"There he is," said Mrs. Solders, when I called to inquire. "That's how he's been for three weeks. He hardly eats anything, and takes no rest, whilst his business is so neglected that I don't know what is going to happen to me and the five children. All day long—and night too—there he is, figuring and figuring, and tearing his hair like a mad thing. It's worrying me into an early grave."

I persuaded Mrs. Solders to explain matters to me. It seems that he had received an order from a customer to make two rectangular zinc cisterns, one with a top and the other without a top. Each was to hold exactly 1,000 cubic feet of water when filled to the brim. The price was to be a certain amount per cistern, including cost of labour. Now Mr. Solders is a thrifty man, so he naturally desired to make the two cisterns of such dimensions that the smallest possible quantity of metal should be required. This was the little question that was so worrying him.



Can my ingenious readers find the dimensions of the most economical cistern with a top, and also the exact proportions of such a cistern without a top, each to hold 1,000 cubic feet of water? By "economical" is meant the method that requires the smallest possible quantity of metal. No margin need be allowed for what ladies would call "turnings." I shall show how I helped Mr. Solders out of his dilemma. He says: "That little wrinkle you gave me would be useful to others in my trade."



98.—The Nelson Column.

During a Nelson celebration I was standing in Trafalgar Square with a friend of puzzling proclivities. He had for some time been gazing at the column in an abstracted way, and seemed quite unconscious of the casual remarks that I addressed to him.

"What are you dreaming about?" I said at last.



"Two feet——" he murmured.

"Somebody's Trilbys?" I inquired.

"Five times round——"

"Two feet, five times round! What on earth are you saying?"

"Wait a minute," he said, beginning to figure something out on the back of an envelope. I now detected that he was in the throes of producing a new problem of some sort, for I well knew his methods of working at these things.

"Here you are!" he suddenly exclaimed. "That's it! A very interesting little puzzle. The height of the shaft of the Nelson column being 200 feet and its circumference 16 feet 8 inches, it is wreathed in a spiral garland which passes round it exactly five times. What is the length of the garland? It looks rather difficult, but is really remarkably easy."

He was right. The puzzle is quite easy if properly attacked. Of course the height and circumference are not correct, but chosen for the purposes of the puzzle. The artist has also intentionally drawn the cylindrical shaft of the column of equal circumference throughout. If it were tapering, the puzzle would be less easy.



99.—The Two Errand Boys.

A country baker sent off his boy with a message to the butcher in the next village, and at the same time the butcher sent his boy to the baker. One ran faster than the other, and they were seen to pass at a spot 720 yards from the baker's shop. Each stopped ten minutes at his destination and then started on the return journey, when it was found that they passed each other at a spot 400 yards from the butcher's. How far apart are the two tradesmen's shops? Of course each boy went at a uniform pace throughout.



100.—On the Ramsgate Sands.

Thirteen youngsters were seen dancing in a ring on the Ramsgate sands. Apparently they were playing "Round the Mulberry Bush." The puzzle is this. How many rings may they form without any child ever taking twice the hand of any other child—right hand or left? That is, no child may ever have a second time the same neighbour.



101.—The Three Motor-Cars.



Pope has told us that all chance is but "direction which thou canst not see," and certainly we all occasionally come across remarkable coincidences—little things against the probability of the occurrence of which the odds are immense—that fill us with bewilderment. One of the three motor men in the illustration has just happened on one of these queer coincidences. He is pointing out to his two friends that the three numbers on their cars contain all the figures 1 to 9 and 0, and, what is more remarkable, that if the numbers on the first and second cars are multiplied together they will make the number on the third car. That is, 78, 345, and 26,910 contain all the ten figures, and 78 multiplied by 345 makes 26,910. Now, the reader will be able to find many similar sets of numbers of two, three, and five figures respectively that have the same peculiarity. But there is one set, and one only, in which the numbers have this additional peculiarity—that the second number is a multiple of the first. In other words, if 345 could be divided by 78 without a remainder, the numbers on the cars would themselves fulfil this extra condition. What are the three numbers that we want? Remember that they must have two, three, and five figures respectively.



102.—A Reversible Magic Square.

Can you construct a square of sixteen different numbers so that it shall be magic (that is, adding up alike in the four rows, four columns, and two diagonals), whether you turn the diagram upside down or not? You must not use a 3, 4, or 5, as these figures will not reverse; but a 6 may become a 9 when reversed, a 9 a 6, a 7 a 2, and a 2 a 7. The 1, 8, and 0 will read the same both ways. Remember that the constant must not be changed by the reversal.



103.—The Tube Railway.



The above diagram is the plan of an underground railway. The fare is uniform for any distance, so long as you do not go twice along any portion of the line during the same journey. Now a certain passenger, with plenty of time on his hands, goes daily from A to F. How many different routes are there from which he may select? For example, he can take the short direct route, A, B, C, D, E, F, in a straight line; or he can go one of the long routes, such as A, B, D, C, B, C, E, D, E, F. It will be noted that he has optional lines between certain stations, and his selections of these lead to variations of the complete route. Many readers will find it a very perplexing little problem, though its conditions are so simple.



104.—The Skipper and the Sea-Serpent.

Mr. Simon Softleigh had spent most of his life between Tooting Bec and Fenchurch Street. His knowledge of the sea was therefore very limited. So, as he was taking a holiday on the south coast, he thought this was a splendid opportunity for picking up a little useful information. He therefore proceeded to "draw" the natives.



"I suppose," said Mr. Softleigh one morning to a jovial, weather-beaten skipper, "you have seen many wonderful sights on the rolling seas?"

"Bless you, sir, yes," said the skipper. "P'raps you've never seen a vanilla iceberg, or a mermaid a-hanging out her things to dry on the equatorial line, or the blue-winged shark what flies through the air in pursuit of his prey, or the sea-sarpint——"

"Have you really seen a sea-serpent? I thought it was uncertain whether they existed."

"Uncertin! You wouldn't say there was anything uncertin about a sea-sarpint if once you'd seen one. The first as I seed was when I was skipper of the Saucy Sally. We was a-coming round Cape Horn with a cargo of shrimps from the Pacific Islands when I looks over the port side and sees a tremenjus monster like a snake, with its 'ead out of the water and its eyes flashing fire, a-bearing down on our ship. So I shouts to the bo'sun to let down the boat, while I runs below and fetches my sword—the same what I used when I killed King Chokee, the cannibal chief as eat our cabin-boy—and we pulls straight into the track of that there sea-sarpint. Well, to make a long story short, when we come alongside o' the beast I just let drive at him with that sword o' mine, and before you could say 'Tom Bowling' I cut him into three pieces, all of exactually the same length, and afterwards we hauled 'em aboard the Saucy Sally. What did I do with 'em? Well, I sold 'em to a feller in Rio Janeiro. And what do you suppose he done with 'em? He used 'em to make tyres for his motor-car—takes a lot to puncture a sea-sarpint's skin."

"What was the length of the creature?" asked Simon.

"Well, each piece was equal in length to three-quarters the length of a piece added to three-quarters of a cable. There's a little puzzle for you to work out, young gentleman. How many cables long must that there sea-sarpint 'ave been?"

Now, it is not at all to the discredit of Mr. Simon Softleigh that he never succeeded in working out the correct answer to that little puzzle, for it may confidently be said that out of a thousand readers who attempt the solution not one will get it exactly right.



105.—The Dorcas Society.

At the close of four and a half months' hard work, the ladies of a certain Dorcas Society were so delighted with the completion of a beautiful silk patchwork quilt for the dear curate that everybody kissed everybody else, except, of course, the bashful young man himself, who only kissed his sisters, whom he had called for, to escort home. There were just a gross of osculations altogether. How much longer would the ladies have taken over their needlework task if the sisters of the curate referred to had played lawn tennis instead of attending the meetings? Of course we must assume that the ladies attended regularly, and I am sure that they all worked equally well. A mutual kiss here counts as two osculations.



106.—The Adventurous Snail.



A simple version of the puzzle of the climbing snail is familiar to everybody. We were all taught it in the nursery, and it was apparently intended to inculcate the simple moral that we should never slip if we can help it. This is the popular story. A snail crawls up a pole 12 feet high, ascending 3 feet every day and slipping back 2 feet every night. How long does it take to get to the top? Of course, we are expected to say the answer is twelve days, because the creature makes an actual advance of 1 foot in every twenty-four hours. But the modern infant in arms is not taken in in this way. He says, correctly enough, that at the end of the ninth day the snail is 3 feet from the top, and therefore reaches the summit of its ambition on the tenth day, for it would cease to slip when it had got to the top.

Let us, however, consider the original story. Once upon a time two philosophers were walking in their garden, when one of them espied a highly respectable member of the Helix Aspersa family, a pioneer in mountaineering, in the act of making the perilous ascent of a wall 20 feet high. Judging by the trail, the gentleman calculated that the snail ascended 3 feet each day, sleeping and slipping back 2 feet every night.

"Pray tell me," said the philosopher to his friend, who was in the same line of business, "how long will it take Sir Snail to climb to the top of the wall and descend the other side? The top of the wall, as you know, has a sharp edge, so that when he gets there he will instantly begin to descend, putting precisely the same exertion into his daily climbing down as he did in his climbing up, and sleeping and slipping at night as before."

