Radio Astronomy Systems Research Group

3121

Solution: (adapted from THE ULTIMATE PUZZLE SITE)

Each pirate leaves 4/5(n-1) coins of a pile of n coins. This results in an awful formula for the complete process (because every time one coin must be taken away to make the pile divisible by 5):

4/5(4/5(4/5(4/5(4/5(p-1)-1)-1)-1)-1), where p is the integer number of coins in the original pile.

The trick is to make the number of coins in the pile divisible by 5, by adding 4 coins. This is possible because you can take away those 4 coins again after taking away one fifth part of the pile: normally, 4/5(n-1) coins are left of a pile of n coins; now 4/5(n+4)=4/5(n-1)+4 coins are left of a pile of n+4 coins. Because of this, the number of coins 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×(p+4)=(4⁵/5⁵)×(p+4), where p is the number of coins in the original pile.

The smallest (p+4) for which the above holds, is 5⁵. So there were p=56-4=3121 coins in the original pile.