B = n. + C = (63 - 5n) + m. C = 1 + m. + D = (128 + 4n) - 4m. D = (376 - 16n) - 4m. + E = 3 + 3m. E = (15n - 183) + 3m. + +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 - 4n) inclusive. The first table thus gives (32 + n) answers for +every value of n; and the second table gives (94 - 4n) answers for every +value of n. The former, therefore, produces 462 and the latter 242 +answers, which together make 704, as already stated. + +Let us take Table I., and say n = 5 and m = 2; also in Table II. take n += 13 and m = 0. Then we at once get these two answers:-- + + A = 6 A = 6 + B = 5 B = 13 + C = 40 C = 1 + D = 140 D = 168 + E = 9 E = 12 + --- --- + 200 doubloons 200 doubloons. + +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. + +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 + 34 + 35 + 36 + ... + 43 + 44, +and the other 42 + 38 + 34 + 30 + ... + 6 + 2. The sum of the first +series is 462, and of the second 242--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. + + +134.--THE BANKER'S PUZZLE. + +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. + + +135.--THE STONEMASON'S PROBLEM. + +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 cubed + 24 cubed + 25 cubed = +204 squared. But it admits the answer, 25 cubed + 26 cubed + 27 cubed + 28 cubed + 29 cubed = 315 squared. The +correct answer, however, requires more heaps, but a smaller aggregate +number of blocks. Here it is: 14 cubed + 15 cubed + ... up to 25 cubed 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 x 312. I will just remark that +one key to the solution lies in what are called triangular numbers. (See +pp. 13, 25, and 166.) + + +136.--THE SULTAN'S ARMY. + +The smallest primes of the form 4n + 1 are 5, 13, 17, 29, and 37, and +the smallest of the form 4n - 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 = 4 + 1; 13 = 9 + 4; 17 = 16 + 1; 29 = 25 + 4; 37 += 36 + 1. But primes of the second form can never be expressed as the +sum of two squares in any way whatever. + +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 x 13), can +be expressed in two ways, 1,105, (5 x 13 x 17), in four ways, 32,045, (5 +x 13 x 17 x 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 _new_ factor, for the _repetition_ of factors is subject to another +law. We cannot express 25, (5 x 5), in two ways, but only in one; yet +125, (5 x 5 x 5), can be given in two ways, and so can 625, (5 x 5 x 5 x +5); while if we take in yet another 5 we can express the number as the +sum of two squares in three different ways. + +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 x 5), will not +work, nor will 135, (3 x 3 x 3 x 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 x 3 x 5, or 9 x +5) = 36 + 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 + 1 and +25 + 25. + +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 x 5 x 13, and consequently +that, as 65 can be expressed in two ways (64 + 1 and 49 + 16), 130 can +also be expressed in two ways, the factor 2 not affecting the question. + +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 x 5 x 13 x 17 x 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. + + +137.--A STUDY IN THRIFT. + +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. + +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 squared + 1 = a square number, x squared 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:-- + + 8 x 1 squared + 1 = 3 squared + 8 x 6 squared + 1 = 17 squared + 8 x 35 squared + 1 = 99 squared + 8 x 204 squared + 1 = 577 squared + 8 x 1189 squared + 1 = 3363 squared + 8 x 6930 squared + 1 = 19601 squared + 8 x 40391 squared + 1 = 114243 squared + +The successive pairs of numbers are found in this way:-- + + (1 x 3) + (3 x 1) = 6 (8 x 1) + (3 x 3) = 17 + (1 x 17) + (3 x 6) = 35 (8 x 6) + (3 x 17) = 99 + (1 x 99) + (3 x 35) = 204 (8 x 35) + (3 x 99) = 577 + +and so on. Look for the numbers in the table above, and the method will +explain itself. + +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. + +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. + +[Illustration + + +-----------+ + +---------+ |. . . . ./.| + |. . . ./.| |. . . ./. .| + |. . ./. .| |. . ./. . .| + |. ./. . .| |. ./. . . .| + |./. . . .| |./. . . . .| + /. . . . .| /. . . . . .| + +---------+ +-----------+ + +] + +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--1, 3, 6, +10, 15, 21, 28--we find that by adding all the consecutive pairs in turn +we get the series of square numbers--9, 16, 25, 36, 49, etc. + +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 stating 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. + + +----------------------+-----------+---------------+-----------------------+ + | Number | Side of | Side of | Sides of Two | Sides of Three | + | | Square. | Triangle. | Triangles. | Triangles. | + +------------+---------+-----------+---------------+-----------------------+ + | 36 | 6 | 8 | 6 + 5 | 5 + 5 + 3 | + | 1225 | 35 | 49 | 36 + 34 | 33 + 32 + 16 | + | 41616 | 204 | 288 | 204 + 203 | 192 + 192 + 95 | + | 1413721 | 1189 | 1681 | 1189 + 1188 | 1121 + 1120 + 560 | + | 48024900 | 6930 | 9800 | 6930 + 6929 | 6533 + 6533 + 3267 | + | 1631432881 | 40391 | 57121 | 40391 + 40390 | 38081 + 38080 + 19040 | + +------------+---------+-----------+---------------+-----------------------+ + +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 L5 present she must +save the respectable sum of L1,631,432,881. + + +138.--THE ARTILLERYMEN'S DILEMMA. + +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. + + 1 2 3 4 5 6 7 + 1 3 6 10 15 21 28 + 1 4 10 20 35 56 84 + 1 5 14 30 55 91 140 + +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. + +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. + +For the benefit of more advanced mathematicians I will add that the +general expression for square pyramid numbers is (2n cubed + 3n squared + n)/6. +For this expression to be also a square number (the special case of 1 +excepted) it is necessary that n = p squared - 1 = 6t squared, where 2p squared - 1 = q squared +(the "Pellian Equation"). In the case of our solution above, n = 24, p = +5, t = 2, q = 7. + + +139.--THE DUTCHMEN'S WIVES. + +The money paid in every case was a square number of shillings, because +they bought 1 at 1s., 2 at 2s., 3 at 3s., and so on. But every husband +pays altogether 63s. 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 Gurtruen bought 8 and 1; Elas and Katruen bought 12 +and 9; Hendrick and Anna bought 32 and 31. And these pairs represent +correctly the three married couples. + +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 2 squared.3 cubed.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--that is, 8. To find these eight squares, as +it is an _even_ number, we first divide by 4 and get 1485, the eight +pairs of factors of which are 1 x 1485, 3 x 495, 5 x 297, 9 x 165, 11 x +135, 15 x 99, 27 x 55, and 33 x 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 +_odd_; so we factorize at once, 1 x 63, 3 x 21, 7 x 9. Then we find that +_half_ 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. + +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 Frenicle, 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. + + +140.--FIND ADA'S SURNAME. + +The girls' names were Ada Smith, Annie Brown, Emily Jones, Mary +Robinson, and Bessie Evans. + + +141.--SATURDAY MARKETING. + +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. + + +142.--THE SILK PATCHWORK. + +[Illustration] + +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. + +[Illustration: Fig. 1.] + +[Illustration: Fig. 2.] + +143.--TWO CROSSES FROM ONE. + +It will be seen that one cross is cut out entire, as A in Fig. 1, while +the four pieces marked 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. + +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. + + +144.--THE CROSS AND THE TRIANGLE. + +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? + +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. + +[Illustration] + +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. + + +145.--THE FOLDED CROSS. + +[Illustration: FIG. 1., FIG 2.] + +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. + +[Illustration: FIG. 3., FIG. 4.] + + +146.--AN EASY DISSECTION PUZZLE. + +[Illustration + + +===========+===========+- + | . | . : \ + | . | . : \ + | . | . : \ + | . | . : \ + | . | . : \ + +-----------+===========+===========+ + | / : . | . : \ + | / : . | . : \ + | / : . | . : \ + | / : . | . : \ + | / : . | . : \ + +===========+===========+===========+===========+ + +] + +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. + + +147.--AN EASY SQUARE PUZZLE. + +[Illustration + + +-----------------------------------------+ + | . /| + | . / | + | . / | + | / / | + | / . / | + | / . / | + | / . / | + | / ./ | + | +--------------------+ | + | / / | + | / / | + | / / | + | / . / | + | / . / | + | / . / | + | / . / | + | / . | + | / . | + | / . | + |/ . | + +-----------------------------------------+ + +] + +The diagram explains itself, one of the five pieces having been cut in +two to form a square. + + +148.--THE BUN PUZZLE. + +[Illustration + + . . + . . + _ . . + . |\ A . + | \ + . C | \ | + | \ + . | \ / + . |______________________\/ + | | + . . + . B . + . . + . . + - + + + _ + . | . + . | . + . | . + | + | + | D | E | + | + | + . | . + . | . + . | . + _ + + _ + . | . + . -+- . + . . . . + - - + . + | G| F | | + + - - + . . . . + . - _ _ - . + . | . + - + + -+- + . . + - - + . + | H | + + - - + . . + - _ _ - + + +] + +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--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--half the +rim of the other. Thus each boy gets an exactly equal share, and there +are only five pieces necessary. + + +149.--THE CHOCOLATE SQUARES. + +[Illustration] + +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. + + +150.--DISSECTING A MITRE. + +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--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. + +This puzzle--with the added condition that it shall be cut into four +parts of the same size and shape--I have not been able to trace to an +earlier date than 1835. Strictly speaking, it is, in that form, +impossible of solution; but I give the answer that is always presented, +and that seems to satisfy most people. + +[Illustration] + +We are asked to assume that the two portions containing the same +letter--AA, BB, CC, DD--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. + +[Illustration] + + +151.--THE JOINER'S PROBLEM. + +[Illustration] + +Nothing could be easier than the solution of this puzzle--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--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. + + +152.--ANOTHER JOINER'S PROBLEM. + +[Illustration] + +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. + +The precise relative proportions of the square and triangle are of no +consequence whatever. It is only necessary to cut the wood or material +into five pieces. + +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--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. + +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. + +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. + + +153.--A CUTTING-OUT PUZZLE. + +[Illustration] + +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 9 x 1, or 16 x 1, or 25 x 1, 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 squared times the breadth, and not more than (n+1) squared +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 24 x 1, 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. + + +154.--MRS. HOBSON'S HEARTHRUG. + +[Illustration] + +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 12 x 6, giving an +area of 72 (inches or yards, as we please), and as the original complete +rug measured 36 x 27, 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 30 x 30 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. + +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. + +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. + + +155.--THE PENTAGON AND SQUARE. + +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--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. + +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 _height_ of the parallelogram. This distance we mark +off from C at K, then draw CK, 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. + +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. + +[Illustration] + +Another correspondent made the side of his square 11/4 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--inch, foot, or anything else--the +side of the square of equal area is 1.3117 nearly, or say roughly +1+3/10. So we can only hope to solve the puzzle by geometrical methods. + + +156.--THE DISSECTED TRIANGLE. + +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. + +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. + +[Illustration] + + +157.--THE TABLE-TOP AND STOOLS. + +[Illustration] + +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 _ovum_, 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--a conical ellipse--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 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." + +[Illustration: THE TWO STOOLS.] + +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--also 7 and +8--making only _six pieces_ in all. But I wished to keep the same number +as in the original story. + +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." + + +158.--THE GREAT MONAD. + +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. + +[Illustration: 1 2 3 4] + +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 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. + + +159.--THE SQUARE OF VENEER. + +[Illustration: + + +---+---+---+---+---+---+---+---+---+---+---+---+---+ + | | | :| | |: | | :| | |: | || | + | | | :| | |: | | :| | |: | || | + +---+---+---+---+---+---+---+---+---+---+---+---+---+ + | | | :| | |: | | :| | |: | || | + | | | :| | |: | | :| | |: | || | + +---+---+---+---+---+---+---+---+---+---+---+---+---+ + | | | :| | |: | | :| | |: | || | + |_ _|___|__:|___|___|:__|___|__:|___|___|:__|__||___| + +---+---+---+---+---+---+---+---+---+---+---+---+---+ + | | | :| | |: | | :| | |: | || | + | | | :| | |: | | :| | |: | || | + +---+---+---+---+---+---+---+---+---+---+===+===+---+ + | | | :| | |: | | :| | ||: | | | + | | | :| | |: | | :| | ||: | | | + +---+---+---+---+---+---+---+---+---+---+---+---+---+ + |---|---|--:|---|---|:--|---|--:|---|--||:--|---|---| + | | | :| | |: | | :| | ||: | | | + +---+---+---+---+---+---+---+---+---+---+---+---+---+ + | | | :| | |: | | :| | ||: | | | + | | | :| | |: | B | :| | ||: | C | | + +---+---+---+---+---+---+---+---+---+---+---+---+---+ + | | | :| | |: | | :| | ||: | | | + |_ _|___|__:|___|___|:__|___|__:|___|__||:__|___|___| + +---+---+---+---+---+---+---+---+===+===+===+===+===+ + | | | :| | |: | | :|| | |: | | | + | | | :| | |: | | :|| | |: | | | + +---+---+---+---+---+---+---+---+---+---+---+---+---+ + | | | :| | |: | | :|| | |: | | | + | | | :| | |: | | :|| | |: | A | | + +---+---+---+---+---+---+---+---+---+---+---+---+---+ + |---|---|--:|---|---|:--|---|--:||--|---|:--|---|---| + | | | :| | |: | | :|| | |: | | | + +---+---+---+---+---+---+---+---+---+---+---+---+---+ + | | | :| | |: | | :|| | |: | | | + | | | :| | |: | | :|| | |: | | | + +===+===+===+===+===+===+===+===+---+---+---+---+---+ + | | | :| | |: | | :|| | |: | | | + | | | :| | D |: | | :|| | |: | | | + +---+---+---+---+---+---+---+---+===+===+===+===+===+ + +] + +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. + +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 12 x 12 and 5 x 5; 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 5 x 5 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. + +Now, n is clearly 5/13 of an inch. Consequently our larger square must +be 60/13 in. x 60/13 in., and our smaller square 25/13 in. x 25/13 in. +The square of 60/13 added to the square of 25/13 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. + +Here is a general formula for finding two squares whose sum shall equal +a given square, say a squared. In the case of the solution of our puzzle p = 3, +q = 2, and a = 5. + + ________________________ + 2pqa \/ a squared( p squared + q squared) squared - (2pqa) squared + --------- = x; --------------------------- = y + p squared + q squared p squared + q squared + + Here x squared + y squared = a squared. + + +160.--THE TWO HORSESHOES. + +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. 158.) + +[Illustration] + +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 +different 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--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. + + +161.--THE BETSY ROSS PUZZLE. + +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. + +[Illustration] + + +162.--THE CARDBOARD CHAIN. + +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 21/2 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. (See next page.) 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. + +It is an interesting variant of the puzzle to cut out two keys on a +ring--in the same manner without join. + +[Illustration] + + +164.--THE POTATO PUZZLE. + +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. + +The general formula is that with n cuts we can always produce (n(n + +1) + 1)/2 . One of the problems proposed by the late Sam Loyd was to +produce the maximum number of pieces by n 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 n cuts we may produce +((n - 1)n(n + 1))/6 + n + 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 n. + + +165.