This is the true version of the puzzle, and my readers will perhaps be interested in working out the exact number of days. Of course, in a puzzle of this kind the day is always supposed to be equally divided into twelve hours' daytime and twelve hours' night.



107.—The Four Princes.

The dominions of a certain Eastern monarch formed a perfectly square tract of country. It happened that the king one day discovered that his four sons were not only plotting against each other, but were in secret rebellion against himself. After consulting with his advisers he decided not to exile the princes, but to confine them to the four corners of the country, where each should be given a triangular territory of equal area, beyond the boundaries of which they would pass at the cost of their lives. Now, the royal surveyor found himself confronted by great natural difficulties, owing to the wild character of the country. The result was that while each was given exactly the same area, the four triangular districts were all of different shapes, somewhat in the manner shown in the illustration. The puzzle is to give the three measurements for each of the four districts in the smallest possible numbers—all whole furlongs. In other words, it is required to find (in the smallest possible numbers) four rational right-angled triangles of equal area.



108.—Plato and the Nines.

Both in ancient and in modern times the number nine has been considered to possess peculiarly mystic qualities. We know, for instance, that there were nine Muses, nine rivers of Hades, and that Vulcan was nine days falling down from heaven. Then it has been confidently held that nine tailors make a man; while we know that there are nine planets, nine days' wonders, and that a cat has nine lives—and sometimes nine tails.

Most people are acquainted with some of the curious properties of the number nine in ordinary arithmetic. For example, write down a number containing as many figures as you like, add these figures together, and deduct the sum from the first number. Now, the sum of the figures in this new number will always be a multiple of nine.

There was once a worthy man at Athens who was not only a cranky arithmetician, but also a mystic. He was deeply convinced of the magic properties of the number nine, and was perpetually strolling out to the groves of Academia to bother poor old Plato with his nonsensical ideas about what he called his "lucky number." But Plato devised a way of getting rid of him. When the seer one day proposed to inflict on him a lengthy disquisition on his favourite topic, the philosopher cut him short with the remark, "Look here, old chappie" (that is the nearest translation of the original Greek term of familiarity): "when you can bring me the solution of this little mystery of the three nines I shall be happy to listen to your treatise, and, in fact, record it on my phonograph for the benefit of posterity."



Plato then showed, in the manner depicted in our illustration, that three nines may be arranged so as to represent the number eleven, by putting them into the form of a fraction. The puzzle he then propounded was so to arrange the three nines that they will represent the number twenty.

It is recorded of the old crank that, after working hard at the problem for nine years, he one day, at nine o'clock on the morning of the ninth day of the ninth month, fell down nine steps, knocked out nine teeth, and expired in nine minutes. It will be remembered that nine was his lucky number. It was evidently also Plato's.

In solving the above little puzzle, only the most elementary arithmetical signs are necessary. Though the answer is absurdly simple when you see it, many readers will have no little difficulty in discovering it. Take your pencil and see if you can arrange the three nines to represent twenty.



109.—Noughts and Crosses.

Every child knows how to play this game. You make a square of nine cells, and each of the two players, playing alternately, puts his mark (a nought or a cross, as the case may be) in a cell with the object of getting three in a line. Whichever player first gets three in a line wins with the exulting cry:—

"Tit, tat, toe, My last go; Three jolly butcher boys All in a row."

It is a very ancient game. But if the two players have a perfect knowledge of it, one of three things must always happen. (1) The first player should win; (2) the first player should lose; or (3) the game should always be drawn. Which is correct?



110.—Ovid's Game.

Having examined "Noughts and Crosses," we will now consider an extension of the game that is distinctly mentioned in the works of Ovid. It is, in fact, the parent of "Nine Men's Morris," referred to by Shakespeare in A Midsummer Night's Dream (Act ii., Scene 2). Each player has three counters, which they play alternately on to the nine points shown in the diagram, with the object of getting three in a line and so winning. But after the six counters are played they then proceed to move (always to an adjacent unoccupied point) with the same object. In the example below White played first, and Black has just played on point 7. It is now White's move, and he will undoubtedly play from 8 to 9, and then, whatever Black may do, he will continue with 5 to 6, and so win. That is the simple game. Now, if both players are equally perfect at the game what should happen? Should the first player always win? Or should the second player win? Or should every game be a draw? One only of these things should always occur. Which is it?



111.—The Farmer's Oxen.

A child may propose a problem that a sage cannot answer. A farmer propounded the following question: "That ten-acre meadow of mine will feed twelve bullocks for sixteen weeks or eighteen bullocks for eight weeks. How many bullocks could I feed on a forty-acre field for six weeks, the grass growing regularly all the time?"

It will be seen that the sting lies in the tail. That steady growth of the grass is such a reasonable point to be considered, and yet to some readers it will cause considerable perplexity. The grass is, of course, assumed to be of equal length and uniform thickness in every case when the cattle begin to eat. The difficulty is not so great as it appears, if you properly attack the question.



112.—The Great Grangemoor Mystery.

Mr. Stanton Mowbray was a very wealthy man, a reputed millionaire, residing in that beautiful old mansion that has figured so much in English history, Grangemoor Park. He was a bachelor, spent most of the year at home, and lived quietly enough.

According to the evidence given, on the day preceding the night of the crime he received by the second post a single letter, the contents of which evidently gave him a shock. At ten o'clock at night he dismissed the servants, saying that he had some important business matters to look into, and would be sitting up late. He would require no attendance. It was supposed that after all had gone to bed he had admitted some person to the house, for one of the servants was positive that she had heard loud conversation at a very late hour.

Next morning, at a quarter to seven o'clock, one of the man-servants, on entering the room, found Mr. Mowbray lying on the floor, shot through the head, and quite dead. Now we come to the curious circumstance of the case. It was clear that after the bullet had passed out of the dead man's head it had struck the tall clock in the room, right in the very centre of the face, and actually welded together the three hands; for the clock had a seconds hand that revolved round the same dial as the hour and minute hands. But although the three hands had become welded together exactly as they stood in relation to each other at the moment of impact, yet they were free to revolve round the swivel in one piece, and had been stupidly spun round several times by the servants before Mr. Wiley Slyman was called upon the spot. But they would not move separately.



Now, inquiries by the police in the neighbourhood led to the arrest in London of a stranger who was identified by several persons as having been seen in the district the day before the murder, but it was ascertained beyond doubt at what time on the fateful morning he went away by train. If the crime took place after his departure, his innocence was established. For this and other reasons it was of the first importance to fix the exact time of the pistol shot, the sound of which nobody in the house had heard. The clock face in the illustration shows exactly how the hands were found. Mr. Slyman was asked to give the police the benefit of his sagacity and experience, and directly he was shown the clock he smiled and said:

"The matter is supremely simple. You will notice that the three hands appear to be at equal distances from one another. The hour hand, for example, is exactly twenty minutes removed from the minute hand—that is, the third of the circumference of the dial. You attach a lot of importance to the fact that the servants have been revolving the welded hands, but their act is of no consequence whatever; for although they were welded instantaneously, as they are free on the swivel, they would swing round of themselves into equilibrium. Give me a few moments, and I can tell you beyond any doubt the exact time that the pistol was fired."

Mr. Wiley Slyman took from his pocket a notebook, and began to figure it out. In a few minutes he handed the police inspector a slip of paper, on which he had written the precise moment of the crime. The stranger was proved to be an old enemy of Mr. Mowbray's, was convicted on other evidence that was discovered; but before he paid the penalty for his wicked act, he admitted that Mr. Slyman's statement of the time was perfectly correct.

Can you also give the exact time?



113.—Cutting a Wood Block.

An economical carpenter had a block of wood measuring eight inches long by four inches wide by three and three-quarter inches deep. How many pieces, each measuring two and a half inches by one inch and a half by one inch and a quarter, could he cut out of it? It is all a question of how you cut them out. Most people would have more waste material left over than is necessary. How many pieces could you get out of the block?



114.—The Tramps and the Biscuits.

Four merry tramps bought, borrowed, found, or in some other manner obtained possession of a box of biscuits, which they agreed to divide equally amongst themselves at breakfast next morning. In the night, while the others were fast asleep under the greenwood tree, one man approached the box, devoured exactly a quarter of the number of biscuits, except the odd one left over, which he threw as a bribe to their dog. Later in the night a second man awoke and hit on the same idea, taking a quarter of what remained and giving the odd biscuit to the dog. The third and fourth men did precisely the same in turn, taking a quarter of what they found and giving the odd biscuit to the dog. In the morning they divided what remained equally amongst them, and again gave the odd biscuit to the animal. Every man noticed the reduction in the contents of the box, but, believing himself to be alone responsible, made no comments. What is the smallest possible number of biscuits that there could have been in the box when they first acquired it?



SOLUTIONS

THE CANTERBURY PUZZLES



1.—The Reve's Puzzle.

The 8 cheeses can be removed in 33 moves, 10 cheeses in 49 moves, and 21 cheeses in 321 moves. I will give my general method of solution in the cases of 3, 4, and 5 stools.

