 If there are 5 divisions, then multiples of 55=3125 are required; the
smallest such number is 3125 - 4 = 3121.After 5 divisions, there are 1020 coconuts left
over, a number divisible by 5 as required by the problem.In fact, after "n" divisions,
it can be proven that the remaining pile is divisible by "n", a property made convenient
use of by the creator of the problem.