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Tuesday, July 29, 2008

Coconut Chaos

Five sailors survive a shipwreck and swim to a tiny island where there is nothing but a coconut tree and a monkey. The sailors gather all the coconuts and put them in a big pile under the tree. Exhausted, they agree to go to wait until the next morning to divide up the coconuts.

At one o'clock in the morning, the first sailor wakes up. He realizes that he can't trust the others, and decides to take his share now. He divides the coconuts into five equal piles, but there is one coconut left over. He gives that coconut to the monkey, hides his coconuts (one of the five piles), and puts the rest of the coconuts (the other four piles) back under the tree.

At two o'clock, the second sailor wakes up. Not realizing that the first sailor has already taken his share, he too divides the coconuts up into five piles, leaving one coconut over which he gives to the monkey. He then hides his share (one of the five piles), and puts the remainder (the other four piles) back under the tree.

At three, four, and five o'clock in the morning, the third, fourth, and fifth sailors each wake up and carry out the same actions.

In the morning, all the sailors wake up, and try to look innocent. No one makes a remark about the diminished pile of coconuts, and no one decides to be honest and admit that they've already taken their share. Instead, they divide the pile up into five piles, for the sixth time, and find that there is yet again one coconut left over, which they give to the monkey.

The Question: What is the smallest amount of coconuts that there could have been in the original pile?


Answer: Every sailor leaves 4/5(n-1) coconuts of a pile of n coconuts. This results in an awful formula for the complete process (because every time one coconut must be taken away to make the pile divisible by 5):

4/5(4/5(4/5(4/5(4/5(4/5(p-1)-1)-1)-1)-1)-1), where p is the number of coconuts in the original pile, must be a whole number.

The trick is to make the number of coconuts in the pile divisible by 5, by adding 4 coconuts. This is possible because you can take away those 4 coconuts again after taking away one fifth part of the pile: normally, 4/5(n-1) coconuts are left of a pile of n coconuts; now 4/5(n+4)=4/5(n-1)+4 coconuts are left of a pile of n+4 coconuts. And because of this, the number of coconuts in the pile stays divisible by 5 during the whole process. So we are now looking for a p for which the following holds:

4/5×4/5×4/5×4/5×4/5×4/5×(p+4)=(46/56)×(p+4), where p is the number of coconuts in the original pile, must be a whole number.

The smallest (p+4) for which the above holds, is 56. So there were p=56-4=15621 coconuts in the original pile.

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