Write out the following table to any required length:—

Stools. Number of Cheeses.

3 1 2 3 4 5 6 7 Natural Numbers. 4 1 3 6 10 15 21 28 Triangular Numbers. 5 1 4 10 20 35 56 84 Triangular Pyramids.

Number of Moves.

3 1 3 7 15 31 63 127 4 1 5 17 49 129 321 769 5 1 7 31 111 351 1023 2815

The first row contains the natural numbers. The second row is found by adding the natural numbers together from the beginning. The numbers in the third row are obtained by adding together the numbers in the second row from the beginning. The fourth row contains the successive powers of 2, less 1. The next series is found by doubling in turn each number of that series and adding the number that stands above the place where you write the result. The last row is obtained in the same way. This table will at once give solutions for any number of cheeses with three stools, for triangular numbers with four stools, and for pyramidal numbers with five stools. In these cases there is always only one method of solution—that is, of piling the cheeses.

In the case of three stools, the first and fourth rows tell us that 4 cheeses may be removed in 15 moves, 5 in 31, 7 in 127. The second and fifth rows show that, with four stools, 10 may be removed in 49, and 21 in 321 moves. Also, with five stools, we find from the third and sixth rows that 20 cheeses require 111 moves, and 35 cheeses 351 moves. But we also learn from the table the necessary method of piling. Thus, with four stools and 10 cheeses, the previous column shows that we must make piles of 6 and 3, which will take 17 and 7 moves respectively—that is, we first pile the six smallest cheeses in 17 moves on one stool; then we pile the next 3 cheeses on another stool in 7 moves; then remove the largest cheese in 1 move; then replace the 3 in 7 moves; and finally replace the 6 in 17: making in all the necessary 49 moves. Similarly we are told that with five stools 35 cheeses must form piles of 20, 10, and 4, which will respectively take 111, 49, and 15 moves.

If the number of cheeses in the case of four stools is not triangular, and in the case of five stools pyramidal, then there will be more than one way of making the piles, and subsidiary tables will be required. This is the case with the Reve's 8 cheeses. But I will leave the reader to work out for himself the extension of the problem.



2.—The Pardoner's Puzzle.

The diagram on page 165 will show how the Pardoner started from the large black town and visited all the other towns once, and once only, in fifteen straight pilgrimages.

See No. 320, "The Rook's Tour," in A. in M.



3.—The Miller's Puzzle.

The way to arrange the sacks of flour is as follows:—2, 78, 156, 39, 4. Here each pair when multiplied by its single neighbour makes the number in the middle, and only five of the sacks need be moved. There are just three other ways in which they might have been arranged (4, 39, 156, 78, 2; or 3, 58, 174, 29, 6; or 6, 29, 174, 58, 3), but they all require the moving of seven sacks.



4.—The Knight's Puzzle.

The Knight declared that as many as 575 squares could be marked off on his shield, with a rose at every corner. How this result is achieved may be realized by reference to the accompanying diagram:—Join A, B, C, and D, and there are 66 squares of this size to be formed; the size A, E, F, G gives 48; A, H, I, J, 32; B, K, L, M, 19; B, N, O, P, 10; B, Q, R, S, 4; E, T, F, C, 57; I, U, V, P, 33; H, W, X, J, 15; K, Y, Z, M, 3; E, a, b, D, 82; H, d, M, D, 56; H, e, f, G, 42; K, g, f, C, 32; N, h, z, F, 24; K, h, m, b, 14; K, O, S, D, 16; K, n, p, G, 10; K, q, r, J, 6; Q, t, p, C, 4; Q, u, r, i, 2. The total number is thus 575. These groups have been treated as if each of them represented a different sized square. This is correct, with the one exception that the squares of the form B, N, O, P are exactly the same size as those of the form K, h, m, b.



5.—The Wife of Bath's Riddles.

The good lady explained that a bung that is made fast in a barrel is like another bung that is falling out of a barrel because one of them is in secure and the other is also insecure. The little relationship poser is readily understood when we are told that the parental command came from the father (who was also in the room) and not from the mother.



6.—The Host's Puzzle.

The puzzle propounded by the jovial host of the "Tabard" Inn of Southwark had proved more popular than any other of the whole collection. "I see, my merry masters," he cried, "that I have sorely twisted thy brains by my little piece of craft. Yet it is but a simple matter for me to put a true pint of fine old ale in each of these two measures, albeit one is of five pints and the other of three pints, without using any other measure whatever."

The host of the "Tabard" Inn thereupon proceeded to explain to the pilgrims how this apparently impossible task could be done. He first filled the 5-pint and 3-pint measures, and then, turning the tap, allowed the barrel to run to waste—a proceeding against which the company protested; but the wily man showed that he was aware that the cask did not contain much more than eight pints of ale. The contents, however, do not affect the solution of the puzzle. He then closed the tap and emptied the 3-pint into the barrel; filled the 3-pint from the 5-pint; emptied the 3-pint into the barrel; transferred the two pints from the 5-pint to the 3-pint; filled the 5-pint from the barrel, leaving one pint now in the barrel; filled 3-pint from 5-pint; allowed the company to drink the contents of the 3-pint; filled the 3-pint from the 5-pint, leaving one pint now in the 5-pint; drank the contents of the 3-pint; and finally drew off one pint from the barrel into the 3-pint. He had thus obtained the required one pint of ale in each measure, to the great astonishment of the admiring crowd of pilgrims.



7.—Clerk of Oxenford's Puzzle.

The illustration shows how the square is to be cut into four pieces, and how these pieces are to be put together again to make a magic square. It will be found that the four columns, four rows, and two long diagonals now add up to 34 in every case.



8.—The Tapiser's Puzzle.

The piece of tapestry had to be cut along the lines into three pieces so as to fit together and form a perfect square, with the pattern properly matched. It was also stipulated in effect that one of the three pieces must be as small as possible. The illustration shows how to make the cuts and how to put the pieces together, while one of the pieces contains only twelve of the little squares.



9.—The Carpenter's Puzzle.

The carpenter said that he made a box whose internal dimensions were exactly the same as the original block of wood—that is, 3 feet by 1 foot by 1 foot. He then placed the carved pillar in this box and filled up all the vacant space with a fine, dry sand, which he carefully shook down until he could get no more into the box. Then he removed the pillar, taking great care not to lose any of the sand, which, on being shaken down alone in the box, filled a space equal to one cubic foot. This was, therefore, the quantity of wood that had been cut away.



10.—The Puzzle of the Squire's Yeoman.

The illustration will show how three of the arrows were removed each to a neighbouring square on the signboard of the "Chequers" Inn, so that still no arrow was in line with another. The black dots indicate the squares on which the three arrows originally stood.



11.—The Nun's Puzzle.



As there are eighteen cards bearing the letters "CANTERBURY PILGRIMS," write the numbers 1 to 18 in a circle, as shown in the diagram. Then write the first letter C against 1, and each successive letter against the second number that happens to be vacant. This has been done as far as the second R. If the reader completes the process by placing Y against 2, P against 6, I against 10, and so on, he will get the letters all placed in the following order:—CYASNPTREIRMBLUIRG, which is the required arrangement for the cards, C being at the top of the pack and G at the bottom.



12.—The Merchant's Puzzle.

This puzzle amounts to finding the smallest possible number that has exactly sixty-four divisors, counting 1 and the number itself as divisors. The least number is 7,560. The pilgrims might, therefore, have ridden in single file, two and two, three and three, four and four, and so on, in exactly sixty-four different ways, the last manner being in a single row of 7,560.

The Merchant was careful to say that they were going over a common, and not to mention its size, for it certainly would not be possible along an ordinary road!

To find how many different numbers will divide a given number, N, let N = a^p b^q c^r ..., where a, b, c ... are prime numbers. Then the number of divisors will be (p + 1) (q + 1) (r + 1) ..., which includes as divisors 1 and N itself. Thus in the case of my puzzle—

7,560 = 2^3 x 3^3 x 5 x 7 Powers = 3 3 1 1 Therefore 4 x 4 x 2 x 2 = 64 divisors.

To find the smallest number that has a given number of divisors we must proceed by trial. But it is important sometimes to note whether or not the condition is that there shall be a given number of divisors and no more. For example, the smallest number that has seven divisors and no more is 64, while 24 has eight divisors, and might equally fulfil the conditions. The stipulation as to "no more" was not necessary in the case of my puzzle, for no smaller number has more than sixty-four divisors.



13.—The Man of Law's Puzzle.

The fewest possible moves for getting the prisoners into their dungeons in the required numerical order are twenty-six. The men move in the following order:—1, 2, 3, 1, 2, 6, 5, 3, 1, 2, 6, 5, 3, 1, 2, 4, 8, 7, 1, 2, 4, 8, 7, 4, 5, 6. As there are never more than one vacant dungeon to be moved into, there can be no ambiguity in the notation.



The diagram may be simplified by my "buttons and string" method, fully explained in A. in M., p. 230. It then takes one of the simple forms of A or B, and the solution is much easier. In A we use counters; in B we can employ rooks on a corner of a chessboard. In both cases we have to get the order



in the fewest possible moves.

See also solution to No. 94.



