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. Conclusion: there were p = 5⁶ - 4 = 15621 coconuts in the original pile.

Back to the puzzle