squeak
This is not as innocent as it looks.
Let the original pile = y and the final pile (before it’s divided into 5) = x.
We know that x must be divisible by 4 and 5 (the last person split it into 4 even groups and the 5 people split it into 5 even groups).
y = ((((((x)*5/4+1)*5/4+1)*5/4+1)*5/4+1)*5/4+1) which simplifies to (3125*x)/1024 + (8404/1024).
(8404/1024) = 8 + 53/256 therefore the fractional part of (3125*x)/1024 must equal 203/256 (so we are not left with a fraction at the end).
=> (3125*x)/1024 = an integer (henceforth denoted as “I”) + 203/256
=> (3125*x)/1024 = I + 203/256
=> 3125*x = 1024*I + 812
So we need MOD(3125*x,1024) = 812
MOD(3125,1024) = 53
MOD(812,53) = 17
MOD(53,17) = 2
=> x = 4 * 5 * MOD(812,53) * (MOD(53,17) + 1) = 4 * 5 * 17 * 3 = 1020
=> x = 1020 and y = 3121
If we weren’t looking for the minimum value of x the correct answer would be 1020 + 1024c where c is a non-negative constant.