14.—The Weaver's Puzzle.

The illustration shows clearly how the Weaver cut his square of beautiful cloth into four pieces of exactly the same size and shape, so that each piece contained an embroidered lion and castle unmutilated in any way.



15.—The Cook's Puzzle.

There were four portions of warden pie and four portions of venison pasty to be distributed among eight out of eleven guests. But five out of the eleven will only eat the pie, four will only eat the pasty, and two are willing to eat of either. Any possible combination must fall into one of the following groups. (i.) Where the warden pie is distributed entirely among the five first mentioned; (ii.) where only one of the accommodating pair is given pie; (iii.) where the other of the pair is given pie; (iv.) where both of the pair are given pie. The numbers of combinations are: (i.) = 75, (ii.) = 50, (iii.) = 10, (iv.) = 10—making in all 145 ways of selecting the eight participants. A great many people will give the answer as 185, by overlooking the fact that in forty cases in class (iii.) precisely the same eight guests would be sharing the meal as in class (ii.), though the accommodating pair would be eating differently of the two dishes. This is the point that upset the calculations of the company.



16.—The Sompnour's Puzzle.

The number that the Sompnour confided to the Wife of Bath was twenty-nine, and she was told to begin her count at the Doctor of Physic, who will be seen in the illustration standing the second on her right. The first count of twenty-nine falls on the Shipman, who steps out of the ring. The second count falls on the Doctor, who next steps out. The remaining three counts fall respectively on the Cook, the Sompnour, and the Miller. The ladies would, therefore, have been left in possession had it not been for the unfortunate error of the good Wife. Any multiple of 2,520 added to 29 would also have served the same purpose, beginning the count at the Doctor.



17.—The Monk's Puzzle.

The Monk might have placed dogs in the kennels in two thousand nine hundred and twenty-six different ways, so that there should be ten dogs on every side. The number of dogs might vary from twenty to forty, and as long as the Monk kept his animals within these limits the thing was always possible.

The general solution to this puzzle is difficult. I find that for n dogs on every side of the square, the number of different ways is (n^4 + 10n^3 + 38n^2 + 62n + 33) / 48, where n is odd, and ((n^4 + 10n^3 + 38n^2 + 68n) / 48) + 1, where n is even, if we count only those arrangements that are fundamentally different. But if we count all reversals and reflections as different, as the Monk himself did, then n dogs (odd or even) may be placed in ((n^4 + 6n^3 + 14n^2 + 15n) / 6) + 1 ways. In order that there may be n dogs on every side, the number must not be less than 2n nor greater than 4n, but it may be any number within these limits.

An extension of the principle involved in this puzzle is given in No. 42, "The Riddle of the Pilgrims." See also "The Eight Villas" and "A Dormitory Puzzle" in A. in M.



18.—The Shipman's Puzzle.

There are just two hundred and sixty-four different ways in which the ship Magdalen might have made her ten annual voyages without ever going over the same course twice in a year. Every year she must necessarily end her tenth voyage at the island from which she first set out.



19.—The Puzzle of the Prioress.

The Abbot of Chertsey was quite correct. The curiously-shaped cross may be cut into four pieces that will fit together and form a perfect square. How this is done is shown in the illustration.

See also p. 31 in A. in M.



20.—The Puzzle of the Doctor of Physic.

Here we have indeed a knotty problem. Our text-books tell us that all spheres are similar, and that similar solids are as the cubes of corresponding lengths. Therefore, as the circumferences of the two phials were one foot and two feet respectively and the cubes of one and two added together make nine, what we have to find is two other numbers whose cubes added together make nine. These numbers clearly must be fractional. Now, this little question has really engaged the attention of learned men for two hundred and fifty years; but although Peter de Fermat showed in the seventeenth century how an answer may be found in two fractions with a denominator of no fewer than twenty-one figures, not only are all the published answers, by his method, that I have seen inaccurate, but nobody has ever published the much smaller result that I now print. The cubes of (415280564497 / 348671682660) and (676702467503 / 348671682660) added together make exactly nine, and therefore these fractions of a foot are the measurements of the circumferences of the two phials that the Doctor required to contain the same quantity of liquid as those produced. An eminent actuary and another correspondent have taken the trouble to cube out these numbers, and they both find my result quite correct.

If the phials were one foot and three feet in circumference respectively, then an answer would be that the cubes of (63284705 / 21446828) and (28340511 / 21446828) added together make exactly 28. See also No. 61, "The Silver Cubes."

Given a known case for the expression of a number as the sum or difference of two cubes, we can, by formula, derive from it an infinite number of other cases alternately positive and negative. Thus Fermat, starting from the known case 1^{3} + 2^{3} = 9 (which we will call a fundamental case), first obtained a negative solution in bigger figures, and from this his positive solution in bigger figures still. But there is an infinite number of fundamentals, and I found by trial a negative fundamental solution in smaller figures than his derived negative solution, from which I obtained the result shown above. That is the simple explanation.

We can say of any number up to 100 whether it is possible or not to express it as the sum of two cubes, except 66. Students should read the Introduction to Lucas's Theorie des Nombres, p. xxx.

Some years ago I published a solution for the case of

6 = (17/21)^3 + (37/21)^3,

of which Legendre gave at some length a "proof" of impossibility; but I have since found that Lucas anticipated me in a communication to Sylvester.



21.—The Ploughman's Puzzle.

The illustration shows how the sixteen trees might have been planted so as to form as many as fifteen straight rows with four trees in every row. This is in excess of what was for a long time believed to be the maximum number of rows possible; and though with our present knowledge I cannot rigorously demonstrate that fifteen rows cannot be beaten, I have a strong "pious opinion" that it is the highest number of rows obtainable.



22.—The Franklin's Puzzle.

The answer to this puzzle is shown in the illustration, where the numbers on the sixteen bottles all add up to 30 in the ten straight directions. The trick consists in the fact that, although the six bottles (3, 5, 6, 9, 10, and 15) in which the flowers have been placed are not removed, yet the sixteen need not occupy exactly the same position on the table as before. The square is, in fact, formed one step further to the left.



23.—The Squire's Puzzle.

The portrait may be drawn in a single line because it contains only two points at which an odd number of lines meet, but it is absolutely necessary to begin at one of these points and end at the other. One point is near the outer extremity of the King's left eye; the other is below it on the left cheek.



24.—The Friar's Puzzle.

The five hundred silver pennies might have been placed in the four bags, in accordance with the stated conditions, in exactly 894,348 different ways. If there had been a thousand coins there would be 7,049,112 ways. It is a difficult problem in the partition of numbers. I have a single formula for the solution of any number of coins in the case of four bags, but it was extremely hard to construct, and the best method is to find the twelve separate formulas for the different congruences to the modulus 12.



25.—The Parson's Puzzle.



A very little examination of the original drawing will have shown the reader that, as he will have at first read the conditions, the puzzle is quite impossible of solution. We have therefore to look for some loophole in the actual conditions as they were worded. If the Parson could get round the source of the river, he could then cross every bridge once and once only on his way to church, as shown in the annexed illustration. That this was not prohibited we shall soon find. Though the plan showed all the bridges in his parish, it only showed "part of" the parish itself. It is not stated that the river did not take its rise in the parish, and since it leads to the only possible solution, we must assume that it did. The answer would be, therefore, as shown. It should be noted that we are clearly prevented from considering the possibility of getting round the mouth of the river, because we are told it "joined the sea some hundred miles to the south," while no parish ever extended a hundred miles!



26.—The Haberdasher's Puzzle.



The illustration will show how the triangular piece of cloth may be cut into four pieces that will fit together and form a perfect square. Bisect AB in D and BC in E; produce the line AE to F making EF equal to EB; bisect AF in G and describe the arc AHF; produce EB to H, and EH is the length of the side of the required square; from E with distance EH, describe the arc HJ, and make JK equal to BE; now, from the points D and K drop perpendiculars on EJ at L and M. If you have done this accurately, you will now have the required directions for the cuts.



I exhibited this problem before the Royal Society, at Burlington House, on 17th May 1905, and also at the Royal Institution in the following month, in the more general form:—"A New Problem on Superposition: a demonstration that an equilateral triangle can be cut into four pieces that may be reassembled to form a square, with some examples of a general method for transforming all rectilinear triangles into squares by dissection." It was also issued as a challenge to the readers of the Daily Mail (see issues of 1st and 8th February 1905), but though many hundreds of attempts were sent in there was not a single solver. Credit, however, is due to Mr. C. W. M'Elroy, who alone sent me the correct solution when I first published the problem in the Weekly Dispatch in 1902.

I add an illustration showing the puzzle in a rather curious practical form, as it was made in polished mahogany with brass hinges for use by certain audiences. It will be seen that the four pieces form a sort of chain, and that when they are closed up in one direction they form the triangle, and when closed in the other direction they form the square.



27.—The Dyer's Puzzle.

The correct answer is 18,816 different ways. The general formula for six fleurs-de-lys for all squares greater than 2^{2} is simply this: Six times the square of the number of combinations of n things, taken three at a time, where n represents the number of fleurs-de-lys in the side of the square. Of course where n is even the remainders in rows and columns will be even, and where n is odd the remainders will be odd.

