Solution for #26

The original number of coins must be a number such that you can subtract one and multiply by four fifths and get an integer. These numbers are 6, 11, 16, 21, 26, and so on.

But the pile remaining after the first pirate has taken his gold must also have this property. So the possibilities for the original number are 16, 36, 56, 76, 96, and so on.

The pile remaining after the second pirate has taken his gold must also have this property. So the possibilities for the original number are 76, 156, 236, 316, 396, and so on.

The pile remaining after the third pirate has taken his gold must also have this property. So the possibilities for the original number are 316, 636, 956, 1276, 1596, and so on.

The pile remaining after the fourth pirate has taken his gold must also have this property. The smallest possibility for this is 1276.

This number is the number of gold pieces in the chest the fourth pirate left behind (for the fifth pirate to divide). The fourth pirate hid a quarter of this number, plus one extra, just before the fifth pirate got there. So the third pirate left behind 1276 + 1276/4 + 1 = 1596 gold pieces.

The third pirate hid a quarter of this number, plus one extra, just before the fourth pirate got there. So the second pirate left behind 1596 + 1596/4 + 1 = 1996 gold pieces.

The second pirate hid a quarter of this number, plus one extra, just before the third pirate got there. So the first pirate left behind 1996 + 1996/4 + 1 = 2496 gold pieces.

The first pirate hid a quarter of this number, plus one extra. So the original number of coins must have been 2496 + 2496/4 + 1 = 3121 gold pieces.