--THE SEVEN PIGS. + +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. + +[Illustration: THE SEVEN PIGS.] + + +166.--THE LANDOWNER'S FENCES. + +Four fences only are necessary, as follows:-- + +[Illustration] + + +167.--THE WIZARD'S CATS. + +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. + +[Illustration: THE WIZARDS' CATS.] + + +168.--THE CHRISTMAS PUDDING. + +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. + +[Illustration] + + +169.--A TANGRAM PARADOX. + +The diagrams will show how the figures are constructed--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. 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. + +[Illustration] + + +170.--THE CUSHION COVERS. + +[Illustration] + +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. + + +171.--THE BANNER PUZZLE. + +The illustration explains itself. Divide the bunting into 25 squares +(because this number is the sum of two other squares--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. + +[Illustration] + + +172.--MRS. SMILEY'S CHRISTMAS PRESENT. + +[Illustration] + +[Illustration] + +The first step is to find six different square numbers that sum to 196. +For example, 1 + 4 + 25 + 36 + 49 + 81 = 196; 1 + 4 + 9 + 25 + 36 + 121 += 196; 1 + 9 + 16 + 25 + 64 + 81 = 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. + + +173.--MRS. PERKINS'S QUILT. + +The following diagram shows how the quilt should be constructed. + +[Illustration] + +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. + + +174.--THE SQUARES OF BROCADE. + +[Illustration: Diagram 1] + +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 _two_ 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 _quarter-turn_ 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 _half-turn_--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. + +[Illustration: Diagram 2] + +[Illustration: Diagram 3] + +[Illustration: Diagram 4] + +[Illustration: Diagram 5] + + +175.--ANOTHER PATCHWORK PUZZLE. + +The lady need only unpick the stitches along the dark lines in the +larger portion of patchwork, when the four pieces will fit together and +form a square, as shown in our illustration. + +[Illustration] + + +176.--LINOLEUM CUTTING. + +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 x 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. + +[Illustration] + + +177.--ANOTHER LINOLEUM PUZZLE. + +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. + +[Illustration: ANOTHER LINOLEUM PUZZLE.] + + +178.--THE CARDBOARD BOX. + +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 x 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--10 in. and 8 in. respectively. + + +179.--STEALING THE BELL-ROPES. + +Whenever we have one side (a) of a right-angled triangle, and know the +difference between the second side and the hypotenuse (which difference +we will call b), then the length of the hypotenuse will be + + a squared b + --- + -. + 2b 2 + +In the case of our puzzle this will be + + 48 x 48 + ------- + 11/2 in. = 32 ft. 11/2 in., + 6 + +which is the length of the rope. + + +180-- THE FOUR SONS. + +[Illustration] + +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 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--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. + + +181.--THE THREE RAILWAY STATIONS. + +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. + + +182.--THE GARDEN PUZZLE. + +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. + + +183.--DRAWING A SPIRAL. + +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 _true_ spiral, but the puzzle was to produce the _particular_ +spiral that was shown, and that was drawn in this simple manner. + +[Illustration] + + +184.--HOW TO DRAW AN OVAL. + +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. + + +185.--ST. GEORGE'S BANNER. + +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 (21/2 ft.) from a quarter of the distance all +round the edge of the flag (31/2 ft.)--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. + + +186.--THE CLOTHES LINE PUZZLE. + +Multiply together, and also add together, the heights of the two poles +and divide one result by the other. That is, if the two heights are a +and b respectively, then ab/(a + b) 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. + + +187.--THE MILKMAID PUZZLE. + +[Illustration: + + A + |\ + | \ + | \ + | \ B RIVER + +----+-------------- + | / \ + | / \ + | / \ + |/ DOOR + STOOL + +] + +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. + + +188.--THE BALL PROBLEM. + +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. + +As for the second question, the ratio of the diameter of a circle to its +circumference we call _pi_; 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 +_pi_ at all. Because, to find the area of the surface of a sphere we +multiply the square of the diameter by _pi_; to find the volume of a +sphere we multiply the cube of the diameter by one-sixth of _pi_. +Therefore we may ignore _pi_, 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 _pi_ in square feet, and its volume also 36 times _pi_ +in cubic feet. + + +189.--THE YORKSHIRE ESTATES. + +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 x 370 x 116) +-(370 + 116 - 74) squared--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. + +[Illustration: + + A + |\ + | \. + | \ . + |5 \ . + | 7 \ . + E +--------- +C . + | | ` . . + | | `. . + |4 |4 ` . . + | 7 | ` .. + D----------+----------------- B + F + +] + +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. + +The area of our triangle ADB is clearly half of 9 x 17, or 761/2 acres. +The area of AEC is half of 5 x 7, or 171/2 acres; the area of CFB is half +of 4 x 10, or 20 acres; and the area of the oblong EDFC is obviously 4 x +7, or 28 acres. Now, if we add together 171/2, 20, and 28 = 651/2, and +deduct this sum from the area of the large triangle ADB (which we have +found to be 761/2 acres), what remains must clearly be the area of ABC. +That is to say, the area we want must be 761/2 - 651/2 = 11 acres exactly. + + +190.--FARMER WURZEL'S ESTATE. + +The area of the complete estate is exactly one hundred acres. To find +this answer I use the following little formula, + + __________________ + \/4ab - (a + b - c) squared + -------------------- + 4 + +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 + 20 + 18 + 9 + 9 + 9 + 9 = 100 acres. + +[Illustration] + +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 21/2, F is 41/2, 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. + +Or we can get our proof in a still easier way. The complete area of the +squared diagram is 12 x 12 = 144 acres, and the portions 1, 2, 3, 4, not +included in the estate, have the respective areas of 121/2, 171/2, 91/2, and +41/2. These added together make 44, which, deducted from 144, leaves 100 +as the required area of the complete estate. + + +191.--THE CRESCENT PUZZLE. + +Referring to the original diagram, let AC be x, let CD be x - 9, and let +EC be x - 5. Then x - 5 is a mean proportional between x - 9 and x, from +which we find that x equals 25. Therefore the diameters are 50 in. and +41 in. respectively. + + +192.--THE PUZZLE WALL. + +[Illustration] + +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. + + +193.--THE SHEEP-FOLD. + +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 point of view, the +accepted answer to this ancient little puzzle breaks down. And yet +attention has never before been drawn to the absurdity. + +[Illustration + + A 24 + +--------------------------------+ + | 24 |1 + +--------------------------------+ + + B + +--------------------------------+ + | 48 |2 + +--------------------------------+ + 24 + C + +--------------------+ D + | | +----------+ + | | | | + | |12 | 48 |6 + | 156 | | | + | | +----------+ + | | 8 + | | + | | + +--------------------+ + 13 + + + 12 . E 13 + . ' ' . + . ' ' . + ' . . ' + 12 ' . ' 13 + +] + + +194.--THE GARDEN WALLS. + +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. + +[Illustration] + +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. + +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--I might almost say, +if they be made homogeneous--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. + +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. + +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. + + +195.--LADY BELINDA'S GARDEN. + +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. + + +196.--THE TETHERED GOAT. + +[Illustration] + +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 +11/2 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." + + +197.--THE COMPASSES PUZZLE. + +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. + +[Illustration] + +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. + +I am aware that this is not the simplest possible solution. + + +198.--THE EIGHT STICKS. + +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 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. + +[Illustration] + + +199.--PAPA'S PUZZLE. + +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 _did_ 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. + +[Illustration] + +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. + +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 +x .366 = 2.562, or a little more than 21/2 inches, for the distance from B +to A. + +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. + + +200.--A KITE-FLYING PUZZLE. + +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 calculations, involving "_pi_," +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." + +[Illustration] + +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. + +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 1/100 is 1/100000, 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 x 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. + +Whether a kite would fly at such a height, or support such a weight, are +questions that do not enter into the problem. + + +201.--HOW TO MAKE CISTERNS. + +Here is a general formula for solving this problem. Call the two sides +of the rectangle a and b. Then + + a + b - (a squared + b squared - ab)^1/2 + --------------------------- + 6 + +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. + + +202.--THE CONE PUZZLE. + +The simple rule is that the cone must be cut at one-third of its +altitude. + + +203.--CONCERNING WHEELS. + +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. + +[Illustration: 1] + +[Illustration: 2] + +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. + + +204.--A NEW MATCH PUZZLE. + +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 length, Fig. 1 contains two +square inches and Fig. 2 contains six square inches--4 x 11/2. 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. + +[Illustration: Figures 1, 2, 3, 4.] + + +205.--THE SIX SHEEP-PENS. + +[Illustration] Place the twelve matches in the manner shown in the +illustration, and you will have six pens of equal size. + + +206.--THE KING AND THE CASTLES. + +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. + +[Illustration: The King and the Castles] + + +207.--CHERRIES AND PLUMS. + +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. + +[Illustration] + + +208.--A PLANTATION PUZZLE. + +The illustration shows the ten trees that must be left to form five rows +with four trees in every row. The dots represent the positions of the +trees that have been cut down. + +[Illustration] + + +209.--THE TWENTY-ONE TREES. + +I give two pleasing arrangements of the trees. In each case there are +twelve straight rows with five trees in every row. + +[Illustration: Figure 1, Figure 2.] + + +210.--THE TEN COINS. + +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 x 100 = 2,400 different solutions. + +[Illustration] + + +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. + +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. + +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--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 + +[Illustration] + +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--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. + +[Illustration: STAR DART COMPASSES FUNNEL SCISSORS NAIL] + + +211.--THE TWELVE MINCE-PIES. + +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. + +[Illustration: The Twelve Mince Pies.] + + +212.--THE BURMESE PLANTATION. + +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. + + +213.--TURKS AND RUSSIANS. + +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 at the foot of this page +will show exactly how the thing is to be done. + +[Illustration] + +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 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. + + +214.--THE SIX FROGS. + +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--the fewest possible. + +If n, the number of frogs, be even, we require (n squared + n)/2 moves, of +which (n squared - n)/2 will be leaps and n simple moves. If n be odd, we +shall need ((n squared + 3n)/2) - 4 moves, of which (n squared - n)/2 will be leaps +and 2n - 4 simple moves. + +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 1/2n 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 = 105 moves. + +In the odd cases, write the even numbers in ascending order and the odd +numbers in descending order, repeat this series 1/2(n - 1) times, follow +with the even numbers in ascending order (omitting n - 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 n). 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. + +This complete general solution is published here for the first time. + + +215.--THE GRASSHOPPER PUZZLE. + +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. + +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 n, 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--only by the first. + + FIRST METHOD. + +----------+---------------------------+-----------------------+-----------+ + | Total No.| L MOVEMENT. | U MOVEMENT. | | + | of +-------------+-------------+----------+------------+ Total No. | + | Counters.| No. of | No. of | No. of | No. of | of Moves. | + | | Counters. | Moves. |Counters. | Moves. | | + +----------+-------------+-------------+----------+------------+-----------+ + | 4n | n-1 and n |2(n-1) squared+5n-7 | 2n+1 |2n squared+3n+1 |4(n squared+n-1) | + | 4n-2 | n-1 " n |2(n-1) squared+5n-7 | 2n-1 |2(n-1) squared+3n-2|4n squared-5 | + | 4n+1 | n " n+1 |2n squared+5n-2 | 2n |2n squared+3n-4 |2(2n squared+4n-3)| + | 4n-1 | n-1 " n |2(n-1) squared+5n-7 | 2n |2n squared+3n-4 |4n squared+4n-9 | + +----------+-------------+-------------+----------+------------+-----------+ + + SECOND METHOD. + +---------+--------------------------+-------------------------+-----------+ + |Total No.| L MOVEMENT. | U MOVEMENT. | | + | of +-------------+------------+----------+--------------+ Total No. | + |Counters.| No. of | No. of | No. of | No. of | of Moves. | + | | Counters. | Moves. | Counters.| Moves. | | + +---------+-------------+------------+----------+--------------+-----------+ + | 4n | n and n |2n squared+3n-4 | 2n | 2(n-1) squared+5n-2 |4(n squared+n-1) | + | 4n-2 | n-1 " n-1 |2(n-1) squared+3n-7| 2n | 2(n-1) squared+5n-2 |4n squared-5 | + | 4n+1 | n " n |2n squared+3n-4 | 2n+1 | 2n squared+5n-2 |2(2n squared+4n-3)| + | 4n-1 | n " n |2n squared+3n-4 | 2n-1 | 2(n-1) squared+5n-7 |4n squared+4n-9 | + +---------+-------------+------------+----------+--------------+-----------+ + +More generally we may say that with m counters, where m is even and +greater than 4, we require (m squared + 4m - 16)/4 moves; and where m is odd +and greater than 3, (m squared + 6m - 31)/4 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. + + +216.--THE EDUCATED FROGS. + +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. + + +217.--THE TWICKENHAM PUZZLE. + +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. + + +218.--THE VICTORIA CROSS PUZZLE. + +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--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!" + +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. + +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--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. + + +219.--THE LETTER BLOCK PUZZLE. + +This puzzle can be solved in 23 moves--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. + + +220.--A LODGING-HOUSE DIFFICULTY. + +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. + + +221.--THE EIGHT ENGINES. + +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. + +There are two other slightly different solutions. + + +222.--A RAILWAY PUZZLE. + +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. + + +223.--A RAILWAY MUDDLE. + +[Illustration: 1] + +[Illustration: 2] + +[Illustration: 3] + +[Illustration: 4] + +[Illustration: 5] + +[Illustration: 6] + +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 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). + + +224.--THE MOTOR-GARAGE PUZZLE. + +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. + + +225.--THE TEN PRISONERS. + +[Illustration] + +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. + + +226.--ROUND THE COAST. + +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. + +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." + + +227.--CENTRAL SOLITAIRE. + +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. + +I do not think the number of moves can be reduced. + + +228.--THE TEN APPLES. + +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. + + +229.