Let $M:=(m_{ij})$ be a square matrix with entries in $\{ 1,2,3,.., n\}$ ,and non-zero determinant $D$ . Let $M_k$ be its reduction Mod($k$) ; $k=2,3,.., n-1$ , i.e., $M_k:=(m_{ij} \mod k) $. Is the determinant of $M_k$ also non-zero? Is there a formula relating $\det M$ with $\det M_k $? I can see that in the mod $2$ reduction, $M_2$ is a permutation matrix, so that it's invertible.
As motivation, not needed to solve the problem, though, I'm thinking of viewing Sudoku puzzles, as in https://www.websudoku.com/ as matrices. There is a result that the determinant of a $\ 9 \times 9$ Sudoku is divisible by $405$. I'm curious about what to happen to the determinant when reducing the matrix mod $k$; $k=2,3,..,8$ ( and mod each of $2,3,.., n-1$ for general $n$).
Thank You.
If $A$ and $B$ are matrices with integer entries then $C$ is too and $$ AB = C $$ then the same identity holds when you reduce all the elements in all the matrices modulo any integer $m$.
The determinant of the product remains the product of the determinants.