Five pirates raid the ship of a wealthy bureaucrat and steal his trunk of gold pieces. By the time they get the trunk aboard their own ship, dusk has fallen, so they agree to split the gold the next morning.

But the pirates are all very greedy. During the night one of the pirates decides to take some of the gold pieces for himself. He sneaks to the trunk and divides the gold pieces into five equal piles, with one gold piece left over. He puts the gold piece in his pile, hides it, puts the other four piles back in the trunk, and sneaks back to bed.

One by one, the remaining pirates do the same. They sneak to the trunk, divide the coins into five piles, with always one coin left over. Each pirate puts the gold coin in his own pile, hides it, and puts the remaining four piles back in the trunk.

What is the smallest number of coins there could have been in the trunk originally?

No takers? Here's the start of the solution:

I started working on it by trial and error taking numbers divisible by 5 and adding one (my algebra is too rusty for the easier route), and may return to it...