--THE NINE ALMONDS. + +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. + + +230.--THE TWELVE PENNIES. + +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. + + +231.--PLATES AND COINS. + +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. + +This is "The Riddle of the Fishpond" (No. 41, _Canterbury Puzzles_) in a +different dress. + + +232.--CATCHING THE MICE. + +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)--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. + +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. + +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. + + +233.--THE ECCENTRIC CHEESEMONGER. + +To leave the three piles at the extreme ends of the rows, the cheeses +may be moved as follows--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. + + +234.--THE EXCHANGE PUZZLE. + +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. + + +235.--TORPEDO PRACTICE. + +[Illustration: + + 10 6 7 + \ |/ + 4 u u 2 + \ u / + 3-u u u u + u u + u u u u -----9--- + / u + 8 u u + / \ + 1 5 + +] + +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. + + +236.--THE HAT PUZZLE. + +[Illustration: + + 1 2 3 4 5 6 7 8 9 10 11 12 + +--+--+--+--+--+--+--+--+--+--+--+--+ + | *| o| *| O| *| O| *| O| *| O| | | + +--+--+--+--+--+--+--+--+--+--+--+--+ + | *| | | O| *| O| *| O| *| O| O| *| + +--+--+--+--+--+--+--+--+--+--+--+--+ + | *| *| O| O| *| O| | | *| O| O| *| + +--+--+--+--+--+--+--+--+--+--+--+--+ + | *| *| O| | | O| O| *| *| O| O| *| + +--+--+--+--+--+--+--+--+--+--+--+--+ + | *| *| O| O| O| O| O| *| *| | | *| + +--+--+--+--+--+--+--+--+--+--+--+--+ + | | | O| O| O| O| O| *| *| *| *| *| + +--+--+--+--+--+--+--+--+--+--+--+--+ + +] + +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 successive 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. + + +237.--BOYS AND GIRLS. + +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:-- + + . . 1 2 3 4 5 6 7 8 + 4 3 1 2 . . 5 6 7 8 + 4 3 1 2 7 6 5 . . 8 + 4 3 1 2 7 . . 5 6 8 + 4 . . 2 7 1 3 5 6 8 + 4 8 6 2 7 1 3 5 . . + + +238.--ARRANGING THE JAM POTS. + +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--17. + + +239.--A JUVENILE PUZZLE. + +[Illustration: + + +-----------------+ + | C E | + | | | | + | D F | + +---------------B | + G | + A | | + | H | + +-----------------+ + +] + +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. + +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. + + +240.--THE UNION JACK. + +[Illustration: + + +-------+ +----- + A B | | / + \ | | / + |\ \ | | / /| + | \ \ | | / / | + | \ \| |/ / | + | \ | / / | + | \ |\ /| / | + +-----\-|-\/-|-/-----+ + \| /\ |/ + |/ \/ + |\ /\ + /| \/ |\ + +-----/-|-/\-|-\-----+ + | / / \| \ | + | / | \ \ | + | / /| |\ \ | + | / / | | \ \ | + |/ / | | \ \| + / | | \ + / | | \ + -----+ +----- + +] + +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. + + +241.--THE DISSECTED CIRCLE. + +[Illustration: + + /---------------\ + / \ + / /------B \ + / / | /^\ \ + / / |\ | / \ \ + / / | \ | / \ \ + / / | \ | / A \ \ + / / | \ | / | \ \ + | / | \|/ | \ | + | | -----+-----*-----+----- | | + | | \ | /|\ | / | | + | | \ | / | \ | / | | + | | \ | / | \ | / | | + | | \ | / | \ | / | | + | | \|/ | \|/ | | + D-+------*-----+-----*----E | | + | /|\ | /|\ | | + | / | \ | / | \ | | + | / | \ | / | \ | | + | / | \ | / | \ | | + | / | \|/ | \ | | + | -----+-----*-----+----- | | + \ | /|\ | / | + \ | / | \ | / / + \ | / | \ | / / + \ | / | \ | / / + \ |/ | \| / / + \ | / / + \------+------/ / + | / + C-------/ + +] + +It can be done in twelve continuous strokes, thus: Start at A in the +illustration, and eight 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--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. + +242.--THE TUBE INSPECTOR'S PUZZLE. + +The inspector need only travel nineteen miles if he starts at B and +takes the following route: BADGDEFIFCBEHKLIHGJK. 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. + +243.--VISITING THE TOWNS. + +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--reading the line the other way. Seven roads are not used. + +244.--THE FIFTEEN TURNINGS. + +[Illustration] + +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. + +245.--THE FLY ON THE OCTAHEDRON. + +[Illustration] + +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. + +246.--THE ICOSAHEDRON PUZZLE. + +[Illustration] + +There are thirty edges, of which eighteen were visible in the original +illustration, represented 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. + +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. + +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--simply from one node to the next--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. + +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 x 10,000 miles = 250,000 miles) and 5 of the +lines twice (5 x 20,000 miles = 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. + +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--that is, thirty-five lines +of 10,000 miles each. + + +247.--INSPECTING A MINE. + +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--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. + + +248.--THE CYCLIST'S TOUR. + +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. + + +249.--THE SAILOR'S PUZZLE. + +[Illustration] + +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:-- + +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 + +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. + +The map may be greatly simplified by the "buttons and string" method, +explained in the solution to No. 341, "The Four Frogs." + +250.--THE GRAND TOUR. + +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 No. 341.) +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--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. + +[Illustration: Fig. 1.] + +[Illustration: Fig. 2.] + +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. + +251.--WATER, GAS, AND ELECTRICITY. + +[Illustration] + +According to the conditions, in the strict sense in which one at first +understands them, there 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. + + +252.--A PUZZLE FOR MOTORISTS. + +[Illustration] + +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. + + +253.--A BANK HOLIDAY PUZZLE. + +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--the one +immediately over the letter E. This town was therefore the cyclist's +destination. + +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 (m+n)!/m!n!, +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 m = 11 and n = 4. Then the formula +gives us the answer 1,365 as above. + + +254.-- THE MOTOR-CAR TOUR. + +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--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. + +[Illustration: Figs. 1, 2, and 3] + +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 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. + +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. + + +255.--THE LEVEL PUZZLE. + +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. + + +256.--THE DIAMOND PUZZLE. + +There are 252 different ways. The general formula is that, for words of +n letters (not palindromes, as in the case of the next puzzle), when +grouped in this manner, there are always 2^(n+1) - 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. + + +257.--THE DEIFIED PUZZLE. + +The correct answer is 1,992 different ways. Every F is either a corner F +or a side F--standing next to a corner in its own square of F's. Now, +FIED may be read _from_ a corner F in 16 ways; therefore DEIF may be +read _into_ a corner F also in 16 ways; hence DEIFIED may be read +_through_ a corner F in 16 x 16 = 256 ways. Consequently, the four +corner F's give 4 x 256 = 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 x 121 = 968 ways. Add 968 +to 1,024 and we get the answer, 1,992. + +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. + +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. + + +258.-- THE VOTERS' PUZZLE. + +THE number of readings here is 63,504, as in the case of "WAS IT A RAT I +SAW" (No. 30, _Canterbury Puzzles_). The general formula is that for +palindromic sentences containing 2n + 1 letters there are (4(2^n -1)) squared +readings. + + +259.-- HANNAH'S PUZZLE. + +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. + + +260.--THE HONEYCOMB PUZZLE. + +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. + + +261.-- THE MONK AND THE BRIDGES. + +[Illustration] + +The problem of the Bridges may be reduced to the simple diagram shown +in illustration. The 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 a and b; the only direct ways from I to Y are +by the bridges c and d; and there is a direct way from M to Y by the +bridge e. Now, what we have to do is to count all the routes that will +lead from M to Y, passing over all the bridges, a, b, c, d, and e 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 a, b, we find there are only two ways of completing the +route; with _a, c_, there are only two routes; with a, d, only two +routes; and so on. It will be found that there are sixteen such routes +in all, as in the following list:-- + + a b e c d b c d a e + a b e d c b c e a d + a c d b e b d c a e + a c e b d b d e a c + a d e b c e c a b d + a d c b e e c b a d + b a e c d e d a b c + b a e d c e d b a c + +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. + + +262.--THOSE FIFTEEN SHEEP. + +If we read the exact words of the writer in the cyclopaedia, 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, "_Now_ 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. + + +263.--KING ARTHUR'S KNIGHTS. + +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. + + +264.--THE CITY LUNCHEONS. + +The men may be grouped as follows, where each line represents a day and +each column a table:-- + + AB CD EF GH IJ KL + AE DL GK FI CB HJ + AG LJ FH KC DE IB + AF JB KI HD LG CE + AK BE HC IL JF DG + AH EG ID CJ BK LF + AI GF CL DB EH JK + AC FK DJ LE GI BH + AD KH LB JG FC EI + AL HI JE BF KD GC + AJ IC BG EK HL FD + +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. + + +265.--A PUZZLE FOR CARD-PLAYERS. + +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. + + A B -- I L | E J -- G K | F H -- C D + A C -- J B | F K -- H L | G I -- D E + A D -- K C | G L -- I B | H J -- E F + A E -- L D | H B -- J C | I K -- F G + A F -- B E | I C -- K D | J L -- G H + A G -- C F | J D -- L E | K B -- H I + A H -- D G | K E -- B F | L C -- I J + A I -- E H | L F -- C G | B D -- J K + A J -- F I | B G -- D H | C E -- K L + A K -- G J | C H -- E I | D F -- L B + A L -- H K | D I -- F J | E G -- B C + +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. + +266.--A TENNIS TOURNAMENT. + +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:-- + + First Court. Second Court. + 1st Day | A d against B e | D a against E b + 2nd Day | A e " D b | E a " B d + 3rd Day | A b " E d | B a " D e + +It will be seen that no man ever plays with or against his own wife--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. + + +267.--THE WRONG HATS. + +The number of different ways in which eight persons, with eight hats, +can each take the wrong hat, is 14,833. + +Here are the successive solutions for any number of persons from one to +eight:-- + + 1 = 0 + 2 = 1 + 3 = 2 + 4 = 9 + 5 = 44 + 6 = 265 + 7 = 1,854 + 8 = 14,833 + +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 x 1 - 1 = 2; 4 x +2 + 1 = 9; 5 x 9 - 1 = 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 + 9) = 44; 5(9 + 44) = 265; and so on. + + +268.--THE PEAL OF BELLS. + +The bells should be rung as follows:-- + + 1 2 3 4 + 2 1 4 3 + 2 4 1 3 + 4 2 3 1 + 4 3 2 1 + 3 4 1 2 + 3 1 4 2 + 1 3 2 4 + 3 1 2 4 + 1 3 4 2 + 1 4 3 2 + 4 1 2 3 + 4 2 1 3 + 2 4 3 1 + 2 3 4 1 + 3 2 1 4 + 2 3 1 4 + 3 2 4 1 + 3 4 2 1 + 4 3 1 2 + 4 1 3 2 + 1 4 2 3 + 1 2 4 3 + 2 1 3 4 + +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. + + +269.--THREE MEN IN A BOAT. + +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 +455^7. 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--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. + +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:-- + + 1 2 3 4 5 + 1st Day (ABC) (DBF) (GHI) (JKL) (MNO) + 8 6 7 9 10 + 2nd Day (ADG) (BKN) (COL) (JEI) (MHF) + 3 5 4 1 2 + 3rd Day (AJM) (BEH) (CFI) (DKO) (GNL) + 7 6 8 9 1 + 4th Day (AEK) (CGM) (BOI) (DHL) (JNF) + 4 5 3 10 2 + 5th Day (AHN) (CDJ) (BFL) (GEO) (MKI) + 6 7 8 10 1 + 6th Day (AFO) (BGJ) (CKH) (DNI) (MEL) + 5 4 3 9 2 + 7th Day (AIL) (BDM) (CEN) (GKF) (JHO) + +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. + +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 _Educational Times +Reprints_, 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 6n+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 _Mathematical Recreations_, +5th edition. + + +270.--THE GLASS BALLS. + +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--order is here +of no importance. The breaking of the balls on A may occupy any 4 out of +these 16 places--that is, the combinations of 16 things, taken 4 +together, will be + + 13 x 14 x 15 x 16 + ----------------- = 1,820 + 1 x 2 x 3 x 4 + +ways for A. In every one of these cases B may occupy any 4 out of the +remaining 12 places, making + + 9 x 10 x 11 x 12 + ----------------- = 495 + 1 x 2 x 3 x 4 + +ways. Thus 1,820 x 495 = 900,900 different placings are open to A and B. +But for every one of these cases C may occupy + + 5 x 6 x 7 x 8 + ------------- = 70 + 1 x 2 x 3 x 4 + +different places; so that 900,900 x 70 = 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. + + +271.--FIFTEEN LETTER PUZZLE. + +The following will be found to comply with the conditions of grouping:-- + + ALE MET MOP BLM + BAG CAP YOU CLT + IRE OIL LUG LNR + NAY BIT BUN BPR + AIM BEY RUM GMY + OAR GIN PLY CGR + PEG ICY TRY CMN + CUE COB TAU PNT + ONE GOT PIU + +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. + +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. + + +272.--THE NINE SCHOOLBOYS. + +The boys can walk out as follows:-- + + 1st Day. 2nd Day. 3rd Day. + A B C B F H F A G + D E F E I A I D B + G H I C G D H C E + + 4th Day. 5th Day. 6th Day. + A D H G B I D C A + B E G C F D E H B + F I C H A E I G F + +Every boy will then have walked by the side of every other boy once and +once only. + +Dealing with the problem generally, 12n+9 boys may walk out in triplets +under the conditions on 9n+6 days, where n may be nought or any integer. +Every possible pair will occur once. Call the number of boys m. Then +every boy will pair m-1 times, of which (m-1)/4 times he will be in the +middle of a triplet and (m-1)/2 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. + + +273.--THE ROUND TABLE. + +The history of this problem will be found in _The Canterbury Puzzles_ +(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. + +The solution for the case of 3 persons seated on 1 occasion needs no +remark. + +A solution for the case of 4 persons on 3 occasions is as follows:-- + + 1 2 3 4 + 1 3 4 2 + 1 4 2 3 + +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. + +The case of 5 persons on 6 occasions may be solved as follows:-- + + 1 2 3 4 5 + 1 2 4 5 3 + 1 2 5 3 4 + --------- + 1 3 2 5 4 + 1 4 2 3 5 + 1 5 2 4 3 + +The case for 6 persons on 10 occasions is solved thus:-- + + 1 2 3 6 4 5 + 1 3 4 2 5 6 + 1 4 5 3 6 2 + 1 5 6 4 2 3 + 1 6 2 5 3 4 + ----------- + 1 2 4 5 6 3 + 1 3 5 6 2 4 + 1 4 6 2 3 5 + 1 5 2 3 4 6 + 1 6 3 4 5 2 + +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) 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. + +The difficult case of 7 persons on 15 occasions is solved as follows, +and was given by me in _The Canterbury Puzzles_:-- + + 1 2 3 4 5 7 6 + 1 6 2 7 5 3 4 + 1 3 5 2 6 7 4 + 1 5 7 4 3 6 2 + 1 5 2 7 3 4 6 + +In this case the 1 is a repeater, and there are _two_ 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. + +A solution for 8 persons on 21 occasions is as follows:-- + + 1 8 6 3 4 5 2 7 + 1 8 4 5 7 2 3 6 + 1 8 2 7 3 6 4 5 + +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. + +Here is my solution for 9 persons on 28 occasions:-- + + 2 1 9 7 4 5 6 3 8 + 2 9 5 1 6 8 3 4 7 + 2 9 3 1 8 4 7 5 6 + 2 9 1 5 6 4 7 8 3 + +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. + +The case of 10 persons on 36 occasions is solved as follows:-- + + 1 10 8 3 6 5 4 7 2 9 + 1 10 6 5 2 9 7 4 3 8 + 1 10 2 9 3 8 6 5 7 4 + 1 10 7 4 8 3 2 9 5 6 + +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. + +My solution for 11 persons on 45 occasions is as follows:-- + + 2 11 9 4 7 6 5 1 8 3 10 + 2 1 11 7 6 3 10 8 5 4 9 + 2 11 10 3 9 4 8 5 1 7 6 + 2 11 5 8 1 3 10 6 7 9 4 + 2 11 1 10 3 4 9 6 7 5 8 + +There are two repeaters, 1 and 2, and the cycle is, 3, 4, 5,... 11. We +thus get 5 groups of 9 lines each. + +The case of 12 persons on 55 occasions is solved thus:-- + + 1 2 3 12 4 11 5 10 6 9 7 8 + 1 2 4 11 6 9 8 7 10 5 12 3 + 1 2 5 10 8 7 11 4 3 12 6 9 + 1 2 6 9 10 5 3 12 7 8 11 4 + 1 2 7 8 12 3 6 9 11 4 5 10 + +Here 1 is a repeater, and the cycle is 2, 3, 4, 5,... 12. We thus get 5 +groups of 11 lines each. + + +274.--THE MOUSE-TRAP PUZZLE. + +If we interchange cards 6 and 13 and begin our count at 14, we may take +up all the twenty-one cards--that is, make twenty-one "catches"--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. + + +275.--THE SIXTEEN SHEEP. + +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. + + +276.