This is a variation on a classic number theory problem, originally submitted by Ravi Raja here.

There are n men and k monkeys shipwrecked on an island. The men have collected a pile of coconuts which they plan to divide equally among themselves the next morning. Not trusting the rest of the group, one of the men wakes up during the night and divides the coconuts into n equal parts leaving k left over, which he gives to the monkeys. He then hides his portion of the pile. During the night, each of the other men does exactly the same thing by dividing the pile he finds into n equal parts leaving k coconuts for the monkeys and hiding his portion. In the morning, the men gather and split the remaining pile of coconuts into n parts and k is left over for the monkeys. What is the minimum number of coconuts, C, the men could have collected for their original pile?

Sorry guys...your solution of n^(n+1) -k(n-1) isn't quite there.

The assumption that there are at least two men on the island is a correct one to make, but nowhere in the problem does it imply that the number of monkeys must be less than the number of men.

This is where that answer breaks down. For example, if there are 2 men and 10 monkeys, your proposed solution returns -2 coconuts...

Posted by yocko on 2005-05-09 02:10:29