Let $M:=(m_{ij})$ be a square ($ n \times n $) 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 Mod2 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.
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1Note that since det is a polynomial in the entries $,a_{ij},$ we have $\bmod k!:\ \det(a_{ij})\equiv \det(a_{ij}\bmod k),,$ by the Polynomial Congruence Rule. – Bill Dubuque Dec 01 '21 at 23:56