--THE EIGHT VILLAS. + +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. + +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 + +[Illustration: + + +---+---+ +-+-----+ +---+---+ + |O OHO O| |OHO O O| |O OHO O| + | H | | + | | +=+ | + |O OHO O| |OHO O O| |O OHOHO| + +-+ +-+-+ +-+-----+ +---+ + | + |O|O O|O| |O|O O O| |O O O|O| + | +---+ | | +-+-+ | | +-+ | + |O O O O| |O O OHO| |O O|O O| + +-------+ +-------+ +-------+ + 2 3 4 + + +-----+-+ +-+-----+ +-------+ + |O O OHO| |OHO O O| |O O O O| + | +=+ | | +=+ | | +=+=+=+ + |O OHO O| |OHOHO O| |OHOHO O| + | +-+-+ + | + +-+ | + + + | + |O|O O|O| |O|O O|O| |O|OHO O| + +=+ +=+ | + +=+ +=+ + | + |O O O O| |OHO O O| |O O|O O| + +-------+ +-+-----+ +---+---+ + 5 6 7 + THE SIXTEEN SHEEP + +] + +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 + (3 x 9) = 37 different +solutions. If we then try 7 in the corner, the result will be 10 + 27 + +40, or 77 solutions. With 6 we get 10 + 27 + 40 + 49 = 126; with 5, 10 + +27 + 40 + 49 + 54 = 180; with 4, the same as with 5, + 55 = 235 ; with +3, the same as with 4, + 52 = 287; with 2, the same as with 3, + 45 = +332; with 1, the same as with 2, + 34 = 366, and with nought in the top +left-hand corner the number of solutions will be found to be 10 + 27 + +40 + 49 + 54 + 55 + 52 + 45 + 34 + 19 = 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 + 37 + 77 + 126 + 180 + 235 + 287 + 332 + 366 + +385 = 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. + +[Illustration: + + A B C D E + +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ + |9| |0| |8| |0| |8| |1| |8| |0| |8| |1| + +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ + | |*| | | |*| | | |*| | | |*| | | |*| | + +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ + |0| | | |0| | | |1| | | |1| | | |0| | | + +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ + +] + +Here is a general formula for solving the puzzle: (n squared + 3n + 2)(n squared + +3n + 3)/6. Whatever may be the stipulated number of residents along +each of the sides (which number is represented by n), the total number +of different arrangements may be thus ascertained. In our particular +case the number of residents was nine. Therefore (81 + 27 + 2) x (81 + +27 + 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. + + +277.--COUNTER CROSSES. + +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:-- + + 12978 13968 14958 + 34956 24957 23967 + + 23958 13769 14759 + 14967 24758 23768 + + 12589 23759 13579 + 34567 14768 24568 + + 14569 23569 14379 + 23578 14578 25368 + + 15369 24369 23189 + 24378 15378 45167 + + 24179 25169 34169 + 35168 34178 25178 + +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 x 2 x 3 +x 4 = 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 x 24 = 576 variations for every form; therefore, as there are 18 +forms, we get 18 x 576 = 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. + +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 x 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. + + +278.--A DORMITORY PUZZLE. + +[Illustration: + + MON. TUES. WED. + +---+---+---+ +---+---+---+ +---+---+---+ + | 1 | 2 | 1 | | 1 | 3 | 1 | | 1 | 4 | 1 | + +---+---+---+ +---+---+---+ +---+---+---+ + | 2 | | 2 | | 1 | | 1 | | 1 | | 1 | + +---+---+---+ +---+---+---+ +---+---+---+ + | 1 | 22| 1 | | 3 | 19| 3 | | 4 | 16| 4 | + +---+---+---+ +---+---+---+ +---+---+---+ + + THURS. FRI. SAT. + +---+---+---+ +---+---+---+ +---+---+---+ + | 1 | 5 | 1 | | 2 | 6 | 2 | | 4 | 4 | 4 | + +---+---+---+ +---+---+---+ +---+---+---+ + | 2 | | 2 | | 1 | | 1 | | 4 | | 4 | + +---+---+---+ +---+---+---+ +---+---+---+ + | 4 | 13| 4 | | 7 | 6 | 7 | | 4 | 4 | 4 | + +---+---+---+ +---+---+---+ +---+---+---+ + +] + +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. + + +279.--THE BARRELS OF BALSAM. + +This is quite easy to solve for any number of barrels--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. + +The general solution to the problem is, in fact, this: + + n + C + 2n + ----- + n + 1 + +where 2n equals the number of barrels. The symbol C, of course, implies +that we have to find how many combinations, or selections, we can make +of 2n things, taken n at a time. + + +280.--BUILDING THE TETRAHEDRON. + +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--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 x 24 x 16 x 2 = 3,840 as the exact number of ways in which +the pyramid may be constructed. This method excludes all possibility of +error. + +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--except under fourth dimensional conditions! + + +281.--PAINTING A PYRAMID. + +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 _four_ 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 _three_ colours, they may be applied in the 3 ways shown in +Figs. 3, 4, and 5. If we take any _two_ colours, they may be applied in +the 3 ways shown in Figs. 6, 7, and 8. If we take any _single_ 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 = 70; 35 in 3 ways += 105; 21 in 3 ways = 63; and 7 in 1 way = 7. Consequently the pyramid +may be painted in 245 different ways (70 + 105 + 63 + 7), using the +seven colours of the solar spectrum in accordance with the conditions of +the puzzle. + +[Illustration: + + 1 2 + +---------------+ +---------------+ + \ R / \ B / \ B / \ R / + \ / \ / \ / \ / + \ / G \ / \ / G \ / + \-------/ \-------/ + \ / \ / + \ Y / \ Y / + \ / \ / + ' ' + + 3 4 5 + +---------------+ +---------------+ +---------------+ + \ R / \ R / \ R / \ G / \ Y / \ R / + \ / \ / \ / \ / \ / \ / + \ / G \ / \ / G \ / \ / G \ / + \-------/ \-------/ \-------/ + \ / \ / \ / + \ Y / \ Y / \ Y / + \ / \ / \ / + ' ' ' + + 6 7 8 + +---------------+ +---------------+ +---------------+ + \ G / \ Y / \ Y / \ Y / \ G / \ G / + \ / \ / \ / \ / \ / \ / + \ / G \ / \ / G \ / \ / G \ / + \-------/ \-------/ \-------/ + \ / \ / \ / + \ Y / \ Y / \ Y / + \ / \ / \ / + ' ' ' + +] + + +282.--THE ANTIQUARY'S CHAIN. + +[Illustration] + +THE number of ways in which nine things may be arranged in a row without +any restrictions is 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 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 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40,320 x 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. + +We now come to the point to which I directed the reader's +attention--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 _ends_ it +is like every other link in having only two _sides_--that is, you cannot +turn over one end without turning the other at the same time. + +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--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 x 564,480 = +1,128,960; if two links might be placed either way up, the answer would +be 4 x 564,480; if three links, then 8 x 564,480, and so on. Since, +therefore, every link may be placed either side up, the number will be +564,480 multiplied by 2^9, or by 512. This raises our total to +289,013,760. + +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! + + +283.--THE FIFTEEN DOMINOES. + +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 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--13--30--04--42--21--10--02--23--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. + +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 2^5 = 32 +different ways. Therefore when we include the doubles there are 264 x 32 += 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 x 15 gives 126,720 different ways as the +correct answer to the puzzle. + +[Illustration: + + ----- + | | + / | | \ + / ----- \ + / . . \ + ----- . . ----- + | | . . | o o | + | o | -.--------.--- | | + | | . . . | o o | + ----- . . .. ----- + \ . . . . / + ----- .. ----- + | o | . . |o | + | | --------- | o | + | o |. .| o| + ----- ----- + +] + +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. + + +284.--THE CROSS TARGET. + +[Illustration: + + -- -- + (CD)( ) + -- -- + (AE)(A ) + -- -- -- -- -- -- + (CE)(E )(A )(AB)(C )(D ) + -- -- -- -- -- -- + (D )( )(B )(E )(EB)( ) + -- -- -- -- -- -- + (C )(B ) + -- -- + ( )(ED) + -- -- + + +] + +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. + + +285.--THE FOUR POSTAGE STAMPS. + +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. + + +286.--PAINTING THE DIE. + +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--the correct answer. + + +287.--AN ACROSTIC PUZZLE. + +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. + + +288.--CHEQUERED BOARD DIVISIONS. + +There are 255 different ways of cutting the board into two pieces of +exactly the same size 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--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; + +[Illustration] + +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. + +Here is a table of the results:-- + + + a b c Ways. + A = 8 + 17 + 21 = 46 + B = 0 + 17 + 21 = 38 + C = 15 + 31 + 39 = 85 + D = 17 + 29 + 39 = 85 + E = 1 + 0 + 0 = 1 + -- -- -- --- + 41 94 120 255 + +I have not attempted the task of enumerating the ways of dividing a +board 8 x 8--that is, an ordinary chessboard. Whatever the method +adopted, the solution would entail considerable labour. + + +289.--LIONS AND CROWNS. + +[Illustration] + +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. + + +290.--BOARDS WITH AN ODD NUMBER OF SQUARES. + +There are fifteen different ways of cutting the 5 x 5 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 + +[Illustration: Fig. 1. Fig. 2.] + +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. + + 1, 4, 8. + 1, 4, 3, 7, 8. + 1, 4, 3, 7, 10, 9. + 1, 4, 3, 7, 10, 6, 5, 9. + 1, 4, 5, 9. + 1, 4, 5, 6, 10, 9. + 1, 4, 5, 6, 10, 7, 8. + 2, 3, 4, 8. + 2, 3, 4, 5, 9. + 2, 3, 4, 5, 6, 10, 9. + 2, 3, 4, 5, 6, 10, 7, 8. + 2, 3, 7, 8. + 2, 3, 7, 10, 9. + 2, 3, 7, 10, 6, 5, 9. + 2, 3, 7, 10, 6, 5, 4, 8. + +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. + + +291.--THE GRAND LAMA'S PROBLEM. + +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. + +The reader may find it interesting to compare this puzzle with that of +the "Weaver" (No. 14, _Canterbury Puzzles_). + +[Illustration: THE GRAND LAMA'S PROBLEM. + + +===+===+===+===+===+===+===+===+ + |:o:| : : : : : : : + I...I...+===+===+===+===+===+===+ + |:::| o |:::::::::::::::::::::::| + I...I...I...+===+===+===+===+...I + |:::| |:o:| : : : |:::| + I...I...I...I...I===+===+...I...I + |:::| |:::| o |:::::::| |:::| + I...I...I...+===I===+...I...I...I + |:::| |:::::::| |:::| |:::| + I...I...+===+===+...+...I...I...I + |:::| : : : |:::| |:::| + I...+===+===+===+===I...I...I...I + |:::::::::::::::::::::::| |:::| + +===+===+===+===+===+===+...I...I + | : : : : : : |:::| + +===+===+===+===+===+===+===+===+ + +] + + +292.--THE ABBOT'S WINDOW. + +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:-- + + +[Illustration: + + +===+===+===+===+===+===+===+===+ + | : : : : : : : | + I...+===+...+...+...+...+===+...I + | IIIII : : : IIIII | + I...+===+===+...+...+===+===+...I + | : IIIII : IIIII : | + I...+...+===+===+===+===+...+...I + | : : IIIIIIIII : : | + I...+...+...+===+===+...+...+...I + | : : IIIIIIIII : : | + I...+...+===+===+===+===+...+...I + | : IIIII : IIIII : | + I...+===+===+...+...+===+===+...I + | IIIII : : : IIIII | + I...+===+...+...+...+...+===+...I + | : : : : : : : | + +===+===+===+===+===+===+===+===+ + +] + +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 _line_ with itself than one bird can go +into a corner and flock in solitude. The Abbot's condition was that no +diagonal _lines_ should contain an odd number of lights." + +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. + + +293.--THE CHINESE CHESSBOARD. + + +===I===+===+===+===I===+===+===+ + | |:::: 2 ::::| 3 |:::| 5 |:6:| + I...+===+...+===+...I...I...+===I + |:::: 1 |:::| ::::| 4 |:::| 7 | + I...+===+===+...I===I...I===+===I + | |:::: |:::| ::::| 9 |:::| + I===I...I===============I...I...I + |:::: 11|:::: ::::: 10|:::| 8 | + I=======I===I===========I...I...I + | ::::: 12|:::: 13::::| |:::| + I=======+...I...+===+===|===+===I + |:::: 14|:::| |:::| 16::::| 17| + I...+...I===I===+...+...+===+...I + | ::::| ::::: 15|:::| ::::| + I=======+===========+===+=======I + |:::: ::::: 18::::: ::::: | + +===+===+===+===+===+===+===+===+ + + +===+===I===I===+===I===+===+===+ + | ::::| |:::: |:::| ::::| + I...+===I...I=======I...I===+...I + |:::| |:::: |:::: |:::| | + I...I===I===============I===I...I + | |:::: ::::| ::::: |:::| + I===I=======I=======I=======I===I + |:::| ::::| ::::| ::::| | + I...I===+...I...+...I...+===+...I + | ::::| |:::: |:::| ::::| + I...+===I...+===I===+...I===+...I + |:::| |:::: |:::: |:::| | + I===I...+=======I=======+...I===I + | |:::: ::::| ::::: |:::| + I...+=======+...I...+=======+...I + |:::: ::::| |:::| ::::: | + +===+===+===+===+===+===+===+===+ + + +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--eight squares--that is possible under the conditions. The second +diagram was prepared under the added condition that no piece should +contain more than five squares. + +No. 74 in _The Canterbury Puzzles_ shows how to cut the board into +twelve pieces, all different, each containing five squares, with one +square piece of four squares. + + +294.--THE CHESSBOARD SENTENCE. + + +===I===I===I===I=======I=======+ + | |:::| |:::| ::::| ::::| + I===I...I===I...I...+===I...+===I + |:::| ::::: |:::| ::::: | + |...|...+===I...I...+===+...+===I + | |:::| |:::| ::::| ::::| + |...+===+...+===I===I===I=======I + |:::: ::::: |:::| ::::: | + I===========I===I...I===I===+...| + | ::::: |:::| |:::| |:::| + |...+===+...|...|...|...I===+...| + |:::| |:::| |:::| |:::: | + |...|...|...|...I===+...+===+...| + | |:::| |:::| ::::: |:::| + I===+...+===I...+=======I===+...| + |:::: ::::| ::::: |:::: | + +===========I===================+ + +The pieces may be fitted together, as shown in the illustration, to form +a perfect chessboard. + + +295.--THE EIGHT ROOKS. + +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 x 7 x 6 x 5 x 4 +x 3 x 2 x 1 = 40,320 (that is 8!), which is the correct answer. + +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. + + +296.--THE FOUR LIONS. + + +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. + + +297.--BISHOPS--UNGUARDED. + + +...+...+...+...+...+...+...+...+ + : ::::: ::::: ::::: ::::: + +...+...+...+...+...+...+...+...+ + ::::: ::::: ::::: ::::: : + +...+...+...+...+...+...+...+...+ + : ::::: ::::: ::::: ::::: + +...+...+...+...+...+...+...+...+ + ::B:: B ::B:: B ::B:: B ::B:: B : + +...+...+...+...+...+...+...+...+ + : ::::: ::::: ::::: ::::: + +...+...+...+...+...+...+...+...+ + ::::: ::::: ::::: ::::: : + +...+...+...+...+...+...+...+...+ + ::::: ::::: ::::: ::::: : + +...+...+...+...+...+...+...+...+ + : ::::: ::::: ::::: ::::: + +...+...+...+...+...+...+...+...+ + + +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. + + +298.--BISHOPS--GUARDED. + + +...+...+...+...+.......+.......+ + : ::::: ::::: ::::: ::::: + +...+...+...+...+...+...+...+...+ + ::::: ::::: ::::: ::::: : + +...+...+...+...+...+...+...+...+ + : ::::: ::::: ::::: ::::: + +...+...+...+...+...+...+.......+ + ::::: B ::B:: B ::::: B ::B:: : + +...........+...+...+...+...+...+ + : ::B:: B ::B:: ::B:: B ::::: + +...+...+...+...+...+...+...+...+ + ::::: ::::: ::::: ::::: : + +...+...+...+...+...+...+...+...+ + : ::::: ::::: ::::: ::::: + +...+...+...+...+.......+...+...+ + ::::: ::::: ::::: ::::: : + +...+...+...+...+.......+...+...+ + +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. + +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 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. + + +299.--BISHOPS IN CONVOCATION. + +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--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 n squared squares 2n - 2 +bishops (the maximum number) may always be placed in 2^n ways without +attacking. On an ordinary chessboard n 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. + +[Illustration] + + +300.--THE EIGHT QUEENS. + +[Illustration] + +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. + + +301.--THE EIGHT STARS. + +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. + +[Illustration] + + +302.--A PROBLEM IN MOSAICS. + +[Illustration] + +The diagram shows how the tiles may be rearranged. As before, one yellow +and one 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. + + +303.--UNDER THE VEIL. + +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. + +[Illustration: + + _ _ I V E L _ _ + E V L _ _ I _ _ + L _ _ I _ _ V E + I _ V E _ _ _ L + _ E _ _ L V _ I + _ L I _ _ I _ E V + /V _ E L _ _ I _ + _ I _ _ V E L _\ + +] + +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--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. + +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--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--that is, the +letters can be so arranged that no word can be read in any of the +directions. + + +304.--BACHET'S SQUARE. + +[Illustration: 1] + +[Illustration: 2] + +[Illustration: 3] + +[Illustration: 4] + +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--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 24x24 = 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 x 576 = 1,152, counting reflections and reversals as +different. Put in another manner, the pairs in the top row may be +written in 16 x 9 x 4 x 1 = 576 different ways, and the square then +completed in 2 ways, making 1,152 ways in all. + + +305.--THE THIRTY-SIX LETTER BLOCKS. + +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. + +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. + +[Illustration: 1] + +[Illustration: 2] + +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. + + +306.