For further solution, see No. 358 in A. in M.



28.—The Great Dispute between the Friar and the Sompnour.

In this little problem we attempted to show how, by sophistical reasoning, it may apparently be proved that the diagonal of a square is of precisely the same length as two of the sides. The puzzle was to discover the fallacy, because it is a very obvious fallacy if we admit that the shortest distance between two points is a straight line. But where does the error come in?

Well, it is perfectly true that so long as our zigzag path is formed of "steps" parallel to the sides of the square that path must be of the same length as the two sides. It does not matter if you have to use the most powerful microscope obtainable; the rule is always true if the path is made up of steps in that way. But the error lies in the assumption that such a zigzag path can ever become a straight line. You may go on increasing the number of steps infinitely—that is, there is no limit whatever theoretically to the number of steps that can be made—but you can never reach a straight line by such a method. In fact it is just as much a "jump" to a straight line if you have a billion steps as it is at the very outset to pass from the two sides to the diagonal. It would be just as absurd to say we might go on dropping marbles into a basket until they become sovereigns as to say we can increase the number of our steps until they become a straight line. There is the whole thing in a nutshell.



29.—Chaucer's Puzzle.

The surface of water or other liquid is always spherical, and the greater any sphere is the less is its convexity. Hence the top diameter of any vessel at the summit of a mountain will form the base of the segment of a greater sphere than it would at the bottom. This sphere, being greater, must (from what has been already said) be less convex; or, in other words, the spherical surface of the water must be less above the brim of the vessel, and consequently it will hold less at the top of a mountain than at the bottom. The reader is therefore free to select any mountain he likes in Italy—or elsewhere!



30.—The Puzzle of the Canon's Yeoman.

The number of different ways is 63,504. The general formula for such arrangements, when the number of letters in the sentence is 2n + 1, and it is a palindrome without diagonal readings, is [4(2^n - 1)]^2.

I think it will be well to give here a formula for the general solution of each of the four most common forms of the diamond-letter puzzle. By the word "line" I mean the complete diagonal. Thus in A, B, C, and D, the lines respectively contain 5, 5, 7, and 9 letters. A has a non-palindrome line (the word being BOY), and the general solution for such cases, where the line contains 2n + 1 letters, is 4(2^n - 1). Where the line is a single palindrome, with its middle letter in the centre, as in B, the general formula is [4(2^n - 1)]^{2}. This is the form of the Rat-catcher's Puzzle, and therefore the expression that I have given above. In cases C and D we have double palindromes, but these two represent very different types. In C, where the line contains 4^n-1 letters, the general expression is 4^(2^{2n}-2). But D is by far the most difficult case of all.

I had better here state that in the diamonds under consideration (i.) no diagonal readings are allowed—these have to be dealt with specially in cases where they are possible and admitted; (ii.) readings may start anywhere; (iii.) readings may go backwards and forwards, using letters more than once in a single reading, but not the same letter twice in immediate succession. This last condition will be understood if the reader glances at C, where it is impossible to go forwards and backwards in a reading without repeating the first O touched—a proceeding which I have said is not allowed. In the case D it is very different, and this is what accounts for its greater difficulty. The formula for D is this:



where the number of letters in the line is 4n+1. In the example given there are therefore 400 readings for n = 2.

See also Nos. 256, 257, and 258 in A. in M.



31.—The Manciple's Puzzle.

The simple Ploughman, who was so ridiculed for his opinion, was perfectly correct: the Miller should receive seven pieces of money, and the Weaver only one. As all three ate equal shares of the bread, it should be evident that each ate 8/3 of a loaf. Therefore, as the Miller provided 15/3 and ate 8/3, he contributed 7/3 to the Manciple's meal; whereas the Weaver provided 9/3, ate 8/3, and contributed only 1/3. Therefore, since they contributed to the Manciple in the proportion of 7 to 1, they must divide the eight pieces of money in the same proportion.



PUZZLING TIMES AT SOLVAMHALL CASTLE

SIR HUGH EXPLAINS HIS PROBLEMS

The friends of Sir Hugh de Fortibus were so perplexed over many of his strange puzzles that at a gathering of his kinsmen and retainers he undertook to explain his posers.



"Of a truth," said he, "some of the riddles that I have put forth would greatly tax the wit of the unlettered knave to rede; yet will I try to show the manner thereof in such way that all may have understanding. For many there be who cannot of themselves do all these things, but will yet study them to their gain when they be given the answers, and will take pleasure therein."



32.—The Game of Bandy-Ball.

Sir Hugh explained, in answer to this puzzle, that as the nine holes were 300, 250, 200, 325, 275, 350, 225, 375, and 400 yards apart, if a man could always strike the ball in a perfectly straight line and send it at will a distance of either 125 yards or 100 yards, he might go round the whole course in 26 strokes. This is clearly correct, for if we call the 125 stroke the "drive" and the 100 stroke the "approach," he could play as follows:—The first hole could be reached in 3 approaches, the second in 2 drives, the third in 2 approaches, the fourth in 2 approaches and 1 drive, the fifth in 3 drives and 1 backward approach, the sixth in 2 drives and 1 approach, the seventh in 1 drive and 1 approach, the eighth in 3 drives, and the ninth hole in 4 approaches. There are thus 26 strokes in all, and the feat cannot be performed in fewer.



33.—Tilting at the Ring.



"By my halidame!" exclaimed Sir Hugh, "if some of yon varlets had been put in chains, which for their sins they do truly deserve, then would they well know, mayhap, that the length of any chain having like rings is equal to the inner width of a ring multiplied by the number of rings and added to twice the thickness of the iron whereof it is made. It may be shown that the inner width of the rings used in the tilting was one inch and two-thirds thereof, and the number of rings Stephen Malet did win was three, and those that fell to Henry de Gournay would be nine."

The knight was quite correct, for 1-2/3 in. x 3 + 1 in. = 6 in., and 1-2/3 in. x 9 + 1 in. = 16 in. Thus De Gournay beat Malet by six rings. The drawing showing the rings may assist the reader in verifying the answer and help him to see why the inner width of a link multiplied by the number of links and added to twice the thickness of the iron gives the exact length. It will be noticed that every link put on the chain loses a length equal to twice the thickness of the iron.



34.—The Noble Demoiselle.



"Some here have asked me," continued Sir Hugh, "how they may find the cell in the Dungeon of the Death's-head wherein the noble maiden was cast. Beshrew me! but 'tis easy withal when you do but know how to do it. In attempting to pass through every door once, and never more, you must take heed that every cell hath two doors or four, which be even numbers, except two cells, which have but three. Now, certes, you cannot go in and out of any place, passing through all the doors once and no more, if the number of doors be an odd number. But as there be but two such odd cells, yet may we, by beginning at the one and ending at the other, so make our journey in many ways with success. I pray you, albeit, to mark that only one of these odd cells lieth on the outside of the dungeon, so we must perforce start therefrom. Marry, then, my masters, the noble demoiselle must needs have been wasting in the other."

The drawing will make this quite clear to the reader. The two "odd cells" are indicated by the stars, and one of the many routes that will solve the puzzle is shown by the dotted line. It is perfectly certain that you must start at the lower star and end at the upper one; therefore the cell with the star situated over the left eye must be the one sought.



35.—The Archery Butt.



"It hath been said that the proof of a pudding is ever in the eating thereof, and by the teeth of Saint George I know no better way of showing how this placing of the figures may be done than by the doing of it. Therefore have I in suchwise written the numbers that they do add up to twenty and three in all the twelve lines of three that are upon the butt."

I think it well here to supplement the solution of De Fortibus with a few remarks of my own. The nineteen numbers may be so arranged that the lines will add up to any number we may choose to select from 22 to 38 inclusive, excepting 30. In some cases there are several different solutions, but in the case of 23 there are only two. I give one of these. To obtain the second solution exchange respectively 7, 10, 5, 8, 9, in the illustration, with 13, 4, 17, 2, 15. Also exchange 18 with 12, and the other numbers may remain unmoved. In every instance there must be an even number in the central place, and any such number from 2 to 18 may occur. Every solution has its complementary. Thus, if for every number in the accompanying drawing we substitute the difference between it and 20, we get the solution in the case of 37. Similarly, from the arrangement in the original drawing, we may at once obtain a solution for the case of 38.



36.—The Donjon Keep Window.

In this case Sir Hugh had greatly perplexed his chief builder by demanding that he should make a window measuring one foot on every side and divided by bars into eight lights, having all their sides equal. The illustration will show how this was to be done. It will be seen that if each side of the window measures one foot, then each of the eight triangular lights is six inches on every side.

"Of a truth, master builder," said De Fortibus slyly to the architect, "I did not tell thee that the window must be square, as it is most certain it never could be."



37.—The Crescent and the Cross.

"By the toes of St. Moden," exclaimed Sir Hugh de Fortibus when this puzzle was brought up, "my poor wit hath never shaped a more cunning artifice or any more bewitching to look upon. It came to me as in a vision, and ofttimes have I marvelled at the thing, seeing its exceeding difficulty. My masters and kinsmen, it is done in this wise."



