There is an integer in each square of an $8 \times 8$ chessboard. In one move, you may choose any $4 \times 4$ or $3 \times 3$ square and add $1$ to each integer of the chosen square. Can we always get a table with each entry divisible by (a) $2$, (b) $3$?
I think finding answer revolves around finding an invariant. Let $S$ = sum of all the integers in the square - sum of integers of third row - sum of integers of sixth row Then $S \bmod 2$ is an invariant. Similarly $S \bmod 3$ will be an invariant if instead of integers of 3rd and 6th rows , We subtract integers of 4th and 8th rows.