--THE CROWDED CHESSBOARD. + +[Illustration] + +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. + + +307.--THE COLOURED COUNTERS. + +The counters may be arranged in this order:-- + + R1, B2, Y3, O4, GS. + Y4, O5, G1, R2, B3. + G2, R3, B4, Y5, O1. + B5, Y1, O2, G3, R4. + O3, G4, R5, B1, Y2. + + +308.--THE GENTLE ART OF STAMP-LICKING. + +The following arrangement shows how sixteen stamps may be stuck on the +card, under the conditions, of a total value of fifty pence, or 4s. +2d.:-- + +[Illustration] + +If, after placing the four 5d. stamps, the reader is tempted to place +four 4d. 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 4s. This is the pitfall that was hinted at. +(Compare with No. 43, _Canterbury Puzzles_.) + + +309.--THE FORTY-NINE COUNTERS. + +The counters may be arranged in this order:-- + + A1, B2, C3, D4, E5, F6, G7. + F4, G5, A6, B7, C1, D2, E3. + D7, E1, F2, G3, A4, B5, C6. + B3, C4, D5, E6, F7, G1, A2. + G6, A7, B1, C2, D3, E4, F5. + E2, F3, G4, A5, B6, C7, D1. + C5, D6, E7, F1, G2, A3, B4. + + +310.--THE THREE SHEEP. + +The number of different ways in which the three sheep may be placed so +that every pen shall always be either occupied or in line with at least +one sheep is forty-seven. + +The following table, if used with the key in Diagram 1, will enable the +reader to place them in all these ways:-- + + +------------+---------------------------+----------+ + | | | No. of | + | Two Sheep. | Third Sheep. | Ways. | + +------------+---------------------------+----------+ + | A and B | C, E, G, K, L, N, or P | 7 | + | A and C | I, J, K, or O | 4 | + | A and D | M, N, or J | 3 | + | A and F | J, K, L, or P | 4 | + | A and G | H, J, K, N, O, or P | 6 | + | A and H | K, L, N, or O | 4 | + | A and O | K or L | 2 | + | B and C | N | 1 | + | B and E | F, H, K, or L | 4 | + | B and F | G, J, N, or O | 4 | + | B and G | K, L, or N | 3 | + | B and H | J or N | 2 | + | B and J | K or L | 2 | + | F and G | J | 1 | + | | | ---- | + | | | 47 | + +------------+---------------------------+----------+ + +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. + +If one pen at least is to be _not_ 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. + +[Illustration] + + +311.--THE FIVE DOGS PUZZLE. + +The diagrams show four fundamentally different solutions. In the case of +A we can reverse 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. + +[Illustration] + + +312.--THE FIVE CRESCENTS OF BYZANTIUM. + +[Illustration] + +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. + + +313.--QUEENS AND BISHOP PUZZLE. + +[Illustration: FIG. 1.] + +[Illustration: FIG. 2.] + +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.) + +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 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 _The Strand Magazine_ for August 1908, here reproduced. (Fig. 3.) + +[Illustration: FIG. 3.] + +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.) + +[Illustration: FIG. 4.] + +[Illustration: FIG. 5.] + + +314.--THE SOUTHERN CROSS. + +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": -- + +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. + + +315.--THE HAT-PEG PUZZLE. + +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. + +[Illustration: 1] + +[Illustration: 2] + +[Illustration: 3] + +[Illustration: 4] + + +316.--THE AMAZONS. + +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 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. + +[Illustration] + + +317.--A PUZZLE WITH PAWNS. + +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. + +[Illustration] + +318.--LION-HUNTING. + +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 x 80 = 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 - 816 = 5,664, the required answer. + +The general solution is this: 1/3n(n - 1)(3n squared - n + 2). This is, of +course, equivalent to saying that if we call the number of squares on +the side of a "chessboard" n, 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. + + +319.--THE KNIGHT-GUARDS. + +[Illustration: DIAGRAM 1.] + +[Illustration: DIAGRAM 2.] + +The smallest possible number of knights with which this puzzle can be +solved is fourteen. + +It has sometimes been assumed that there are a great many different +solutions. As a matter of fact, there are only three arrangements--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. + +[Illustration: DIAGRAM 3.] + +[Illustration: DIAGRAM 4.] + +[Illustration: DIAGRAM 5.] + +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 _1a_ and _2a_. Now, by placing the transparent Diagram +_1a_ over 1 you will be able to obtain the solution in Diagram 3, by +placing _2a_ over 2 you will get Diagram 4, and by placing _2a_ 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. + + +320.--THE ROOK'S TOUR. + +[Illustration] + +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*. + +[Illustration: THE ROOK'S TOUR.] + + +321.--THE ROOK'S JOURNEY. + +[Illustration] + +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." + + +322.--THE LANGUISHING MAIDEN. + +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 "The Grand Tour," p. 200.) + + +323.--A DUNGEON PUZZLE. + +If the prisoner takes the route shown in the diagram--where for +clearness the doorways are omitted--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. + +[Illustration: THE LANGUISHING MAIDEN] + +[Illustration: A DUNGEON PUZZLE.] + + +324.--THE LION AND THE MAN. + +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--"they occasionally +got glimpses of one another." Now, if we take one route for the 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! + +[Illustration] + + +325.--AN EPISCOPAL VISITATION. + +[Illustration] + +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. + + +326.--A NEW COUNTER PUZZLE. + +Play as follows: 2--3, 9--4, 10--7, 3--8, 4--2, 7--5, 8--6, 5--10, 6--9, +2--5, 1--6, 6--4, 5--3, 10--8, 4--7, 3--2, 8--1, 7--10. The white +counters have now changed places with the red ones, in eighteen moves, +without breaking the conditions. + + +327.--A NEW BISHOP'S PUZZLE. + +[Illustration: A] + +[Illustration: B] + +Play as follows, using the notation indicated by the numbered squares in +Diagram A:-- + + White. | Black. | White. | Black. + 1. 18--15 | 1. 3--6 | 10. 20--10 | 10. 1--11 + 2. 17--8 | 2. 4--13 | 11. 3--9 | 11. 18--12 + 3. 19--14 | 3. 2--7 | 12. 10--13 | 12. 11--8 + 4. 15--5 | 4. 6--16 | 13. 19--16 | 13. 2--5 + 5. 8--3 | 5. 13-18 | 14. 16--1 | 14. 5--20 + 6. 14--9 | 6. 7--12 | 15. 9--6 | 15. 12--15 + 7. 5--10 | 7. 16-11 | 16. 13-7 | 16. 8--14 + 8. 9--19 | 8. 12--2 | 17. 6--3 | 17. 15-18 + 9. 10--4 | 9. 11-17 | 18. 7--2 | 18. 14--19 + +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. + +328.--THE QUEEN'S TOUR. + +[Illustration] + +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. + + +329.--THE STAR PUZZLE. + +The illustration explains itself. The stars are all struck out in +fourteen straight strokes, starting and ending at a white star. + +[Illustration] + + +330.--THE YACHT RACE. + +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. + +[Illustration] + + +331.--THE SCIENTIFIC SKATER. + +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. + +Here is one way of performing the feat:-- + +[Illustration] + +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. + + +332.--THE FORTY-NINE STARS. + +The illustration shows how all the stars may be struck out in twelve +straight strokes, beginning and ending at a black star. + +[Illustration] + + +333.--THE QUEEN'S JOURNEY. + +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. + +[Illustration] + + +334.--ST. GEORGE AND THE DRAGON. + +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. + +[Illustration: St. George and the Dragon.] + + +335.--FARMER LAWRENCE'S CORNFIELDS. + +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. + +[Illustration: Farmer Lawrence's Cornfields.] + + +336.--THE GREYHOUND PUZZLE. + +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. + +The next point is that the first half-string must end in one of the +central rows, and the 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--that is, circled or +non-circled as the case may be. The half-strings can, therefore, only be +linked in the two central rows. + +[Illustration] + +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. + +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--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. + + +337.--THE FOUR KANGAROOS. + +[Illustration] + +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. + + +338.--THE BOARD IN COMPARTMENTS. + +[Illustration] + +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 determining 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. + +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. + + +339.--THE FOUR KNIGHTS' TOURS. + +[Illustration] + +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. + + +340.--THE CUBIC KNIGHT'S TOUR. + +[Illustration] + +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. + + +341.--THE FOUR FROGS. + +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--5), (3--7, 7--1), (8--4, 4--3, 3--7), (6--2, 2--8, 8--4, +4--3), (5--6, 6--2, 2--8), (1--5, 5--6), (7--1). + +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. + +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. + +[Illustrations: A B C D E] + +In Diagram D I give another familiar puzzle that first appeared in a +book published in Brussels in 1789, _Les Petites Aventures de Jerome +Sharp_. 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 _to_ the point that you +last moved _from_." This is not, of course, the only way of placing the +counters, but it is the simplest solution to carry in the mind. + +There are several puzzles in this book that the reader will find lend +themselves readily to this method. + + +342.--THE MANDARIN'S PUZZLE. + +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. + +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. + + +343.--EXERCISE FOR PRISONERS. + +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 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. + + A + +--+--+--+--+ + | 6| 1|10|15| + +--+--+--+--+ + | 9|12| 7| 4| + +--+--+--+--+ + | 2| 5|14|11| + +--+--+--+--+ + |13| 8| 3|**| + +--+--+--+--+ + + B + +--+--+--+--+ + | 6| 1|10|15| + +--+--+--+--+ + |11|14| 7| 4| + +--+--+--+--+ + | 2| 5|12| 9| + +--+--+--+--+ + |13| 8| 3|**| + +--+--+--+--+ + + C + +--+--+--+--+ + | 6| 9| 4|15| + +--+--+--+--+ + | 1|12| 7|10| + +--+--+--+--+ + | 8| 5|14| 3| + +--+--+--+--+ + |13| 2|11|**| + +--+--+--+--+ + + D + +--+--+--+--+ + | 6|11| 4|15| + +--+--+--+--+ + | 1|14| 7|10| + +--+--+--+--+ + | 8| 5|12| 3| + +--+--+--+--+ + |13| 2| 9|**| + +--+--+--+--+ + +[Illustration: A, B, C, D] + +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 = 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. + +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. + + +344.--THE KENNEL PUZZLE. + +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. + + +-----+------+------+------+------+ + |1 |2 |3 |4 |5 | + | | | | | | + | 1 | 18 | 9 | 14 | 5 | + | | | | | | + +-----+------+------+------+------+ + |6 |7 |8 |9 |10 | + | | | | | | + | 8 | 13 | 4 | 19 | 10 | + | | | | | | + +-----+------+------+------+------+ + |11 |12 |13 |14 |15 | + | | | | | | + | 17 | 2 | 11 | 6 | 15 | + | | | | | | + +-----+------+------+------+------+ + |16 |17 |18 |19 |20 | + | | | | | | + | 12 | 7 | 16 | 3 | 20 | + | | | | | | + +-----+------+------+------+------+ + |21 |22 |23 |24 |25 | + | | | | | | + | | | | | | + | | | | | | + +-----+------+------+------+------+ + +[Illustration] + +This position may be arrived at in as few as forty-six moves, as +follows: 16--21, 16--22, 16--23, 17--16, 12--17, 12--22, 12--21,7--12, +7--17, 7--22, 11--12, 11--17, 2--7, 2--12, 6--11, 8--7, 8--6, 13--8, +18--13, 11--18, 2--17, 18--12, 18--7, 18--2, 13--7, 3--8, 3--13, 4--3, +4--8, 9--4, 9--3, 14--9, 14--4, 19--14, 19--9, 3--14, 3--19, 6--12, +6--13, 6--14, 17--11, 12--16, 2--12, 7--17, 11--13, 16--18 = 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. + + +345.--THE TWO PAWNS. + +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 7x8x9x10x11x12 divided by 1x2x3x4x5x6 +gives us the number of variations for this case--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. 270, on p. 204.) + + +346.--SETTING THE BOARD. + +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 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 x +322,560 = 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. + + +347.--COUNTING THE RECTANGLES. + +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 n squared squares +contains ((n squared + n) squared)/4 rectangles, of which (2n cubed + 3n squared + n)/6 are +squares and (3n^4 + 2n cubed - 3n squared - 2n)/12 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 +n. + + +348.--THE ROOKERY. + +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--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--thirty-two. The Black king's moves are all forced, and need not +be given. + + +349.--STALEMATE. + + +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 :-- + + White. Black. + 1. P--Q4 1. P--K4 + 2. Q--Q3 2. Q--R5 + 3. Q--KKt3 3. B--Kt5 ch + 4. Kt--Q2 4. P--QR4 + 5. P--R4 5. P--Q3 + 6. P--R3 6. B--K3 + 7. R--R3 7. P--KB4 + 8. Q--R2 8. P--B4 + 9. R--KKt3 9. B--Kt6 + 10. P--QB4 10. P--B5 + 11. P--B3 11. P--K5 + 12. P--Q5 12. P--K6 + +And White is stalemated. + +We give a diagram of the curious position arrived at. It will be seen +that not one of White's pieces may be moved. + +[Illustration] + + +-+-+-+-+-+-+-+-+ + |r|n| | |k| |n|r| + +-+-+-+-+-+-+-+-+ + | |p| | | | |p|p| + +-+-+-+-+-+-+-+-+ + | | | |p| | | | | + +-+-+-+-+-+-+-+-+ + |p| |p|P| | | | | + +-+-+-+-+-+-+-+-+ + |P|b|P| | |p| |q| + +-+-+-+-+-+-+-+-+ + | |b| | |p|P|R|P| + +-+-+-+-+-+-+-+-+ + | |P| |N|P| |P|Q| + +-+-+-+-+-+-+-+-+ + | | |B| |K|B|N|R| + +-+-+-+-+-+-+-+-+ + + +350.--THE FORSAKEN KING. + +Play as follows:-- + + White. Black. + 1. P to K 4th 1. Any move + 2. Q to Kt 4th 2. Any move except on KB file (a) + 3. Q to Kt 7th 3. K moves to royal row + 4. B to Kt 5th 4. Any move + 5. Mate in two moves + If 3. K other than to royal row + 4. P to Q 4th 4. Any move + 5. Mate in two moves + (a) If 2. Any move on KB file + 3. Q to Q 7th 3. K moves to royal row + 4. P to Q Kt 3rd 4. Any move + 5. Mate in two moves + If 3. K other than to royal row + 4. P to Q 4th 4. Any move + 5. Mate in two moves + +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. + + +351.--THE CRUSADER. + + White. Black. + 1. Kt to QB 3rd 1. P to Q 4th + 2. Kt takes QP 2. Kt to QB 3rd + 3. Kt takes KP 3. P to KKt 4th + 4. Kt takes B 4. Kt to KB 3rd + 5. Kt takes P 5. Kt to K 5th + 6. Kt takes Kt 6. Kt to B 6th + 7. Kt takes Q 7. R to KKt sq + 8. Kt takes BP 8. R to KKt 3rd + 9. Kt takes P 9. R to K 3rd + 10. Kt takes P 10. Kt to Kt 8th + 11. Kt takes B 11. R to R 6th + 12. Kt takes R 12. P to Kt 4th + 13. Kt takes P (ch) 13. K to B 2nd + 14. Kt takes P 14. K to Kt 3rd + 15. Kt takes R 15. K to R 4th + 16. Kt takes Kt 16. K to R 5th + White now mates in three moves. + 17. P to Q 4th 17. K to R 4th + 18. Q to Q 3rd 18. K moves + 19. Q to KR 3rd (mate) + If 17. K to Kt 5th + 18. P to K 4th (dis. ch) 18. K moves + 19. P to KKt 3rd (mate) + +The position after the sixteenth move, with the mate in three moves, was +first given by S. Loyd in _Chess Nuts_. + + +352.--IMMOVABLE PAWNS. + + 1. Kt to KB 3 + 2. Kt to KR 4 + 3. Kt to Kt 6 + 4. Kt takes R + 5. Kt to Kt 6 + 6. Kt takes B + 7. K takes Kt + 8. Kt to QB 3 + 9. Kt to R 4 + 10. Kt to Kt 6 + 11. Kt takes R + 12. Kt to Kt 6 + 13. Kt takes B + 14. Kt to Q 6 + 15. Q to K sq + 16. Kt takes Q + 17. K takes Kt, and the position is reached. + +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. + + +353.--THIRTY-SIX MATES. + +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:-- + + By discovery from Q 8 + By discovery from R at Q 6th 13 + By discovery from B at R 8th 11 + Given by Kt at R 5th 2 + Given by pawns 2 + -- + Total 36 + +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. + + +354.--AN AMAZING DILEMMA. + +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. + +The late Sam Loyd first pointed out the peculiarity on which this puzzle +is based. + + +355.--CHECKMATE! + +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 _en passant_, checkmate. This was +therefore White's last move, and leaves the position given. It is the +only possible solution. + + +356.--QUEER CHESS. + + +-+-+-+-+-+-+-+-+ + | | | | | | | | | + +-+-+-+-+-+-+-+-+ + | | |R|k|R|N| | | + +-+-+-+-+-+-+-+-+ + | | | | | | | | | + +-+-+-+-+-+-+-+-+ + +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. + +I believe the position was first published by the late S. Loyd. + + +357.--ANCIENT CHINESE PUZZLE. + +Play as follows:-- + + 1. R--Q 6 + 2. K--R 7 + 3. R (R 6)--B 6 (mate). + +Black's moves are forced, so need not be given. + + +358.