## 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 = N^{3} coins. One chest with N coins goes to the barber.
For the six brothers, N^{3} - N coins remain.
We can write this as: N(N^{2} - 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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