The worthy knight then pointed out that the crescent was of a particular and somewhat irregular form—the two distances a to b and c to d being straight lines, and the arcs ac and bd being precisely similar. He showed that if the cuts be made as in Figure 1, the four pieces will fit together and form a perfect square, as shown in Figure 2, if we there only regard the three curved lines. By now making the straight cuts also shown in Figure 2, we get the ten pieces that fit together, as in Figure 3, and form a perfectly symmetrical Greek cross. The proportions of the crescent and the cross in the original illustration were correct, and the solution can be demonstrated to be absolutely exact and not merely approximate.

I have a solution in considerably fewer pieces, but it is far more difficult to understand than the above method, in which the problem is simplified by introducing the intermediate square.



38.—The Amulet.

The puzzle was to place your pencil on the A at the top of the amulet and count in how many different ways you could trace out the word "Abracadabra" downwards, always passing from a letter to an adjoining one.

A B B R R R A A A A C C C C C A A A A A A D D D D D D D A A A A A A A A B B B B B B B B B R R R R R R R R R R A A A A A A A A A A A

"Now, mark ye, fine fellows," said Sir Hugh to some who had besought him to explain, "that at the very first start there be two ways open: whichever B ye select, there will be two several ways of proceeding (twice times two are four); whichever R ye select, there be two ways of going on (twice times four are eight); and so on until the end. Each letter in order from A downwards may so be reached in 2, 4, 8, 16, 32, etc., ways. Therefore, as there be ten lines or steps in all from A to the bottom, all ye need do is to multiply ten 2's together, and truly the result, 1024, is the answer thou dost seek."



39.—The Snail on the Flagstaff.

Though there was no need to take down and measure the staff, it is undoubtedly necessary to find its height before the answer can be given. It was well known among the friends and retainers of Sir Hugh de Fortibus that he was exactly six feet in height. It will be seen in the original picture that Sir Hugh's height is just twice the length of his shadow. Therefore we all know that the flagstaff will, at the same place and time of day, be also just twice as long as its shadow. The shadow of the staff is the same length as Sir Hugh's height; therefore this shadow is six feet long, and the flagstaff must be twelve feet high. Now, the snail, by climbing up three feet in the daytime and slipping back two feet by night, really advances one foot in a day of twenty-four hours. At the end of nine days it is three feet from the top, so that it reaches its journey's end on the tenth day.

The reader will doubtless here exclaim, "This is all very well; but how were we to know the height of Sir Hugh? It was never stated how tall he was!" No, it was not stated in so many words, but it was none the less clearly indicated to the reader who is sharp in these matters. In the original illustration to the donjon keep window Sir Hugh is shown standing against a wall, the window in which is stated to be one foot square on the inside. Therefore, as his height will be found by measurement to be just six times the inside height of the window, he evidently stands just six feet in his boots!



40.—Lady Isabel's Casket.

The last puzzle was undoubtedly a hard nut, but perhaps difficulty does not make a good puzzle any the less interesting when we are shown the solution. The accompanying diagram indicates exactly how the top of Lady Isabel de Fitzarnulph's casket was inlaid with square pieces of rare wood (no two squares alike) and the strip of gold 10 inches by a quarter of an inch. This is the only possible solution, and it is a singular fact (though I cannot here show the subtle method of working) that the number, sizes, and order of those squares are determined by the given dimensions of the strip of gold, and the casket can have no other dimensions than 20 inches square. The number in a square indicates the length in inches of the side of that square, so the accuracy of the answer can be checked almost at a glance.

Sir Hugh de Fortibus made some general concluding remarks on the occasion that are not altogether uninteresting to-day.



"Friends and retainers," he said, "if the strange offspring of my poor wit about which we have held pleasant counsel to-night hath mayhap had some small interest for ye, let these matters serve to call to mind the lesson that our fleeting life is rounded and beset with enigmas. Whence we came and whither we go be riddles, and albeit such as these we may never bring within our understanding, yet there be many others with which we and they that do come after us will ever strive for the answer. Whether success do attend or do not attend our labour, it is well that we make the attempt; for 'tis truly good and honourable to train the mind, and the wit, and the fancy of man, for out of such doth issue all manner of good in ways unforeseen for them that do come after us."



THE MERRY MONKS OF RIDDLEWELL



41.—The Riddle of the Fish-pond.

Number the fish baskets in the illustration from 1 to 12 in the direction that Brother Jonathan is seen to be going. Starting from 1, proceed as follows, where "1 to 4" means, take the fish from basket No. 1 and transfer it to basket 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.



42.—The Riddle of the Pilgrims.

If it were not for the Abbot's conditions that the number of guests in any room may not exceed three, and that every room must be occupied, it would have been possible to accommodate either 24, 27, 30, 33, 36, 39, or 42 pilgrims. But to accommodate 24 pilgrims so that there shall be twice as many sleeping on the upper floor as on the lower floor, and eleven persons on each side of the building, it will be found necessary to leave some of the rooms empty. If, on the other hand, we try to put up 33, 36, 39 or 42 pilgrims, we shall find that in every case we are obliged to place more than three persons in some of the rooms. Thus we know that the number of pilgrims originally announced (whom, it will be remembered, it was possible to accommodate under the conditions of the Abbot) must have been 27, and that, since three more than this number were actually provided with beds, the total number of pilgrims was 30. The accompanying diagram shows how they might be arranged, and if in each instance we regard the upper floor as placed above the lower one, it will be seen that there are eleven persons on each side of the building, and twice as many above as below.



43.—The Riddle of the Tiled Hearth.

The correct answer is shown in the illustration on page 196. No tile is in line (either horizontally, vertically, or diagonally) with another tile of the same design, and only three plain tiles are used. If after placing the four lions you fall into the error of placing four other tiles of another pattern, instead of only three, you will be left with four places that must be occupied by plain tiles. The secret consists in placing four of one kind and only three of each of the others.



44.—The Riddle of the Sack of Wine.

The question was: Did Brother Benjamin take more wine from the bottle than water from the jug? Or did he take more water from the jug than wine from the bottle? He did neither. The same quantity of wine was transferred from the bottle as water was taken from the jug. Let us assume that the glass would hold a quarter of a pint. There was a pint of wine in the bottle and a pint of water in the jug. After the first manipulation the bottle contains three-quarters of a pint of wine, and the jug one pint of water mixed with a quarter of a pint of wine. Now, the second transaction consists in taking away a fifth of the contents of the jug—that is, one-fifth of a pint of water mixed with one-fifth of a quarter of a pint of wine. We thus leave behind in the jug four-fifths of a quarter of a pint of wine—that is, one-fifth of a pint—while we transfer from the jug to the bottle an equal quantity (one-fifth of a pint) of water.



45.—The Riddle of the Cellarer.

There were 100 pints of wine in the cask, and on thirty occasions John the Cellarer had stolen a pint and replaced it with a pint of water. After the first theft the wine left in the cask would be 99 pints; after the second theft the wine in the cask would be 9801/100 pints (the square of 99 divided by 100); after the third theft there would remain 970299/10000 (the cube of 99 divided by the square of 100); after the fourth theft there would remain the fourth power of 99 divided by the cube of 100; and after the thirtieth theft there would remain in the cask the thirtieth power of 99 divided by the twenty-ninth power of 100. This by the ordinary method of calculation gives us a number composed of 59 figures to be divided by a number composed of 58 figures! But by the use of logarithms it may be quickly ascertained that the required quantity is very nearly 73-97/100 pints of wine left in the cask. Consequently the cellarer stole nearly 26.03 pints. The monks doubtless omitted the answer for the reason that they had no tables of logarithms, and did not care to face the task of making that long and tedious calculation in order to get the quantity "to a nicety," as the wily cellarer had stipulated.

By a simplified process of calculation, I have ascertained that the exact quantity of wine stolen would be

26.0299626611719577269984907683285057747323737647323555652999

pints. A man who would involve the monastery in a fraction of fifty-eight decimals deserved severe punishment.



46.—The Riddle of the Crusaders.

The correct answer is that there would have been 602,176 Crusaders, who could form themselves into a square 776 by 776; and after the stranger joined their ranks, they could form 113 squares of 5,329 men—that is, 73 by 73. Or 113 x 73^2 - 1 = 776^2. This is a particular case of the so-called "Pellian Equation," respecting which see A. in M., p. 164.



47.—The Riddle of St. Edmondsbury.

The reader is aware that there are prime numbers and composite whole numbers. Now, 1,111,111 cannot be a prime number, because if it were the only possible answers would be those proposed by Brother Benjamin and rejected by Father Peter. Also it cannot have more than two factors, or the answer would be indeterminate. As a matter of fact, 1,111,111 equals 239 x 4649 (both primes), and since each cat killed more mice than there were cats, the answer must be 239 cats. See also the Introduction, p. 18.

Treated generally, this problem consists in finding the factors, if any, of numbers of the form (10^n - 1)/9.