--THE SIX PAWNS. + +The general formula for six pawns on all squares greater than 2 squared is +this: Six times the square of the number of combinations of n things +taken three at a time, where n represents the number of squares on the +side of the board. Of course, where n is even the unoccupied squares +in the rows and columns will be even, and where n is odd the number of +squares will be odd. Here n is 8, so the answer is 18,816 different +ways. This is "The Dyer's Puzzle" (_Canterbury Puzzles_, 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. + +[Illustration] + +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 a and b placed in the first and +second columns the pair c may be disposed in any one of the remaining +six columns, and so give six solutions. Now slide pair b into the +third column, and there are five possible positions for c. Slide b +into the fourth column, and c may produce four new solutions. And so +on, until (still leaving a in the first column) you have b in the +seventh column, and there is only one place for c--in the eighth +column. Then you may put a in the second column, b in the third, and c +in the fourth, and start sliding c and b as before for another series +of solutions. + +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 (8 x 7 x 6)/(1 x 2 x 3) = 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 x 6 = 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 x 6 = 18,816, +which is the total number of ways, as we have already stated. + + +359.--COUNTER SOLITAIRE. + +Play as follows: 3--11, 9--10, 1--2, 7--15, 8--16, 8--7, 5--13, 1--4, +8--5, 6--14, 3--8, 6--3, 6--12, 1--6, 1--9, and all the counters will +have been removed, with the exception of No. 1, as required by the +conditions. + + +360.--CHESSBOARD SOLITAIRE. + +Play as follows: 7--15, 8--16, 8--7, 2--10, 1--9, 1--2, 5--13, 3--4, +6--3, 11--1, 14--8, 6--12, 5--6, 5--11, 31--23, 32--24, 32--31, 26--18, +25--17, 25--26, 22--32, 14--22, 29--21, 14--29, 27--28, 30--27, 25--14, +30--20, 25--30, 25--5. The two counters left on the board are 25 and +19--both belonging to the same group, as stipulated--and 19 has never +been moved from its original place. + +I do not think any solution is possible in which only one counter is +left on the board. + + +361.--THE MONSTROSITY. + + White Black, + 1. P to KB 4 P to QB 3 + 2. K to B 2 Q to R 4 + 3. K to K 3 K to Q sq + 4. P to B 5 K to B 2 + 5. Q to K sq K to Kt 3 + 6. Q to Kt 3 Kt to QR 3 + 7. Q to Kt 8 P to KR 4 + 8. Kt to KB 3 R to R 3 + 9. Kt to K 5 R to Kt 3 + 10. Q takes B R to Kt 6, ch + 11. P takes R K to Kt 4 + 12. R to R 4 P to B 3 + 13. R to Q 4 P takes Kt + 14. P to QKt 4 P takes R, ch + 15. K to B 4 P to R 5 + 16. Q to K 8 P to R 6 + 17. Kt to B 3, ch P takes Kt + 18. B to R 3 P to R 7 + 19. R to Kt sq P to R 8 (Q) + 20. R to Kt 2 P takes R + 21. K to Kt 5 Q to KKt 8 + 22. Q to R 5 K to R 5 + 23. P to Kt 5 R to B sq + 24. P to Kt 6 R to B 2 + 25. P takes R P to Kt 8 (B) + 26. P to B 8 (R) Q to B 2 + 27. B to Q 6 Kt to Kt 5 + 28. K to Kt 6 K to R 6 + 29. R to R 8 K to Kt 7 + 30. P to R 4 Q (Kt 8) to Kt 3 + 31. P to R 5 K to B 8 + 32. P takes Q K to Q 8 + 33. P takes Q K to K 8 + 34. K to B 7 Kt to KR 3, ch + 35. K to K 8 B to R 7 + 36. P to B 6 B to Kt sq + 37. P to B 7 K takes B + 38. P to B 8 (B) Kt to Q 4 + 39. B to Kt 8 Kt to B 3, ch + 40. K to Q 8 Kt to K sq + 41. P takes Kt (R) Kt to B 2, ch + 42. K to B 7 Kt to Q sq + 43. Q to B 7, ch K to Kt 8 + +And the position is reached. + +The order of the moves is immaterial, and this order may be greatly +varied. But, although many attempts have been made, nobody has succeeded +in reducing the number of my moves. + + +362.--THE WASSAIL BOWL. + +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. + + Barrel. 5-pint. 3-pint. X. Y. Z. + + 7 .. 5 .. 0 .. 0 .. 0 .. 0 + 7 .. 2 .. 3 .. 0 .. 0 .. 0 + 7 .. 0 .. 3 .. 2 .. 0 .. 0 + 7 .. 3 .. 0 .. 2 .. 0 .. 0 + 4 .. 3 .. 3 .. 2 .. 0 .. 0 + 0 .. 3 .. 3 .. 2 .. 4 .. 0 + 0 .. 5 .. 1 .. 2 .. 4 .. 0 + 0 .. 5 .. 0 .. 2 .. 4 .. 1 + 0 .. 2 .. 3 .. 2 .. 4 .. 1 + 0 .. 0 .. 3 .. 4 .. 4 .. 1 + 0 .. 0 .. 0 .. 4 .. 4 .. 4 + +And each man has received his four pints of ale. + + +363.--THE DOCTOR'S QUERY. + +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. + + + +364.--THE BARREL PUZZLE. + +[Illustration: Figs. 1, 2, and 3] + +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 a at the same +time that it touches the edge of the bottom b, 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 a it had risen to the +point c 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 d, then it would be less than half full. + +This method applies to all symmetrically constructed vessels. + + + +365.--NEW MEASURING PUZZLE. + +The following solution in eleven manipulations shows the contents of +every vessel at the start and after every manipulation:-- + + 10-quart. 10-quart. 5-quart. 4-quart. + + 10 .. 10 .. 0 .. 0 + 5 .. 10 .. 5 .. 0 + 5 .. 10 .. 1 .. 4 + 9 .. 10 .. 1 .. 0 + 9 .. 6 .. 1 .. 4 + 9 .. 7 .. 0 .. 4 + 9 .. 7 .. 4 .. 0 + 9 .. 3 .. 4 .. 4 + 9 .. 3 .. 5 .. 3 + 9 .. 8 .. 0 .. 3 + 4 .. 8 .. 5 .. 3 + 4 .. 10 .. 3 .. 3 + + + +366.--THE HONEST DAIRYMAN. + +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. + + + +367.--WINE AND WATER. + +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. + + + +368.--THE KEG OF WINE. + +The capacity of the jug must have been a little less than three gallons. +To be more exact, it was 2.93 gallons. + + + +369.--MIXING THE TEA. + +There are three ways of mixing the teas. Taking them in the order of +quality, 2s. 6d., 2s. 3d., 1s. 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 2s. 41/2d. per pound; but the last +case requires the smallest quantity of the best tea, therefore it is +the correct answer. + + +370.--A PACKING PUZZLE. + +On the side of the box, 14 by 22+4/5, 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+9/10 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. + + +371.--GOLD PACKING IN RUSSIA. + +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. + + +372.--THE BARRELS OF HONEY. + +The only way in which the barrels could be equally divided among the +three brothers, so that each should receive his 31/2 barrels of honey +and his 7 barrels, is as follows:-- + + Full. Half-full. Empty. + A 3 1 3 + B 2 3 2 + C 2 3 2 + +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. + + +373.--CROSSING THE STREAM. + +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. + +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. 375 or 376) 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. + + +374.--CROSSING THE RIVER AXE. + +Here is the solution:-- + + | {J 5) | G T8 3 + 5 | ( J } | G T8 3 + 5 | {G 3) | JT8 + 53 | ( G } | JT8 + 53 | {J T) | G 8 + J 5 | (T 3} | G 8 + J 5 | {G 8) | T 3 + G 8 | (J 5} | T + G 8 | {J T) | 53 + JT8 | ( G } | 53 + JT8 | {G 3) | 5 + G T8 3 | ( J } | 5 + G T8 3 | {J 5) | + +G, J, and T stand for Giles, Jasper, and Timothy; and 8, 5, 3, for L800, +L500, and L300 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. + +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. + + +375.--FIVE JEALOUS HUSBANDS. + +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. + +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. + + ABCDE abcde *|..| + | | + 1. ABCDE de |..|* abc + 2. ABCDE bcde *|..| a + 3. ABCDE e |..|* abcd + 4. ABCDE de *|..| abc + 5. DE de |,,|* ABC abc + 6. CDE cde *|..| AB ab + 7. cde |..|* ABCDE ab + 8. bcde *|..| ABCDE a + 9. e |..|* ABCDE abcd + 10. bc e *|..| ABCDE a d + 11. |..|* ABCDE abcde + +There is a little subtlety concealed in the words "show the _quickest_ +way." + +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. + +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. + +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. + + +376.--THE FOUR ELOPEMENTS. + +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. + + Lawn. | Island. | Shore. + | | + ABCDabcd * | | + ABCD cd | | ab * + ABCD bcd * | | a + ABCD d | bc * | a + ABCD cd * | b | a + CD cd | b | AB a * + BCD cd * | b | A a + BCD | bcd * | A a + BCD d * | bc | A a + D d | bc | ABC a * + D d | abc * | ABC + D d | b | ABC a c * + B D d * | b | A C a c + d | b | ABCD a c * + d | bc * | ABCD a + d | | ABCD abc * + cd * | | ABCD ab + | | ABCD abcd * + +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. + + +377.--STEALING THE CASTLE TREASURE. + +Here is the best answer, in eleven manipulations:-- + + Treasure down. + Boy down--treasure up. + Youth down--boy up. + Treasure down. + Man down--youth and treasure up. + Treasure down. + Boy down--treasure up. + Treasure down. + Youth down--boy up. + Boy down--treasure up. + Treasure down. + + +378.--DOMINOES IN PROGRESSION. + +There are twenty-three different ways. You may start with any domino, +except the 4--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:-- + +With a common difference of 1, the first domino may be either of these: +0--0, 0--1, 1--0, 0--2, 1--1, 2--0, 0--3, 1--2, 2--1, 3--0, 0--4, 1--3, +2--2, 3--1, 1--4, 2--3, 3--2, 2--4, 3--3, 3--4. With a difference of 2, +the first domino may be 0--0, 0--2, or 0--1. Take the last case of all +as an example. Having played the 0--1, and the difference being 2, we +are compelled to continue with 1--2, 2--3, 3--4. 4--5, 5--6. There are +three dominoes that can never be used at all. These are 0--5, 0--6, and +1--6. If we used a box of dominoes extending to 9--9, there would be +forty different ways. + + +379.--THE FIVE DOMINOES. + +There are just ten different ways of arranging the dominoes. Here is one +of them:-- + +(2--0) (0--0) (0--1) (1--4) (4--0). + +I will leave my readers to find the remaining nine for themselves. + + +380.--THE DOMINO FRAME PUZZLE. + + +[Illustration: + + +---+-------+-------+-------+-------+-------+-------+-------+ + | 2 | 2 | 5 | 5 | 6 | 6 | 6 | 6 | 1 | 1 | | | | | 4 | + | - +-------+-------+-------+-------+-------+-------+---+---+ + | 2 | | 4 | + +---+ | - | + | 2 | | 3 | + | - | +---+ + | 6 | | 3 | + +---+ T H E | - | + | 6 | | 3 | + | - | +---+ + | 3 | | 3 | + +---+ | - | + | 3 | | 1 | + | - | D O M I N O F R A M E +---+ + | | | 1 | + +---+ | - | + | | | 1 | + | - | +---+ + | 5 | | 1 | + +---+ -S-O-L-U-T-I-O-N- | - | + | 5 | | 4 | + | - | +---+ + | 3 | | 4 | + +---+ | - | + | 3 | | 6 | + | - | +---+ + | 2 | | 6 | + +---+---+-------+-------+-------+-------+-------+-------+ - | + | 2 | 1 | 1 | 5 | 5 | 5 | 5 | 4 | 4 | 4 | 4 | 2 | 2 | | | + +-------+-------+-------+-------+-------+-------+-------+---+ + +] + +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--2 down to 3--2 may be reversed, or from +2--6 to 3--2, or from 3--0 to 5--3. Also, on the right-hand side we may +reverse from 4--3 to 1--4. These changes will not affect the correctness +of the solution. + + +381.--THE CARD FRAME PUZZLE. + +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:-- + +[Illustration: + + +-------+-------+-------+ + | 2 | 10 | 6 | + +---+---+------ +---+---+ + | | | | + | 3 | | 7 | + | | | | + +---+ +---+ + | | | | + | 8 | | 1 | + | | | | + +---+---+-------+--+----+ + | 5 | 9 | 4 | + +-------+-------+-------+ + +] + +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--ten arrangements in all. Readers may like to find all these for +themselves. + + +382.--THE CROSS OF CARDS. + +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. + + 5 6 1 7 4 2 4 5 6 8 + 3 5 1 6 8 3 4 5 6 7 + 3 4 1 7 8 1 4 7 6 8 + 2 5 1 7 8 2 3 7 6 8 + 2 5 3 6 8 2 4 7 5 8 + 1 5 3 7 8 3 4 9 5 6 + 2 4 3 7 8 2 4 9 5 7 + 1 4 5 7 8 1 4 9 6 7 + 2 3 5 7 8 2 3 9 6 7 + +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. + + +383.--THE "T" CARD PUZZLE. + +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 x 24 = 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. + + +384.--CARD TRIANGLES. + +The following arrangements of the cards show (1) the smallest possible +sum, 17; and (2) the largest possible, 23. + + 1 7 + 9 6 4 2 + 4 8 3 6 + 3 7 5 2 9 5 1 8 + +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. + + +385.--"STRAND" PATIENCE. + +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) = 62 moves in all. This is +my record; perhaps the reader can beat it. + + +386.--A TRICK WITH DICE. + +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. + +The process merely consists in giving 100a + 10b + c + 250, where a, b, +and c represent the three throws. The result is obvious. + + +387.--THE VILLAGE CRICKET MATCH. + +[Illustration: + + | Mr. Dumkins >>--> + |------------------------> | + | <------------------- | + | -------------------> | + 1 |<----------------------- | + | | + | <------------------------| + | -------------------> | + | <------------------- | + | ----------------------->| + | <--<< Mr. Podder | + + + | Mr. Luffey >>--> + |------------------------> | + | <------------------- | + | ----------------------->| + 2 | | + |<----------------------- | + | -------------------> | + | <------------------------| + <--<< Mr. Struggles | + +] + +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 manoeuvre must increase Mr. Struggles's total by +one run. + + +388.--SLOW CRICKET. + +The captain must have been "not out" and scored 21. Thus:-- + + 2 men (each lbw) 19 + 4 men (each caught) 17 + 1 man (run out) 0 + 3 men (each bowled) 9 + 1 man (captain--not out) 21 + -- -- + 11 66 + +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. + + +389.--THE FOOTBALL PLAYERS. + +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. + + +390.--THE HORSE-RACE PUZZLE. + +The answer is: L12 on Acorn, L15 on Bluebottle, L20 on Capsule. + +391.--THE MOTOR-CAR RACE. + +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--the +correct answer. + + +392.--THE PEBBLE GAME. + +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. + + +393.--THE TWO ROOKS. + +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. + + r: black rook + R: white rook + + +-+-+-+-+-+-+-+-+ + |r| | | | | | | | + +-+-+-+-+-+-+-+-+ + | |A| | | | | | | + +-+-+-+-+-+-+-+-+ + | | | | | | | | | + +-+-+-+-+-+-+-+-+ + | | | | | | | | | + +-+-+-+-+-+-+-+-+ + | | | | |B|D|F| | + +-+-+-+-+-+-+-+-+ + | | | | | |R|C|E| + +-+-+-+-+-+-+-+-+ + | | | | | | |I|G| + +-+-+-+-+-+-+-+-+ + | | | | | | | |H| + +-+-+-+-+-+-+-+-+ + +THE TWO ROOKS. + + +394.--PUSS IN THE CORNER. + +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:-- + +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. + + +395.--A WAR PUZZLE GAME. + +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 following 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 _the same way that he went +in_, 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. + +[Illustration] + +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. + +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. + +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 +_not_ 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. + +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. + + 24 20 19 15 11 7 3 1 2 6 10 14 18 19 20 24 + ----------------------------------------------- + 13 9 13 17 21 20 24 23 19 15 19 23 24 25 27 + +The enemy must now go to 25 or B, in either of which towns he is +immediately captured. + + +396.--A MATCH MYSTERY. + +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. + +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 2^0 = 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--12, 11, 7. Expressed in powers of 2 we have-- + + 12 = 8 4 - - + 11 = 8 - 2 1 + 7 = - 4 2 1 + ------- + 2 2 2 2 + ------- + +As there are thus two of every power, you must win. Say your opponent +takes 7 from the 12 heap. He then leaves-- + + 5 = - 4 - 1 + 11 = 8 - 2 1 + 7 = - 4 2 1 + ------- + 1 2 2 3 + ------- + +Here the powers are not all even in number, but by taking 9 from the 11 +heap you immediately restore your winning position, thus-- + + 5 = - 4 - 1 + 2 = - - 2 - + 7 = - 4 2 1 + ------- + - 2 2 2 + ------- + +And so on to the end. This solution is quite 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. + + +397.--THE MONTENEGRIN DICE GAME. + +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. + + +398.--THE CIGAR PUZZLE. + +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--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. + +The theories of some other friends would be quite sound if the shape of +the cigar were that of a torpedo--perfectly symmetrical and pointed at +both ends. + +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. + +[Illustration: 1] + +[Illustration: 2] + +The first player must place his first cigar _on end_ 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 _not_ having played the best possible. + +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--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. + + +399.--THE TROUBLESOME EIGHT. + +[Illustration: + + +---+---+---+ + | 41/2| 8 | 21/2| + +---+---+---+ + | 3 | 5 | 7 | + +---+---+---+ + | 71/2| 2 | 51/2| + +---+---+---+ + +] + +The conditions were to place a different number in each of the nine +cells so that the three rows, 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--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. + + +400.--THE MAGIC STRIPS. + +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:-- + +[Illustration: + + +-------------+ + |1 2 3 4 5 6 7| + +---------+---+ + |3 4 5 6 7|1 2| + +-----+---+---+ + |5 6 7|1 2 3 4| + +-+---+-------+ + |7|1 2 3 4 5 6| + +-+---------+-+ + |2 3 4 5 6 7|1| + +-------+---+-+ + |4 5 6 7|1 2 3| + +---+---+-----+ + |6 7|1 2 3 4 5| + +---+---------+ + +] + +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 +_perpendicular_ 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. + + +401.--EIGHT JOLLY GAOL-BIRDS. + +There are eight ways of forming the magic square--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. + +[Illustration: + + +---+---+---+ +---+---+---+ + | | | | | | | | + | 5 7 | | 7 4 3 | + | | | | | | | | + +- -+- -+- -+ +- -+- -+- -+ + | | | | | | | | + | 6 4 2 | | 4 8 | + | | | | | | | | + +- -+- -+- -+ +- -+- -+- -+ + | | | | | | | | + | 1 8 3 | | 5 6 1 | + | | | | | | | | + +---+---+---+ +---+---+---+ + +] + + +402.