Solution

(related to Problem: 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.

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References

Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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