# Problem 108 Solution

The original pile must be a number such that you can subtract one and multiply by 4/5 and get an integer. Possibilities are 6, 11, 16, 21, 26 ...

After the first person is finished the remaining pile will have 4, 8, 12, 16, 20, ...

The remaining pile after the first person must be a number such that you can subtract one then multiply by 4/5 and get an integer. Possibilities are 16, 36, 56, 76, 96, ...

After the second person is finished the remaining pile will have 12, 28, 44, 60, 76, ...

The remaining pile after the second person must be a number such that you can subtract one then multiply by 4/5 and get an integer. Possibilities are 76, 156, 236, 316, 396, ...

After the third person is finished the remaining pile will have 60, 124, 188, 252, 316 ...

The remaining pile after the third person must be a number such that you can subtract one then multiply by 4/5 and get an integer. Possibilities are 316, 636, 956, 1276, 1596, ...

After the fourth person is finished the remaining pile will have 252, 508, 764, 1020, 1276 ...

The remaining pile after the fourth person must be a number such that you can subtract one then multiply by 4/5 and get an integer. The smallest possibilities is 1276

After the fifth person is finished the remaining pile will have 1020 ...

So the fifth person will get 1276-1020-1 = 255.

The fifth person will leave behind a pile of 1020.

The fourth person will have left behind a pile of 1020+255+1 = 1276.

The fourth person will get 1276/4 = 319.

The third person will have left behind a pile of 1276+319+1 = 1596.

The third person will get 1596/4 = 399.

The second person will have left behind a pile of 399+1596+1 = 1996.

The second person will get 1996/4 = 499.

The first person will have left behind a pile of 499+1996+1 = 2496.

The first person will get 2496/4 = 624.

The original pile must have had 624+2496+1 = 3121.

Michael Shackleford, A.S.A.