Long-Jumping Checkers

Note that after the operation of $(m,n)$ on $(i,j)$ , the value of $i\pmod{3}$ stays the same. Ditto for $j\pmod{3}$ : $3m-2i\equiv -2i\equiv i\pmod{3}$ $3n-2j\equiv -2j\equiv j\pmod{3}$

Thus, a checker $(i,j)$ can only move to spaces $(i_1,j_1)$ where $i_1\equiv i\pmod{3}$ and $j_1\equiv j\pmod{3}$ .

In the diagram below, each different shade of grey corresponds to a different value of $(i,j)\pmod{3}$ . Identical colors means a checker of that color can move to any other spot of the same color.

Imgur

It's clear now that:

$(1,1)\pmod{3}\implies \text{9 different spaces}$ $(1,2)\pmod{3}\implies \text{9 different spaces}$ $(1,3)\pmod{3}\implies \text{6 different spaces}$ $(2,1)\pmod{3}\implies \text{9 different spaces}$ $(2,2)\pmod{3}\implies \text{9 different spaces}$ $(2,3)\pmod{3}\implies \text{6 different spaces}$ $(3,1)\pmod{3}\implies \text{6 different spaces}$ $(3,2)\pmod{3}\implies \text{6 different spaces}$ $(3,3)\pmod{3}\implies \text{4 different spaces}$

In our original setup of $32$ checkers in one half of the board, we have:

$\text{6 checkers }\equiv (1,1)\pmod{3}$ $\text{3 checkers }\equiv (1,2)\pmod{3}$ $\text{3 checkers }\equiv (1,3)\pmod{3}$ $\text{6 checkers }\equiv (2,1)\pmod{3}$ $\text{3 checkers }\equiv (2,2)\pmod{3}$ $\text{3 checkers }\equiv (2,3)\pmod{3}$ $\text{4 checkers }\equiv (3,1)\pmod{3}$ $\text{2 checkers }\equiv (3,2)\pmod{3}$ $\text{2 checkers }\equiv (3,3)\pmod{3}$

Thus the number of different arrangements is: $\dbinom{9}{6}\cdot\dbinom{9}{3}\cdot\dbinom{6}{3}\cdot\dbinom{9}{6}\cdot\dbinom{9}{3}\cdot\dbinom{6}{3}\cdot\dbinom{6}{4}\cdot\dbinom{6}{2}\cdot\dbinom{4}{2}=26885053440000$

Thus $M=4$ and $N=2688505344$ so the last three digits of $M+n$ is $\boxed{348}$ .