Solution to:
Coconut Chaos
Every sailor leaves ^{4}/_{5}(n1) 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}(p1)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 onefifth part of the pile:
normally, ^{4}/_{5}(n1) coconuts are left of a pile of n coconuts;
now ^{4}/_{5}(n+4) = ^{4}/_{5}(n1)+4 coconuts are left of a pile of n+4 coconuts.
In the last step, ^{1}/_{5}(n+4) = ^{1}/_{5}(n1)+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 whole 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) = (4^{5}/5^{6})×(p+4),
where r must be a whole number.
The smallest (p+4) for which the above holds, is 5^{6}.
So there were p = 5^{6}4 = 15621 coconuts in the original pile.
back to the puzzle