--NINE JOLLY GAOL BIRDS. + +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--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 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. + +[Illustration: + + A B + +---+---+---+ +---+---+---+ + | | | | | | | | + | 2 9 4 | | 6 7 2 | + | | | | | | | | + +- -+- -+- -+ +- -+- -+- -+ + | | | | | | | | + | 7 5 3 | | 1 5 9 | + | | | | | | | | + +- -+- -+- -+ +- -+- -+- -+ + | | | | | | | | + | 6 1 8 | | 8 3 4 | + | | | | | | | | + +---+---+---+ +---+---+---+ + +] + + +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:-- + +Place 5 on 1. Then, 6, 9, 8, 6, 4, (6), 2, 4, 9, 3, 4, 9, (6), 7, 6, 1. + +Place 5 on 9. Then, 4, 1, 2, 4, 6, (14), 8, 6, 1, 7, 6, 1, (14), 3, 4, +9. + +Place 5 on 3. Then, 6, (8), 2, 6, 4, 7, 8, 4, 7, 1, 6, 7, (8), 9, 4, 3. + +Place 5 on 7. Then, 4, (12), 8, 4, 6, 3, 2, 6, 3, 9, 4, 3, (12), 1, 6, +7. + +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. + + +403.--THE SPANISH DUNGEON. + +[Illustration] + + +-----+-----+-----+-----+ +-----+-----+-----+-----+ + | | | | | | | | | | + | 1 | 2 | 3 | 4 | | 10 | 9 | 7 | 4 | + |_____|_____|_____|_____| |_____|_____|_____|_____| + | | | | | | | | | | + | 5 | 6 | 7 | 8 | | 6 | 5 | 11 | 8 | + |_____|_____|_____|_____| |_____|_____|_____|_____| + | | | | | | | | | | + | 9 | 10 | 11 | 12 | | 1 | 2 | 12 | 15 | + |_____|_____|_____|_____| |_____|_____|_____|_____| + | | | | | | | | | | + | 13 | 14 | 15 | | | 13 | 14 | | 3 | + | | | | | | | | | | + +-----+-----+-----+-----+ +-----+-----+-----+-----+ + +This can best be solved by working backwards--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. + +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. + + +404.--THE SIBERIAN DUNGEONS. + +[Illustration] + + +-----+-----+-----+-----+ + | | | | | + | 8 | 5 | 10 | 11 | + |_____|_____|_____|_____| + | | | | | + | 16 | 13 | 2 | 3 | + |_____|_____|_____|_____| + | | | | | + | 1 | 12 | 7 | 14 | + |_____|_____|_____|_____| + | | | | | + | 9 | 4 | 15 | 6 | + | | | | | + +-----+-----+-----+-----+ + +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. + +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. + + +405.--CARD MAGIC SQUARES. + +Arrange the cards as follows for the three new squares:-- + + 3 2 4 6 5 7 9 8 10 + 4 3 2 7 6 5 10 9 8 + 2 4 3 5 7 6 8 10 9 + +Three aces and one ten are not used. The summations of the four squares +are thus: 9, 15, 18, and 27--all different, as required. + + +406.--THE EIGHTEEN DOMINOES. + +[Illustration] + +The illustration explains itself. It will be found that the pips in +every column, row, and long diagonal add up 18, as required. + + +407.--TWO NEW MAGIC SQUARES. + +Here are two solutions that fulfil the conditions:-- + +[Illustration: + + SUBTRACTING DIVIDING + 11 4 14 13 36 8 54 27 + 16 7 1 2 216 12 1 2 + 6 5 3 12 6 3 4 72 + 9 19 8 15 9 18 24 108 + +] + +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. + + +408.--MAGIC SQUARES OF TWO DEGREES. + +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. + +[Illustration: + + 7 53 | 41 27 | 2 52 | 48 30 + 12 58 | 38 24 | 13 63 | 35 17 + ------+-------+-------+------ + 51 1 | 29 47 | 54 8 | 28 42 + 64 14 | 18 36 | 57 11 | 23 37 + ------+-------+-------+------ + 25 43 | 55 5 | 32 46 | 50 4 + 22 40 | 60 10 | 19 33 | 61 15 + ------+-------+-------+------ + 45 31 | 3 49 | 44 26 | 6 56 + 34 20 | 16 62 | 39 21 | 9 59 + +] + +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 + 23 ++ 26 + 31 + 48 + 56 + 57 = 260, and the sum of their squares is 11,180. +Therefore 64 + 47 + 42 + 39 + 34 + 17 + 9 + 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 :-- + +[Illustration: + + 1 8 28 29 42 47 51 54 + 2 7 27 30 41 48 52 53 + 3 6 26 31 44 45 49 56 + 4 5 25 32 43 46 50 55 + +] + +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. + + +409.--THE BASKETS OF PLUMS. + +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 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. + +[Illustration] + + A B + +-----+-----+-----+ +-----+-----+-----+ + | | | | | | | | + | 7 | 61 | 43 | | 83 | 29 | 101 | + |_____|_____|_____| |_____|_____|_____| + | | | | | | | | + | 73 | 37 | 1 | | 89 | 71 | 53 | + |_____|_____|_____| |_____|_____|_____| + | | | | | | | | + | 31 | 13 | 67 | | 41 | 113 | 59 | + | | | | | | | | + +-----+-----+-----+ +-----+-----+-----+ + + C D + +-----+-----+-----+ +-----+-----+-----+ + | | | | | | | | + | 103 | 79 | 37 | |1669 | 199 |1249 | + |_____|_____|_____| |_____|_____|_____| + | | | | | | | | + | 7 | 73 | 139 | | 619 |1039 |1459 | + |_____|_____|_____| |_____|_____|_____| + | | | | | | | | + | 109 | 67 | 43 | | 829 |1879 | 409 | + | | | | | | | | + +-----+-----+-----+ +-----+-----+-----+ + +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. + +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--all primes with a common difference of 210. + + +410.--THE MANDARIN'S "T" PUZZLE. + +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. + + +[Illustration: THE MANDARIN'S "T" PUZZLE. + + +-----+-----+-----+-----+-----+ + | | | | | | + | 19 | 23 | 11 | 5 | 7 | + |_____|_____|_____|_____|_____| + | | | | | | + | 1 | 10 | 17 | 24 | 13 | + |_____|_____|_____|_____|_____| + | | | | | | + | 22 | 14 | 3 | 6 | 20 | + |_____|_____|_____|_____|_____| + | | | | | | + | 8 | 16 | 25 | 12 | 4 | + |_____|_____|_____|_____|_____| + | | | | | | + | 15 | 2 | 9 | 18 | 21 | + | | | | | | + +-----+-----+-----+-----+-----+ + + +411.--A MAGIC SQUARE OF COMPOSITES. + +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. + +[Illustration] + + +-----+-----+-----+ + | | | | + | 121 | 114 | 119 | + |_____|_____|_____| + | | | | + | 116 | 118 | 120 | + |_____|_____|_____| + | | | | + | 117 | 122 | 115 | + | | | | + +-----+-----+-----+ + +First write down any consecutive numbers, the smallest being greater +than 1--say, 2, 3, 4, 5, 6, 7, 8, 9, 10. The only factors in these +numbers are 2, 3, 5, and 7. We therefore multiply 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. + +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. + + +412.--THE MAGIC KNIGHT'S TOUR. + + +----+----+----+----+----+----+----+----+ + | 46 | 55 | 44 | 19 | 58 | 9 | 22 | 7 | + +----+----+----+----+----+----+----+----+ + | 43 | 18 | 47 | 56 | 21 | 6 | 59 | 10 | + +----+----+----+----+----+----+----+----+ + | 54 | 45 | 20 | 41 | 12 | 57 | 8 | 23 | + +----+----+----+----+----+----+----+----+ + | 17 | 42 | 53 | 48 | 5 | 24 | 11 | 60 | + +----+----+----+----+----+----+----+----+ + | 52 | 3 | 32 | 13 | 40 | 61 | 34 | 25 | + +----+----+----+----+----+----+----+----+ + | 31 | 16 | 49 | 4 | 33 | 28 | 37 | 62 | + +----+----+----+----+----+----+----+----+ + | 2 | 51 | 14 | 29 | 64 | 39 | 26 | 35 | + +----+----+----+----+----+----+----+----+ + | 15 | 30 | 1 | 50 | 27 | 36 | 63 | 38 | + +----+----+----+----+----+----+----+----+ + +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--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. + + +413.--A CHESSBOARD FALLACY. + +[Illustration] + +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+1/7 in. Consequently the rectangle is really 9+1/7 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. + + +414.--WHO WAS FIRST? + +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. + + +415.--A WONDERFUL VILLAGE. + +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. + + +416.--A CALENDAR PUZZLE. + +The first day of a century can never fall on a Sunday; nor on a +Wednesday or a Friday. + + +417.--THE TIRING-IRONS. + +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^(n + 1) - 2) moves; and since n in our case is 14, all +the rings may be taken off in 10,922 moves. Then I say 10,922 - 9,999 = +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 the binary scale method used by Monsieur L. Gros, for an +account of which see W.W. Rouse Ball's _Mathematical Recreations_. + +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:-- + + O O O OO + ------------------------- + OOOO OO O O O + +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^(n + 1) - 1). + +With n rings (where n is _odd_) there are 2^n positions counting all on +and all off. In (1/3)(2^(n + 1) + 2) positions they are all removed. The +number of positions not used is (1/3)(2^n - 2). + +With n rings (where n is _even_) there are 2^n positions counting all on +and all off. In (2^(n + 1) + 1) positions they are all removed. The +number of positions not used is here (1/3)(2^n - 1). + +It will be convenient to tabulate a few cases. + + +--------+------------+-----------+-----------+ + | No. of | Total | Positions | Positions | + | Rings. | Positions. | used. | not used. | + +--------+------------+-----------+-----------+ + | 1 | 2 | 2 | 0 | + | 3 | 8 | 6 | 2 | + | 5 | 32 | 22 | 10 | + | 7 | 128 | 86 | 42 | + | 9 | 512 | 342 | 170 | + | | | | | + | 2 | 4 | 3 | 1 | + | 4 | 16 | 11 | 5 | + | 6 | 64 | 43 | 21 | + | 8 | 256 | 171 | 85 | + | 10 | 1024 | 683 | 341 | + +--------+------------+-----------+-----------+ + +Note first that the number of _positions used_ is one more than the +number of _moves_ 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 _positions not used_ is the same as the number of _moves +used_ 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. + + +418.--SUCH A GETTING UPSTAIRS. + +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. + + +419.--THE FIVE PENNIES. + +[Illustration] + +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. + + +420.--THE INDUSTRIOUS BOOKWORM. + +The hasty reader will assume that the bookworm, in boring from the first +to the last page 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 91/2 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, 1/2 in.) and the leaves in the second volume (3 in.), +or a distance of 31/2 inches, in order to tunnel from the first page to +the last. + + +421.--A CHAIN PUZZLE. + +To open and rejoin a link costs threepence. Therefore to join the nine +pieces into an endless chain would cost 2s. 3d., whereas a new chain +would cost 2s. 2d. But if we break up the piece of eight links, these +eight will join together the remaining eight pieces at a cost of 2s. 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 1s. 9d. + + +422.--THE SABBATH PUZZLE. + +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, _Three Sundays in a Week_, or +of Jules Verne's _Round the World in Eighty Days_, 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--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. + + +423.--THE RUBY BROOCH. + +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." + +[Illustration] + +The diagram shows the arrangement before the robbery. It will be seen +that it was only necessary to reset one ruby--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." + + +424.--THE DOVETAILED BLOCK. + +[Illustration] + +The mystery is made clear by the illustration. It will be seen at once +how the two pieces slide together in a diagonal direction. + + +425.--JACK AND THE BEANSTALK. + +The serious blunder that the artist made in this drawing was in +depicting the tendrils of + +[Illustration] + +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. + + +426.--THE HYMN-BOARD POSER. + +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. + +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. + +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. + +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. + +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 43/4d., and +the cost 11s. 101/2d. 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 41/2d., thus saving a farthing on each of the previous +thirty, and reducing the cost of the last one from 1s. to 41/2d. + +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. + +Here is the solution of the difficulty that was sent to the vicar of +Chumpley St. Winifred. Where the sign X 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. + + d. L s. d. + 31 plates painted 5 X 9 @ 41/2 = 0 11 71/2 + 30 " 7 X 8 @ 43/4 = 0 11 101/2 + 21 " 1 X 2 @ 7 = 0 12 3 + 21 " 3 X 0 @ 7 = 0 12 3 + 12 " 1 X 3 @ 91/4 = 0 9 3 + 12 " 2 X 4 @ 91/4 = 0 9 3 + 12 " 9 X 4 @ 91/4 = 0 9 3 + 8 " 4 X 0 @ 101/4 = 0 6 10 + 1 " 5 X 4 @ 12 = 0 1 0 + 1 " 5 X 0 @ 12 = 0 1 0 + 149 plates @ 6d. each = 3 14 6 + ---------- + L7 19 1 + +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. + +This puzzle appeared in _Tit-Bits_, and the following remarks, made by +me in the issue for 11th December 1897, may be of interest. + +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 L1,347, 10s., +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 _Tit-Bits_ 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. + + +427.--PHEASANT-SHOOTING. + +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." + + +428.--THE GARDENER AND THE COOK. + +Nobody succeeded in solving the puzzle, so I had to let the cat out of +the bag--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. + +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+1/3 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. + +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 _any_ gardener and cook, but of a +_particular_ 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. + + +429.--PLACING HALFPENNIES. + +Thirteen coins may be placed as shown on page 252. + + +430.--FIND THE MAN'S WIFE. + +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. + +[Illustration: PLACING HALFPENNIES.] + +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--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;--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. + + + + +INDEX. + + + Abbot's Puzzle, The, 20, 161. + ---- Window, The, 87, 213. + + Academic Courtesies, 18, 160. + + Acrostic Puzzle, An, 84, 210. + + Adam and Eve and the Apples, 18. + + Aeroplanes, The Two, 2, 148. + + Age and Kinship Puzzles, 6. + ---- Concerning Tommy's, 7, 153. + ---- Mamma's, 7, 152. + ---- Mrs. Timpkins's, 7, 152. + ---- Rover's, 7, 152. + + Ages, The Family, 7, 152. + ---- Their, 7, 152. + + Alcuin, Abbot, 20, 112. + + Almonds, The Nine, 64, 195. + + Amazons, The, 94, 221. + + Andrews, W.S., 125. + + Apples, A Deal in, 3, 149. + ---- Buying, 6, 151. + ---- The Ten, 64, 195. + + Approximations in Dissection, 28. + + Arithmetical and Algebraical Problems, 1. + ---- Various, 17. + + Arthur's Knights, King, 77, 203. + + Artillerymen's Dilemma, 26, 167. + + Asparagus, Bundles of, 140. + + Aspects all due South, 137. + + Associated Magic Squares, 120. + + Axiom, A Puzzling, 138. + + + Bachet de Meziriac, 90, 109, 112. + + Bachet's Square, 90, 216. + + Ball Problem, The, 51, 183. + + Ball, W.W. Rouse, 109, 204, 248. + + Balls, The Glass, 78, 204. + + Banker's Puzzle, The, 25, 165. + + Bank Holiday Puzzle, A, 73, 201. + + Banner Puzzle, The, 46, 179. + ---- St. George's, 50, 182. + + Barrel Puzzle, The, 109, 235. + + Barrels of Balsam, The, 82, 208. + + Beanfeast Puzzle, A, 2, 148. + + Beef and Sausages, 3, 149. + + Beer, The Barrel of, 13, 155. + + Bell-ropes, Stealing the, 49, 181. + + Bells, The Peal of, 78, 204. + + Bergholt, E., 116, 119, 125. + + Betsy Ross Puzzle, The, 40, 176. + + Bicycle Thief, The, 6, 152. + + Bishops--Guarded, 88, 214. + ---- in Convocation, 89, 215. + ---- Puzzle, A New, 98, 225. + ---- Unguarded, 88, 214. + + Board, The Chess-, 85. + ---- in Compartments, The, 102, 228. + ---- Setting the, 105, 231. + + Boards with Odd Number of Squares, 86, 212. + + Boat, Three Men in a, 78, 204. + + Bookworm, The Industrious, 143, 248. + + Boothby, Guy, 154. + + Box, The Cardboard, 49, 181. + ---- The Paper, 40. + + Boys and Girls, 67, 197. + + Bridges, The Monk and the, 75, 202. + + Brigands, The Five, 25, 164. + + Brocade, The Squares of, 47, 180. + + Bun Puzzle, The, 35, 170. + + Busschop, Paul, 172. + + Buttons and String Method, 230. + + + Cab Numbers, The, 15, 157. + + Calendar Puzzle, A, 142, 247. + + _Canterbury Puzzles, The_, 14, 28, 58, 117, 121, 195, 202, 205, 206, 212, 213, 217, 233. + + Card Frame Puzzle, The, 114, 238. + ---- Magic Squares, 123, 244. + ---- Players, A Puzzle for, 78, 203. + ---- Puzzle, The "T," 115, 239. + ---- Triangles, 115, 239. + + Cards, The Cross of, 115, 238. + + Cardan, 142. + + Carroll, Lewis, 43. + + Castle Treasure, Stealing the, 113, 237. + + Cats, the Wizard's, 42, 178. + + Cattle, Judkins's, 6, 151. + ---- Market, At a, 1, 148. + + Census Puzzle, A, 7, 152. + + Century Puzzle, The, 16, 158. + ---- The Digital, 16, 159. + + Chain Puzzle, A, 144, 249. + ---- The Antiquary's, 83, 209. + ---- The Cardboard, 40, 176. + + Change, Giving, 4, 150. + ---- Ways of giving, 151. + + Changing Places, 10, 154. + + Channel Island, 138. + + Charitable Bequest, A, 2, 148 + + Charity, Indiscriminate, 2, 148. + + Checkmate, 107, 233. + + Cheesemonger, The Eccentric, 66, 196. + + Chequered Board Divisions, 85, 210. + + Cherries and Plums, 56, 189. + + Chess Puzzles, Dynamical, 96. + ---- Statical, 88. + ---- Various, 105. + ---- Queer, 107, 233. + + Chessboard, The, 85. + ---- Fallacy, A, 141, 247. + ---- Guarded, 95. + ---- Non-attacking Arrangements, 96. + ---- Problems, 84. + ---- Sentence, The, 87, 214. + ---- Solitaire, 108, 234. + ---- The Chinese, 87, 213. + ---- The Crowded, 91, 217. + + Chestnuts, Buying, 6, 152. + + Chinese Money, 4, 150. + ---- Puzzle, Ancient, 107, 233. + ---- ---- _The Fashionable_, 43. + + Christmas Boxes, The, 4, 150. + ---- Present, Mrs. Smiley's, 46, 179. + ---- Pudding, The, 43, 178. + + Cigar Puzzle, The, 119, 242. + + Circle, The Dissected, 69, 197. + + Cisterns, How to Make, 54, 188. + + Civil Service "Howler," 154. + + Clare, John, 58. + + Clock Formulae, 154. + ---- Puzzles, 9. + ---- The Club, 10, 