Solution to: Coconut Chaos

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):

r = ¹⁄₅(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, and r is the number of coconuts each sailor gets at the final division (which 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. In the last step, ¹⁄₅(n+4) = ¹⁄₅(n-1)+1 coconuts are left of a pile of n+4 coconuts. In this way, the number of coconuts in the pile stays divisible by 5 during the entire process. So, we are now looking for a p for which the following holds:

r + 1= ¹⁄₅×⁴⁄₅×⁴⁄₅×⁴⁄₅×⁴⁄₅×⁴⁄₅×(p+4) = (4⁵/5⁶)×(p+4), where r 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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