Solution to: Coconut ChaosEvery 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.
|
