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.) 136.--THE SULTAN'S ARMY. The smallest primes of the form 4n + 1 are 5, 13, 17, 29, and 37, and the smallest of the form 4n - 1 are 3, 7, 11, 19, and 23. Now, primes of the first form can always be expressed as the sum of two squares, and in only one way. Thus, 5 = 4 + 1; 13 = 9 + 4; 17 = 16 + 1; 29 = 25 + 4; 37 = 36 + 1. But primes of the second form can never be expressed as the sum of two squares in any way whatever.