Long ago, there was a king who had six sons. The king possessed a huge amount of gold, which he hid carefully in a building consisting of a number of rooms. In each room there were a number of chests; this number of chests was equal to the number of rooms in the building. Each chest contained a number of golden coins that equaled the number of chests per room. When the king died, one chest was given to the royal barber. The remainder of the coins had to be divided fairly between his six sons.

The Question: Is a fair division possible in all situations?

Answer: 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!

## Sunday, July 27, 2008

### The King's Gold

Posted by Covert Bay at 12:59 AM

Labels: Riddles/Puzzles

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