Part 25

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 × 3^3 × 5 × 7 Powers = 3 3 1 1 Therefore 4 × 4 × 2 × 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

{1 2 3} {4 5 6} {7 8 } ]

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.