The Canterbury Puzzles Part 21

There are just two hundred and sixty-four different ways in which the s.h.i.+p _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.

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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 ill.u.s.tration.

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 circ.u.mferences 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 n.o.body 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 circ.u.mferences of the two phials that the Doctor required to contain the same quant.i.ty 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 circ.u.mference 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. x.x.x.

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 antic.i.p.ated me in a communication to Sylvester.

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21.--_The Ploughman's Puzzle._

The ill.u.s.tration 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 ill.u.s.tration, 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.

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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 part.i.tion 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._

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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 ill.u.s.tration. 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 a.s.sume 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._

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The ill.u.s.tration 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.

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I exhibited this problem before the Royal Society, at Burlington House, on 17th May 1905, and also at the Royal Inst.i.tution 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 rea.s.sembled 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 ill.u.s.tration showing the puzzle in a rather curious practical form, as it was made in polished mahogany with bra.s.s 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.