Let us take Table I., and say n = 5 and m = 2; also in Table II. take n = 13 and m = 0.  Then we at once get these two answers:—­

A =   6           A =   6
B =   5           B =  13
C =  40           C =   1
D = 140           D = 168
E =   9           E =  12
—–­               —–­
200 doubloons     200 doubloons.

These will be found to work correctly.  All the rest of the 704 answers, where Alfonso always holds six doubloons, may be obtained in this way from the two tables by substituting the different numbers for the letters m and n.

Put in another way, for every holding of Alfonso the number of answers is the sum of two arithmetical progressions, the common difference in one case being 1 and in the other -4.  Thus in the case where Alfonso holds 6 doubloons one progression is 33 + 34 + 35 + 36 + ... + 43 + 44, and the other 42 + 38 + 34 + 30 + ... + 6 + 2.  The sum of the first series is 462, and of the second 242—­results which again agree with the figures already given.  The problem may be said to consist in finding the first and last terms of these progressions.  I should remark that where Alfonso holds 9, 10, or 11 there is only one progression, of the second form.

134.—­THE BANKER’S PUZZLE.

In order that a number of sixpences may not be divisible into a number of equal piles, it is necessary that the number should be a prime.  If the banker can bring about a prime number, he will win; and I will show how he can always do this, whatever the customer may put in the box, and that therefore the banker will win to a certainty.  The banker must first deposit forty sixpences, and then, no matter how many the customer may add, he will desire the latter to transfer from the counter the square of the number next below what the customer put in.  Thus, banker puts 40, customer, we will say, adds 6, then transfers from the counter 25 (the square of 5), which leaves 71 in all, a prime number.  Try again.  Banker puts 40, customer adds 12, then transfers 121 (the square of 11), as desired, which leaves 173, a prime number.  The key to the puzzle is the curious fact that any number up to 39, if added to its square and the sum increased by 41, makes a prime number.  This was first discovered by Euler, the great mathematician.  It has been suggested that the banker might desire the customer to transfer sufficient to raise the contents of the box to a given number; but this would not only make the thing an absurdity, but breaks the rule that neither knows what the other puts in.

135.—­THE STONEMASON’S PROBLEM.

The puzzle amounts to this.  Find the smallest square number that may be expressed as the sum of more than three consecutive cubes, the cube 1 being barred.  As more than three heaps were to be supplied, this condition shuts out the otherwise smallest answer, 23 cubed + 24 cubed + 25 cubed = 204 squared.  But it admits the answer, 25 cubed + 26 cubed + 27 cubed + 28 cubed + 29 cubed = 315 squared.  The correct answer, however, requires more heaps, but a smaller aggregate number of blocks.  Here it is:  14 cubed + 15 cubed + ... up to 25 cubed inclusive, or twelve heaps in all, which, added together, make 97,344 blocks of stone that may be laid out to form a square 312 x 312.  I will just remark that one key to the solution lies in what are called triangular numbers. (See pp. 13, 25, and 166.)