Question : The legendary king Midas possessed a huge amount of gold. He hid this treasure carefully: in a building consisting of a number of rooms. In each room there were a number of boxes; this number was equal to the number of rooms in the building. Each box contained a number of golden coins that equaled the number of boxes per room. When the king died, one box was given to the royal barber. The remainder of the coins had to be divided fairly between his six sons. Is a fair division possible in all situations?
A fair division of Midas' coins is indeed possible. Let the number of rooms be N. This means that per room there are N boxes with N coins each. In total there are N?? = N3 coins. One box 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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