Amusements in Mathematics Part 31

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.

[Ill.u.s.tration]

+-----+-----+-----+-----+ +-----+-----+-----+-----+ | | | | | | | | | | | 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 a.n.a.lysis 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 ill.u.s.tration 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 pa.s.s from his cell into the next one when his time came round.

404.--THE SIBERIAN DUNGEONS.

[Ill.u.s.tration]

+-----+-----+-----+-----+ | | | | | | 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.

[Ill.u.s.tration]

The ill.u.s.tration 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:-- [Ill.u.s.tration: 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 a.s.sociated pairs all have a difference of 4. The second square, by dividing, has a constant 9, and all the a.s.sociated 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 subst.i.tute, in its allotted place, its square, then the constant will be 11,180. Readers can write out for themselves the second degree square.

[Ill.u.s.tration: 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 :-- [Ill.u.s.tration: 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.

[Ill.u.s.tration]

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 ill.u.s.tration," 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 a.s.sumed (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.

[Ill.u.s.tration: 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.

[Ill.u.s.tration]

+-----+-----+-----+ | | | | | 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 n.o.body can yet say whether a perfect solution is possible or impossible.

413.--A CHESSBOARD FALLACY.

[Ill.u.s.tration]

The explanation of this little fallacy is as follows. The error lies in a.s.suming 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 j.a.pan 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.

[Ill.u.s.tration]

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 a.s.sume 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 9 in., which is a long way from the correct answer. You will find, on examining any three consecutive volumes on your shelves, that the first page of Vol. I. and the last page of Vol. III. are actually the pages that are nearest to Vol. II., so that the worm would only have to penetrate four covers (together, in.) and the leaves in the second volume (3 in.), or a distance of 3 inches, in order to tunnel from the first page to the last.

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 pa.s.sing 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."

[Ill.u.s.tration]

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.

[Ill.u.s.tration]

The mystery is made clear by the ill.u.s.tration. 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 [Ill.u.s.tration]

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 quant.i.ty: 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 4d., and the cost 11s. 10d. 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 4d., thus saving a farthing on each of the previous thirty, and reducing the cost of the last one from 1s. to 4d.

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. s. d. 31 plates painted 5 X 9 @ 4 = 0 11 7 30 " 7 X 8 @ 4 = 0 11 10 21 " 1 X 2 @ 7 = 0 12 3 21 " 3 X 0 @ 7 = 0 12 3 12 " 1 X 3 @ 9 = 0 9 3 12 " 2 X 4 @ 9 = 0 9 3 12 " 9 X 4 @ 9 = 0 9 3 8 " 4 X 0 @ 10 = 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 ---------- 7 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 _t.i.t-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 1,347, 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 compet.i.tor has succeeded in maintaining the reputation of the _t.i.t-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 compet.i.tors 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.

n.o.body 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 ill.u.s.tration. 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: "a.s.suming 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 compet.i.tor 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 s.e.x.

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.

[Ill.u.s.tration: 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 b.u.t.tons 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.