# Problem 509: Divisor Nim

![graphic](img509.gif)

Anton and Bertrand love to play three pile Nim. However, after a lot of
games of Nim they got bored and changed the rules somewhat. They may
only take a number of stones from a pile that is a proper divisor of the
number of stones present in the pile. E.g. if a pile at a certain moment
contains 24 stones they may take only 1,2,3,4,6,8 or 12 stones from that
pile. So if a pile contains one stone they can't take the last stone
from it as 1 isn't a proper divisor of 1. The first player that can't
make a valid move loses the game. Of course both Anton and Bertrand play
optimally. The triple (a,b,c) indicates the number of stones in the
three piles. Let S(n) be the number of winning positions for the next
player for 1 ≤ a, b, c ≤ n.S(10) = 692 and S(100) = 735494. Find
S(123456787654321) modulo 1234567890.