154. + ---- The Railway Station, 11, 155. + + Clocks, The Three, 11, 154. + + Clothes Line Puzzle, The, 50, 182. + + Coast, Round the, 63, 195. + + Coincidence, A Queer, 2, 148. + + Coins, The Broken, 5, 150. + ---- The Ten, 57, 190. + ---- Two Ancient, 140. + + Combination and Group Problems, 76. + + Compasses Puzzle, The, 53, 186. + + Composite Magic Squares, 127, 246. + + Cone Puzzle, The, 55, 188. + + Corn, Reaping the, 20, 161. + + Cornfields, Farmer Lawrence's, 101, 227. + + Costermonger's Puzzle, The, 6, 152. + + Counter Problems, Moving, 58. + ---- Puzzle, A New, 98, 225. + ---- Solitaire, 107, 234. + + Counters, The Coloured, 91, 217. + ---- The Forty-nine, 92, 217. + ---- The Nine, 14, 156. + ---- The Ten, 15, 156. + + Crescent Puzzle, The, 52, 184. + + Crescents of Byzantium, The Five, 92, 219. + + Cricket Match, The Village, 116, 239. + ---- Slow, 116, 239. + + Cross and Triangle, 35, 169. + ---- of Cards, 115, 238. + ---- The Folded, 35, 169. + ---- The Southern, 93, 220. + + Crosses, Counter, 81, 207. + ---- from One, Two, 35, 168. + ---- ---- Three, 169. + + Crossing River Problems, 112. + + Crusader, The, 106, 232. + + Cubes, Sums of, 165. + + Cushion Covers, The, 46, 179. + + Cutting-out Puzzle, A, 37, 172. + + Cyclists' Feast, The, 2, 148. + + + Dairyman, The Honest, 110, 235. + + Definition, A Question of, 23, 163. + + De Fonteney, 112. + + Deified Puzzle, The, 74, 202. + + Delannoy, 112. + + De Morgan, A., 27. + + De Tudor, Sir Edwyn, 12, 155. + + Diabolique Magic Squares, 120. + + Diamond Puzzle, The, 74, 202. + + Dice, A Trick with, 116, 239. + ---- Game, The Montenegrin, 119, 242. + ---- Numbers, The, 17, 160. + + Die, Painting the, 84, 210, + + Digital Analysis, 157, 158. + ---- Division, 16, 158. + ---- Multiplication, 15, 156. + ---- Puzzles, 13. + + Digits, Adding the, 16, 158. + ---- and Squares, 14, 155. + ---- Odd and Even, 14, 156. + + Dilemma, An Amazing, 106, 233. + + Diophantine Problem, 164. + + Dissection Puzzle, An Easy, 35, 170. + ---- Puzzles, 27. + ---- ---- Various, 35. + + Dividing Magic Squares, 124. + + Division, Digital, 16, 158. + ---- Simple, 23, 163. + + Doctor's Query, The, 109, 235. + + Dogs Puzzle, The Five, 92, 218. + + Domestic Economy, 5, 151. + + Domino Frame Puzzle, The, 114, 238. + + Dominoes in Progression, 114, 237. + ---- The Eighteen, 123, 245. + ---- The Fifteen, 83, 209. + ---- The Five, 114, 238. + + Donkey Riding, 13, 155. + + Dormitory Puzzle, A, 81, 208. + + Dovetailed Block, The, 145, 249. + + Drayton's _Polyolbion_, 58. + + Dungeon Puzzle, A, 97, 224. + + Dungeons, The Siberian, 123, 244. + ---- The Spanish, 122, 244. + + Dutchmen's Wives, The, 26, 167. + + Dynamical Chess Puzzles, 96. + + + Earth's Girdle, The, 139. + + _Educational Times Reprints_, 204. + + Eggs, A Deal in, 3, 149. + ---- Obtaining the, 140. + + Election, The Muddletown, 19, 161. + ---- The Parish Council, 19, 161. + + Eleven, The Mystic, 16, 159. + + Elopements, The Four, 113, 237. + + Elrick, E., 231. + + Engines, The Eight, 61, 194. + + Episcopal Visitation, An, 98, 225. + + Estate, Farmer Wurzel's, 51, 184. + + Estates, The Yorkshire, 51, 183. + + Euclid, 31, 138. + + Euler, L., 165. + + Exchange Puzzle, The, 66, 196. + + + Fallacy, A Chessboard, 141, 247. + + Family Party, A, 8, 153. + + Fare, The Passenger's, 13, 155. + + Farmer and his Sheep, The, 22, 163. + + Fence Problem, A, 21, 162. + + Fences, The Landowner's, 42, 178. + + Fermat, 164, 168. + + Find the Man's Wife, 147, 251. + + Fly on the Octahedron, The, 70, 198. + + Fog, Mr. Gubbins in a, 18, 161. + + Football Players, The, 116, 240. + + Fraction, A Puzzling, 138. + + Fractions, More Mixed, 16, 159. + + Frame Puzzle, The Card, 114, 238. + ---- ---- The Domino, 114, 238. + + Frankenstein, E.N., 232. + + Frenicle, B., 119, 168. + + Frogs, The Educated, 59, 194- + ---- The Four, 103, 229. + ---- The Six, 59, 193. + + Frost, A.H., 120. + + + Games, Puzzle, 117. + ---- Problems concerning, 114. + + Garden, Lady Belinda's, 52, 186. + ---- Puzzle, The, 49, 182. + + Gardener and the Cook, The, 146, 251. + + Geometrical Problems, 27. + ---- Puzzles, Various, 49. + + George and the Dragon, St., 101, 227. + + Getting Upstairs, Such a, 143, 248. + + Girdle, the Earth's, 139. + + Goat, The Tethered, 53, 186. + + Grand Lama's Problem, The, 86, 212. + + Grasshopper Puzzle, The, 59, 193. + + Greek Cross Puzzles, 28. + ---- ---- Three from One, 169. + + Greyhound Puzzle, The, 101, 227. + + Grocer and Draper, The, 5, 151. + + Gros, L., 248. + + Group Problems, Combination and, 76. + + Groups, The Three, 14, 156. + + Guarini, 229. + + + Hairdresser's Puzzle, The, 137. + + Halfpennies, Placing, 147, 251. + + Hampton Court Maze solved, 133. + + Hannah's Puzzle, 75, 202. + + Hastings, The Battle of, 23, 164. + + Hatfield Maze solved, 136. + + Hat Puzzle, The, 67, 196. + + Hat-peg Puzzle, The, 93, 221. + + Hats, The Wrong, 78, 203. + + Hay, The Trusses of, 18, 161. + + Heads or Tails, 22, 163. + + Hearthrug, Mrs. Hobson's, 37, 172. + + Helmholtz, Von, 41. + + Honey, The Barrels of, 111, 236. + + Honeycomb Puzzle, The, 75, 202. + + Horse Race Puzzle, The, 117, 240. + + Horseshoes, The Two, 40, 175. + + Houdin, 68. + + Hydroplane Question, The, 12, 155. + + Hymn-board Poser, The, 145, 250. + + + Icosahedron Puzzle, The, 70, 198. + + + Jack and the Beanstalk, 145, 249. + + Jackson, John, 56. + + Jaenisch, C.F. de, 92. + + Jampots, Arranging the, 68, 197. + + Jealous Husbands, Five, 113, 236. + + Joiner's Problem, The, 36, 171. + ---- ---- Another, 37, 171. + + Jolly Gaol-Birds, Eight, 122, 243. + ---- ---- Nine, 122, 243. + + Journey, The Queen's, 100, 227. + ---- The Rook's, 96, 224. + + Junior Clerks' Puzzle, The, 4, 150. + + Juvenile Puzzle, A, 68, 197. + + + Kangaroos, The Four, 102, 228. + + Kelvin, Lord, 41. + + Kennel Puzzle, The, 105, 231. + + King and the Castles, The, 56, 189. + ---- The Forsaken, 106, 232. + + Kite-flying Puzzle, A, 54, 187. + + Knight-guards, The, 95, 222. + + Knights, King Arthur's, 77, 203. + ---- Tour, Magic, 127, 247. + ---- ---- The Cubic, 103, 229. + ---- ---- The Four, 103, 229. + + + Labosne, A., 25, 90, 216. + + Labourer's Puzzle, The, 18, 160. + + _Ladies' Diary_, 26. + + Lagrange, J.L., 9. + + Laisant, C.A., 76. + + Lamp-posts, Painting the, 19, 161. + + Leap Year, 155. + ---- ---- Ladies, The, 19, 161. + + Legacy, A Puzzling, 20, 161. + + Legal Difficulty, A, 23, 163. + + Le Plongeon, Dr., 29. + + Letter Block Puzzle, The, 60, 194. + ---- Blocks, The Thirty-six, 91, 216. + ---- Puzzle, The Fifteen, 79, 205. + + Level Puzzle, The, 74, 202. + + Linoleum Cutting, 48, 181. + ---- Puzzle, Another, 49, 181. + + Lion and the Man, The, 97, 224. + ---- Hunting, 94, 222. + + Lions and Crowns, 85, 212. + ---- The Four, 88, 214. + + Lockers Puzzle, The, 14, 156. + + Locomotion and Speed Puzzles, 11. + + Lodging-house Difficulty, A, 61, 194. + + London and Wise, 131. + + Loyd, Sam, 8, 43, 44, 98, 144, 232, 233. + + Lucas, Edouard, 16, 76, 112, 121. + + Luncheons, The City, 77, 203. + + + MacMahon, Major, 109. + + Magic Knight's Tour, 127, 247. + ---- Square Problems, 119. + ---- ---- Card, 123, 244. + ---- ---- of Composites, 127, 246. + ---- ---- of Primes, 125. + ---- ---- of Two Degrees, 125, 245. + ---- ---- Two New, 125, 245. + ---- Strips, 121, 243. + + Magics, Subtracting, Multiplying, and Dividing, 124. + + Maiden, The Languishing, 97, 224. + + Mandarin's Puzzle, The, 103, 230. + ---- "T" Puzzle, The, 126, 246. + + Marketing, Saturday, 27, 168. + + Market Women, The, 3, 149. + + Mary and Marmaduke, 7, 152. + + Mary, How Old was, 8, 153. + + Massacre of Innocents, 139. + + Match Mystery, A, 118, 241. + ---- Puzzle, A New, 55, 188. + + Mates, Thirty-six, 106, 233. + + Mazes and how to thread Them, 127. + + Measuring, Weighing, and Packing Puzzles, 109. + ---- Puzzle, New, 110, 235. + + Meeting, The Suffragists', 19, 161. + + Mellor, W.M.F., 242. + + Menages, Probleme de, 76. + + Mersenne, M., 168. + + Mice, Catching the, 65, 196. + + Milkmaid Puzzle, The, 50, 183. + + Millionaire's Perplexity, The, 3, 149. + + Mince Pies, The Twelve, 57, 191. + + Mine, Inspecting a, 71, 199. + + Miners' Holiday, The, 23, 163. + + Miser, The Converted, 21, 162. + + Mitre, Dissecting a, 35, 170. + + Monad, The Great, 39, 174. + + Money, A Queer Thing in, 2, 148. + ---- Boxes, The Puzzling, 3, 149. + ----, Pocket, 3, 149. + ---- Puzzles, 1. + ---- Puzzle, A New, 2, 148. + ----, Square, 3, 149. + + _Monist, The_, 125. + + Monk and the Bridges, The, 75, 202. + + Monstrosity, The, 108, 234. + + Montenegrin Dice Game, The, 119, 242. + + Moreau, 76. + + Morris, Nine Men's, 58. + + Mosaics, A Problem in, 90, 215. + + Mother and Daughter, 7, 152. + + Motor-car Race, The, 117, 240. + ---- Tour, The, 74, 201. + ---- Garage Puzzle, The, 62, 195. + + Motorists, A Puzzle for, 73, 201. + + Mouse-trap Puzzle, The, 80, 206. + + Moving Counter Problems, 58. + + Multiplication, Digital, 15, 156. + ---- Queer, 15, 157. + ---- Simple, 23, 163. + + Multiplying Magic Squares, 124. + + Muncey, J.N., 125. + + Murray, Sir James, 44. + + + Napoleon, 43, 44. + + Nasik Magic Squares, 120. + + Neighbours, Next-Door, 8, 153. + + Newton, Sir Isaac, 56. + + Nine Men's Morris, 58. + + Notation, Scales of, 149. + + Noughts and Crosses, 58, 117. + + _Nouvelles Annales de Mathematiques_, 14. + + Number Checks Puzzle, The, 16, 158. + + Numbers, Curious, 20, 162. + + Nuts, The Bag of, 8, 153. + + + Observation, Defective, 4, 150. + + Octahedron, The Fly on the, 70, 198. + + Oval, How to draw an, 50, 182. + + Ovid's Game, 58. + + + Packing in Russia, Gold, 111, 236. + ---- Puzzles, Measuring, Weighing, and, 109. + ---- Puzzle, A, 111, 236. + + Pandiagonal Magic Squares, 120. + + Papa's Puzzle, 53, 187. + + Pappus, 53. + + Paradox Party, The, 137. + + Party, A Family, 8, 153. + + Patchwork Puzzles, 46. + ---- Puzzle, Another, 48, 180. + ---- The Silk, 34, 168. + + Patience, _Strand_, 116, 239. + + Pawns, A Puzzle with, 94, 222. + ---- Immovable, 106, 233. + ---- The Six, 107, 233. + ---- The Two, 105, 231. + + Pearls, The Thirty-three, 18, 160. + + Pebble Game, The, 117, 240. + + Pedigree, A Mixed, 8, 153. + + Pellian Equation, 164, 167. + + Pennies, The Five, 143, 248. + ---- The Twelve, 65, 195. + + Pension, Drawing her, 12, 155. + + Pentagon and Square, The, 37, 172. + ---- Drawing a, 37. + + Pfeffermann, M., 125. + + Pheasant-Shooting, 146, 251. + + Philadelphia Maze solved, 137. + + Pierrot's Puzzle, The, 15, 156. + + Pigs, The Seven, 41, 177. + + Planck, C., 220, 246. + + Plane Paradox, 138. + + Plantation Puzzle, A, 57, 189. + ---- The Burmese, 58, 191. + + Plates and Coins, 65, 195. + + Plums, The Baskets of, 126, 245. + + Poe, E.A., 249. + + Points and Lines Problems, 56. + + Postage Stamps, The Four, 84, 210. + + Post-Office Perplexity, A, 1, 148. + + Potato Puzzle, The, 41, 177. + + Potatoes, The Basket of, 13, 155. + + Precocious Baby, The, 139. + + Presents, Buying, 2, 148. + + Prime Magic Squares, 125. + + Printer's Error, A, 20, 162. + + Prisoners, Exercise for, 104, 230. + ---- The Ten, 62, 195. + + Probabilities, Two Questions in, 5, 150. + + Problems concerning Games, 114. + + Puss in the Corner, 118, 240. + + Puzzle Games, 117. + + Pyramid, Painting a, 83, 208. + + Pyramids, Square and Triangular, 167. + + Pythagoras, 31. + + + "Queen, The," 120. + + Queens and Bishop Puzzle, 93, 219. + ---- The Eight, 89, 215. + + Queen's Journey, The, 100, 227. + ---- Tour, The, 98, 225. + + Quilt, Mrs. Perkins's, 47, 180. + + + Race Puzzle, The Horse-, 117, 240. + ---- The Motor-car, 117, 240. + + Rackbrane's Little Loss, 21, 163. + + Railway Muddle, A, 62, 194. + ---- Puzzle, A, 61, 194. + ---- Stations, The Three, 49, 182. + + _Rational Amusement for Winter Evenings_, 56. + + Rectangles, Counting the, 105, 232. + + Reiss, M., 58. + + Relationships, Queer, 8, 153. + + Reversals, A Puzzle in, 5, 151. + + River Axe, Crossing the, 112, 236. + + River Problems, Crossing, 112. + + Rookery, The, 105, 232. + + Rook's Journey, The, 96, 224. + ---- Tour, The, 96, 223. + + Rooks, The Eight, 88, 214. + ---- The Two, 117, 240. + + Round Table, The, 80, 205. + + Route Problems, Unicursal and, 68. + + Ruby Brooch, The, 144, 249. + + + Sabbath Puzzle, The, 144, 249. + + Sailor's Puzzle, The, 71, 199. + + Sayles, H.A., 125. + + Schoolboys, The Nine, 80, 205. + + Schoolgirls, The Fifteen, 80, 204. + + Scramble, The Great, 19, 161. + + Sculptor's Problem, The, 23, 164. + + Second Day of Week, 139. + + See-Saw Puzzle, The, 22, 163. + + Semi-Nasik Magic Squares, 120. + + Senior and Junior, 140. + + Sevens, The Four, 17, 160. + + Sharp's Puzzle, 230. + + Sheepfold, The, 52, 184. + + Sheep Pens, The Six, 55, 189. + ---- The Sixteen, 80, 206. + ---- The Three, 92, 217. + ---- Those Fifteen, 77, 203. + + Shopping Perplexity, A, 4, 150. + + Shuldham, C.D., 125, 126. + + Siberian Dungeons, The, 123, 244. + + Simpleton, The Village, 11, 155. + + Skater, The Scientific, 100, 226. + + Skeat, Professor, 127. + + Solitaire, Central, 63, 195. + ---- Chessboard, 108, 234. + ---- Counter, 107, 234. + + Sons, The Four, 49, 181. + + Spanish Dungeons, The, 122, 244. + ---- Miser, The, 24, 164. + + Speed and Locomotion Puzzles, 11. + ---- Average, 11, 155. + + Spiral, Drawing a, 50, 182. + + Spot on the Table, The, 17, 160. + + Square Numbers, Check for, 13. + ---- ---- Digital, 16, 159. + ---- of Veneer, The, 39, 175. + ---- Puzzle, An Easy, 35, 170. + + Squares, A Problem in, 23, 163. + ---- Circling the, 21, 162. + ---- Difference of Two, 167. + ---- Magic, 119. + ---- Sum of Two, 165, 175. + ---- The Chocolate, 35, 170. + + Stalemate, 106, 232. + + Stamp-licking, The Gentle Art of, 91, 217. + + Star Puzzle, The, 99, 226. + + Stars, The Eight, 89, 215. + ---- The Forty-nine, 100, 226. + + Statical Chess Puzzles, 88. + + Sticks, The Eight, 53, 186. + + Stonemason's Problem, The, 25, 165. + + Stop-watch, The, 11, 154. + + _Strand Magazine, The_, 44, 116, 220. + + _Strand_ Patience, 116, 239. + + Stream, Crossing the, 112, 236. + + Strutt, Joseph, 59. + + Subtracting Magic Squares, 124. + + Sultan's Army, The, 25, 165. + + Suppers, The New Year's Eve, 3, 149. + + Surname, Find Ada's, 27, 168. + + Swastika, The, 29, 31, 169. + + + "T" Card Puzzle, The, 115, 239. + + Table, The Round, 80, 205. + + Table-top and Stools, The, 38, 173. + + Tangram Paradox, A, 43, 178. + + Target, The Cross, 84, 210. + + Tarry, 112. + + Tartaglia, 25, 109, 112. + + Tea, Mixing the, 111, 235. + + Telegraph Posts, The, 139. + + Tennis Tournament, A, 78, 203. + + Tetrahedron, Building the, 82, 208. + + Thief, Catching the, 19, 161. + + Thrift, A Study in, 25, 166. + + Thompson, W.H., 232. + + Ticket Puzzle, The Excursion, 5, 151. + + Time Puzzle, A, 10, 153. + ---- What was the, 10, 153. + + Tiring Irons, The, 142, 247. + + _Tit-Bits_, 58, 79, 124, 251. + + Torn Number, The, 20, 162. + + Torpedo Practice, 67, 196. + + Tour, The Cyclists', 71, 199. + ---- The Grand, 72, 200. + ---- The Queen's, 98, 225. + ---- The Rook's, 96, 223. + + Towns, Visiting the, 70, 198. + + Trains, The Two, 11, 155. + + Treasure Boxes, The Nine, 24, 164. + + Trees, The Twenty-one, 57, 190. + + Tremaux, M., 133, 135. + + Triangle, The Dissected, 38, 173. + + Triangular Numbers, 13, 25, 166. + ---- ---- Check for, 13. + + Troublesome Eight, The, 121, 242. + + Tube Inspector's Puzzle, The, 69, 198. + ---- Railway, Heard on the, 8, 153. + + Turks and Russians, 58, 191. + + Turnings, The Fifteen, 70, 198. + + Twickenham Puzzle, The, 60, 194. + + Two Pieces Problem, The, 96. + + + Unclassified Puzzles, 142. + + Unicursal and Route Problems, 68. + + Union Jack, The, 50, 69, 197. + + + Vandermonde, A., 58, 103. + + Veil, Under the, 90, 216. + + Verne, Jules, 249. + + Victoria Cross Puzzle, The, 60, 194. + + Village, A Wonderful, 142, 247. + + Villages, The Three, 12, 155. + + Villas, The Eight, 80, 206. + + Vortex Rings, 40. + + Voter's Puzzle, The, 75, 202. + + + Wall, The Puzzle, 52, 184. + + Wallis, J., 142. + ---- (Another), 220. + + Walls, The Garden, 52, 185. + + Wapshaw's Wharf Mystery, The, 10, 153. + + War Puzzle Game, The, 118, 240. + + Wassail Bowl, The, 109, 235. + + Watch, A Puzzling, 10, 153. + + Water, Gas, and Electricity, 73, 200. + + _Weekly Dispatch, The_, 28, 124, 125, 146, 148. + + Weighing Puzzles, Measuring, Packing, and, 109. + + Wheels, Concerning, 55, 188. + + Who was First? 142, 247. + + Whyte, W.T., 147. + + Widow's Legacy, The, 2, 148. + + Wife, Find the Man's, 147, 251. + + Wilkinson, Rev. Mr., 193. + + Wilson, Professor, 29. + + Wilson's Poser, 9, 153. + + Wine and Water, 110, 235. + ---- The Keg of, 110, 235. + + Wotherspoon, G., 244. + + + Yacht race, The, 99, 226. + + Youthful Precocity, 1, 148. + + + Zeno, 139. + + + + +THE END. + + + + + +End of Project Gutenberg's Amusements in Mathematics, by Henry Ernest Dudeney + +*** END OF THIS PROJECT GUTENBERG EBOOK AMUSEMENTS IN MATHEMATICS *** + +***** This file should be named 16713.txt or 16713.zip ***** +This and all associated files of various formats will be found in: + https://www.gutenberg.org/1/6/7/1/16713/ + +Produced by Stephen Schulze, Jonathan Ingram and the Online +Distributed Proofreading Team at https://www.pgdp.net + + +Updated editions will replace the previous one--the old editions +will be renamed. + +Creating the works from public domain print editions means that no +one owns a United States copyright in these works, so the Foundation +(and you!) can copy and distribute it in the United States without +permission and without paying copyright royalties. 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