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Solution to: The King's Gold

A fair division of the coins is indeed possible. Let the number of rooms be N. This means that per room there are N chests with N coins each. In total there are N×N×N = N3 coins. One chest with N coins goes to the barber. For the six brothers, N3 - N coins remain. We can write this as: N(N2 - l), or N(N - 1)(N + l). This last expression is divisible by 6 in all cases, since a number is divisible by 6 when it is both divisible by 3 and even. This is indeed the case here: whatever N may be, the expression N(N - 1)(N + l) always contains three successive numbers. One of those is always divisible by 3, and at least one of the others is even. This even holds when N=1; in that case all the brothers get nothing, which is also a fair division!

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