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 Bractals...if you plug in the numbers and then actually run them through, doing the necessary subtractions and divisions, you'll see that your solution is wrong. For example: for 2 men, 1 monkey, you solution gives C=7.

So the first man gives 1 to the monkey and hides his portion (3), leaving 3. The second man gives one to the monkey and hides his portion (1), leaving 1. Then in the morning, the men gather, give one to the monkey and they are left with none to split among themselves.

Maybe it wasn't clear in the problem that by the end, the men still have some coconuts left...

Posted by yocko on 2005-05-11 21:11:22