Lucas, in his L'Arithmetique Amusante, gives a number of curious tables which he obtained from an arithmetical treatise, called the Talkhys, by Ibn Albanna, an Arabian mathematician and astronomer of the first half of the thirteenth century. In the Paris National Library are several manuscripts dealing with the Talkhys, and a commentary by Alkalacadi, who died in 1486. Among the tables given by Lucas is one giving all the factors of numbers of the above form up to n = 18. It seems almost inconceivable that Arabians of that date could find the factors where n = 17, as given in my Introduction. But I read Lucas as stating that they are given in Talkhys, though an eminent mathematician reads him differently, and suggests to me that they were discovered by Lucas himself. This can, of course, be settled by an examination of Talkhys, but this has not been possible during the war.

The difficulty lies wholly with those cases where n is a prime number. If n = 2, we get the prime 11. The factors when n = 3, 5, 11, and 13 are respectively (3 . 37), (41 . 271), (21,649 . 513,239), and (53 . 79 . 265371653). I have given in these pages the factors where n = 7 and 17. The factors when n= 19, 23, and 37 are unknown, if there are any.[B] When n = 29, the factors are (3,191 . 16,763 . 43,037. 62,003 . 77,843,839,397); when n = 31, one factor is 2,791; and when n = 41, two factors are (83 . 1,231).

[B] Mr. Oscar Hoppe, of New York, informs me that, after reading my statement in the Introduction, he was led to investigate the case of n = 19, and after long and tedious work he succeeded in proving the number to be a prime. He submitted his proof to the London Mathematical Society, and a specially appointed committee of that body accepted the proof as final and conclusive. He refers me to the Proceedings of the Society for 14th February 1918.

As for the even values of n, the following curious series of factors will doubtless interest the reader. The numbers in brackets are primes.

n = 2 = (11)

n = 6 = (11) x 111 x 91

n = 10 = (11) x 11,111 x (9,091)

n = 14 = (11) x 1,111,111 x (909,091)

n = 18 = (11) x 111,111,111 x 90,909,091

Or we may put the factors this way:—

n = 2 = (11)

n = 6 = 111 x 1,001

n = 10 = 11,111 x 100,001

n = 14 = 1,111,111 x 10,000,001

n = 18 = 111,111,111 x 1,000,000,001

In the above two tables n is of the form 4m + 2. When n is of the form 4m the factors may be written down as follows:—

n= 4 = (11) x (101)

n = 8 = (11) x (101) x 10,001

n = 12 = (11) x (101) x 100,010,001

n = 16 = (11) x (101) x 1,000,100,010,001.

When n = 2, we have the prime number 11; when n = 3, the factors are 3 . 37; when n = 6, they are 11 . 3 . 37 . 7. 13; when n = 9, they are 3^2 . 37 . 333,667. Therefore we know that factors of n = 18 are 11. 3^2 . 37 . 7 . 13 . 333,667, while the remaining factor is composite and can be split into 19 . 52579. This will show how the working may be simplified when n is not prime.



48.—The Riddle of the Frogs' Ring.

The fewest possible moves in which this puzzle can be solved are 118. I will give the complete solution. The black figures on white discs move in the directions of the hands of a clock, and the white figures on black discs the other way. The following are the numbers in the order in which they move. Whether you have to make a simple move or a leaping move will be clear from the position, as you never can have an alternative. The moves enclosed in brackets are to be played five times over: 6, 7, 8, 6, 5, 4, 7, 8, 9, 10, 6, 5, 4, 3, 2, 7, 8, 9, 10, 11 (6, 5, 4, 3, 2, 1), 6, 5, 4, 3, 2, 12, (7, 8, 9, 10, 11, 12), 7, 8, 9, 10, 11, 1, 6, 5, 4, 3, 2, 12, 7, 8, 9, 10, 11, 6, 5, 4, 3, 2, 8, 9, 10, 11, 4, 3, 2, 10, 11, 2. We thus have made 118 moves within the conditions, the black frogs have changed places with the white ones, and 1 and 12 are side by side in the positions stipulated.

The general solution in the case of this puzzle is 3n^{2} + 2n - 2 moves, where the number of frogs of each colour is n. The law governing the sequence of moves is easily discovered by an examination of the simpler cases, where n = 2, 3, and 4.

If, instead of 11 and 12 changing places, the 6 and 7 must interchange, the expression is n^{2} + 4n + 2 moves. If we give n the value 6, as in the example of the Frogs' Ring, the number of moves would be 62.

For a general solution of the case where frogs of one colour reverse their order, leaving the blank space in the same position, and each frog is allowed to be moved in either direction (leaping, of course, over his own colour), see "The Grasshopper Puzzle" in A. in M., p. 193.



THE STRANGE ESCAPE OF THE KING'S JESTER

Although the king's jester promised that he would "thereafter make the manner thereof plain to all," there is no record of his having ever done so. I will therefore submit to the reader my own views as to the probable solutions to the mysteries involved.



49.—The Mysterious Rope.

When the jester "divided his rope in half," it does not follow that he cut it into two parts, each half the original length of the rope. No doubt he simply untwisted the strands, and so divided it into two ropes, each of the original length, but one-half the thickness. He would thus be able to tie the two together and make a rope nearly twice the original length, with which it is quite conceivable that he made good his escape from the dungeon.



50.—The Underground Maze.

How did the jester find his way out of the maze in the dark? He had simply to grope his way to a wall and then keep on walking without once removing his left hand (or right hand) from the wall. Starting from A, the dotted line will make the route clear when he goes to the left. If the reader tries the route to the right in the same way he will be equally successful; in fact, the two routes unite and cover every part of the walls of the maze except those two detached parts on the left-hand side—one piece like a U, and the other like a distorted E. This rule will apply to the majority of mazes and puzzle gardens; but if the centre were enclosed by an isolated wall in the form of a split ring, the jester would simply have gone round and round this ring.

See the article, "Mazes, and How to Thread Them," in A. in M.



51.—The Secret Lock.

This puzzle entailed the finding of an English word of three letters, each letter being found on a different dial. Now, there is no English word composed of consonants alone, and the only vowel appearing anywhere on the dials is Y. No English word begins with Y and has the two other letters consonants, and all the words of three letters ending in Y (with two consonants) either begin with an S or have H, L, or R as their second letter. But these four consonants do not appear. Therefore Y must occur in the middle, and the only word that I can find is "PYX," and there can be little doubt that this was the word. At any rate, it solves our puzzle.



52.—Crossing the Moat.

No doubt some of my readers will smile at the statement that a man in a boat on smooth water can pull himself across with the tiller rope! But it is a fact. If the jester had fastened the end of his rope to the stern of the boat and then, while standing in the bows, had given a series of violent jerks, the boat would have been propelled forward. This has often been put to a practical test, and it is said that a speed of two or three miles an hour may be attained. See W. W. Rouse Ball's Mathematical Recreations.



53.—The Royal Gardens.



This puzzle must have struck many readers as being absolutely impossible. The jester said: "I had, of a truth, entered every one of the sixteen gardens once, and never more than once." If we follow the route shown in the accompanying diagram, we find that there is no difficulty in once entering all the gardens but one before reaching the last garden containing the exit B. The difficulty is to get into the garden with a star, because if we leave the B garden we are compelled to enter it a second time before escaping, and no garden may be entered twice. The trick consists in the fact that you may enter that starred garden without necessarily leaving the other. If, when the jester got to the gateway where the dotted line makes a sharp bend, his intention had been to hide in the starred garden, but after he had put one foot through the doorway, upon the star, he discovered it was a false alarm and withdrew, he could truly say: "I entered the starred garden, because I put my foot and part of my body in it; and I did not enter the other garden twice, because, after once going in I never left it until I made my exit at B." This is the only answer possible, and it was doubtless that which the jester intended.

See "The Languishing Maiden," in A. in M.



54.—Bridging the Ditch.



The solution to this puzzle is best explained by the illustration. If he had placed his eight planks, in the manner shown, across the angle of the ditch, he would have been able to cross without much trouble. The king's jester might thus have well overcome all his difficulties and got safely away, as he has told us that he succeeded in doing.



THE SQUIRE'S CHRISTMAS PUZZLE PARTY

HOW THE VARIOUS TRICKS WERE DONE

The record of one of Squire Davidge's annual "Puzzle Parties," made by the old gentleman's young lady relative, who had often spent a merry Christmas at Stoke Courcy Hall, does not contain the solutions of the mysteries. So I will give my own answers to the puzzles and try to make them as clear as possible to those who may be more or less novices in such matters.



55.—The Three Teacups.



Miss Charity Lockyer clearly must have had a trick up her sleeve, and I think it highly probable that it was conceived on the following lines. She proposed that ten lumps of sugar should be placed in three teacups, so that there should be an odd number of lumps in every cup. The illustration perhaps shows Miss Charity's answer, and the figures on the cups indicate the number of lumps that have been separately placed in them. By placing the cup that holds one lump inside the one that holds two lumps, it can be correctly stated that every cup contains an odd number of lumps. One cup holds seven lumps, another holds one lump, while the third cup holds three lumps. It is evident that if a cup contains another cup it also contains the contents of that second cup.

There are in all fifteen different solutions to this puzzle. Here they are:—

1 0 9 1 4 5 9 0 1 3 0 7 7 0 3 7 2 1 1 2 7 5 2 3 5 4 1 5 0 5 3 4 3 3 6 1 3 2 5 1 6 3 1 8 1

The first two numbers in a triplet represent respectively the number of lumps to be placed in the inner and outer of the two cups that are placed one inside the other. It will be noted that the outer cup of the pair may itself be empty.



56.—The Eleven Pennies.

It is rather evident that the trick in this puzzle was as follows:—From the eleven coins take five; then add four (to those already taken away) and you leave nine—in the second heap of those removed!



57.—The Christmas Geese.

Farmer Rouse sent exactly 101 geese to market. Jabez first sold Mr. Jasper Tyler half of the flock and half a goose over (that is, 50-1/2 + 1/2, or 51 geese, leaving 50); he then sold Farmer Avent a third of what remained and a third of a goose over (that is, 16-2/3 + 1/3, or 17 geese, leaving 33); he then sold Widow Foster a quarter of what remained and three-quarters of a goose over (that is, 8-1/4 + 3/4 or 9 geese, leaving 24); he next sold Ned Collier a fifth of what he had left and gave him a fifth of a goose "for the missus" (that is, 4-4/5 + 1/5 or 5 geese, leaving 19). He then took these 19 back to his master.



58.—The Chalked Numbers.

This little jest on the part of Major Trenchard is another trick puzzle, and the face of the roguish boy on the extreme right, with the figure 9 on his back, showed clearly that he was in the secret, whatever that secret might be. I have no doubt (bearing in mind the Major's hint as to the numbers being "properly regarded") that his answer was that depicted in the illustration, where boy No. 9 stands on his head and so converts his number into 6. This makes the total 36—an even number—and by making boys 3 and 4 change places with 7 and 8, we get 1278 and 5346, the figures of which, in each case, add up to 18. There are just three other ways in which the boys may be grouped: 1368—2457, 1467—2358, and 2367—1458.



59.—Tasting the Plum Puddings.

The diagram will show how this puzzle is to be solved. It is the only way within the conditions laid down. Starting at the pudding with holly at the top left-hand corner, we strike out all the puddings in twenty-one straight strokes, taste the steaming hot pudding at the end of the tenth stroke, and end at the second sprig of holly.

Here we have an example of a chess rook's path that is not re-entrant, but between two squares that are at the greatest possible distance from one another. For if it were desired to move, under the condition of visiting every square once and once only, from one corner square to the other corner square on the same diagonal, the feat is impossible.

There are a good many different routes for passing from one sprig of holly to the other in the smallest possible number of moves—twenty-one—but I have not counted them. I have recorded fourteen of these, and possibly there are more. Any one of these would serve our purpose, except for the condition that the tenth stroke shall end at the steaming hot pudding. This was introduced to stop a plurality of solutions—called by the maker of chess problems "cooks." I am not aware of more than one solution to this puzzle; but as I may not have recorded all the tours, I cannot make a positive statement on the point at the time of writing.



60.—Under the Mistletoe Bough.

Everybody was found to have kissed everybody else once under the mistletoe, with the following additions and exceptions: No male kissed a male; no man kissed a married woman except his own wife; all the bachelors and boys kissed all the maidens and girls twice; the widower did not kiss anybody, and the widows did not kiss each other. Every kiss was returned, and the double performance was to count as one kiss. In making a list of the company, we can leave out the widower altogether, because he took no part in the osculatory exercise.

7 Married couples 14 3 Widows 3 12 Bachelors and Boys 12 10 Maidens and Girls 10 Total 39 Persons

Now, if every one of these 39 persons kissed everybody else once, the number of kisses would be 741; and if the 12 bachelors and boys each kissed the 10 maidens and girls once again, we must add 120, making a total of 861 kisses. But as no married man kissed a married woman other than his own wife, we must deduct 42 kisses; as no male kissed another male, we must deduct 171 kisses; and as no widow kissed another widow, we must deduct 3 kisses. We have, therefore, to deduct 42+171+3=216 kisses from the above total of 861, and the result, 645, represents exactly the number of kisses that were actually given under the mistletoe bough.



61.—The Silver Cubes.

There is no limit to the number of different dimensions that will give two cubes whose sum shall be exactly seventeen cubic inches. Here is the answer in the smallest possible numbers. One of the silver cubes must measure 2-23278/40831 inches along each edge, and the other must measure 11663/40831 inch. If the reader likes to undertake the task of cubing each number (that is, multiply each number twice by itself), he will find that when added together the contents exactly equal seventeen cubic inches. See also No. 20, "The Puzzle of the Doctor of Physic."



THE ADVENTURES OF THE PUZZLE CLUB



62.—The Ambiguous Photograph.

One by one the members of the Club succeeded in discovering the key to the mystery of the Ambiguous Photograph, except Churton, who was at length persuaded to "give it up." Herbert Baynes then pointed out to him that the coat that Lord Marksford was carrying over his arm was a lady's coat, because the buttons are on the left side, whereas a man's coat always has the buttons on the right-hand side. Lord Marksford would not be likely to walk about the streets of Paris with a lady's coat over his arm unless he was accompanying the owner. He was therefore walking with the lady.

As they were talking a waiter brought a telegram to Baynes.

"Here you are," he said, after reading the message. "A wire from Dovey: 'Don't bother about photo. Find lady was the gentleman's sister, passing through Paris.' That settles it. You might notice that the lady was lightly clad, and therefore the coat might well be hers. But it is clear that the rain was only a sudden shower, and no doubt they were close to their destination, and she did not think it worth while to put the coat on."



63.—The Cornish Cliff Mystery.

Melville's explanation of the Cornish Cliff Mystery was very simple when he gave it. Yet it was an ingenious trick that the two criminals adopted, and it would have completely succeeded had not our friends from the Puzzle Club accidentally appeared on the scene. This is what happened: When Lamson and Marsh reached the stile, Marsh alone walked to the top of the cliff, with Lamson's larger boots in his hands. Arrived at the edge of the cliff, he changed the boots and walked backwards to the stile, carrying his own boots.

This little manoeuvre accounts for the smaller footprints showing a deeper impression at the heel, and the larger prints a deeper impression at the toe; for a man will walk more heavily on his heels when going forward, but will make a deeper impression with the toes in walking backwards. It will also account for the fact that the large footprints were sometimes impressed over the smaller ones, but never the reverse; also for the circumstance that the larger footprints showed a shorter stride, for a man will necessarily take a smaller stride when walking backwards. The pocket-book was intentionally dropped, to lead the police to discover the footprints, and so be put on the wrong scent.



64.—The Runaway Motor-Car.

Russell found that there are just twelve five-figure numbers that have the peculiarity that the first two figures multiplied by the last three—all the figures being different, and there being no 0—will produce a number with exactly the same five figures, in a different order. But only one of these twelve begins with a 1—namely, 14926. Now, if we multiply 14 by 926, the result is 12964, which contains the same five figures. The number of the motor-car was therefore 14926.

Here are the other eleven numbers:—24651, 42678, 51246, 57834, 75231, 78624, 87435, 72936, 65281, 65983, and 86251.

Compare with the problems in "Digital Puzzles," section of A. in M., and with Nos. 93 and 101 in these pages.



65.—The Mystery of Ravensdene Park.

The diagrams show that there are two different ways in which the routes of the various persons involved in the Ravensdene Mystery may be traced, without any path ever crossing another. It depends whether the butler, E, went to the north or the south of the gamekeeper's cottage, and the gamekeeper, A, went to the south or the north of the hall. But it will be found that the only persons who could have approached Mr. Cyril Hastings without crossing a path were the butler, E, and the man, C. It was, however, a fact that the butler retired to bed five minutes before midnight, whereas Mr. Hastings did not leave his friend's house until midnight. Therefore the criminal must have been the man who entered the park at C.



66.—The Buried Treasure.

The field must have contained between 179 and 180 acres—to be more exact, 179.37254 acres. Had the measurements been 3, 2, and 4 furlongs respectively from successive corners, then the field would have been 209.70537 acres in area.

One method of solving this problem is as follows. Find the area of triangle APB in terms of x, the side of the square. Double the result=xy. Divide by x and then square, and we have the value of y^{2} in terms of x. Similarly find value of z^{2} in terms of x; then solve the equation y^{2}z^{2}=3^{2}, which will come out in the form x^{4}-20x^{2}=-37. Therefore x^{2}=10(sqrt{63})=17.937254 square furlongs, very nearly, and as there are ten acres in one square furlong, this equals 179.37254 acres. If we take the negative root of the equation, we get the area of the field as 20.62746 acres, in which case the treasure would have been buried outside the field, as in Diagram 2. But this solution is excluded by the condition that the treasure was buried in the field. The words were, "The document ... states clearly that the field is square, and that the treasure is buried in it."



THE PROFESSOR'S PUZZLES



67.—The Coinage Puzzle.

The point of this puzzle turns on the fact that if the magic square were to be composed of whole numbers adding up 15 in all ways, the two must be placed in one of the corners. Otherwise fractions must be used, and these are supplied in the puzzle by the employment of sixpences and half-crowns. I give the arrangement requiring the fewest possible current English coins—fifteen. It will be seen that the amount in each corner is a fractional one, the sum required in the total being a whole number of shillings.