Rows of a Matrix is divisible by 19, show that its Determinant is also divisible by 19 [duplicate]

I came across the following problem while self studying:

Let \begin{equation} A = \begin{bmatrix} 2 & 1 & 3 & 7 & 5\\ 3 & 8 & 7 & 9 & 8\\ 3 & 4 & 1 & 6 & 2\\ 4 & 0 & 2 & 2 & 3\\ 7 & 9 & 1 & 5 & 4\\ \end{bmatrix} \end{equation}

Use the fact that 21375, 38798, 34162, 40223, and 79154 are divisible by 19 to show, without evaluating, that $\det[A]$ is divisible by 19.

I noticed that each of these numbers are the entries in the rows of A, but I don't see how that helps me.

Solution 1:

Note that $$10^{10} \det(A) = 10^{4+3+2+1+0} \det(A) = \det \begin{pmatrix} 2 \cdot 10^{4} & 1\cdot 10^3 & 3\cdot 10^2 & 7\cdot 10 & 5 \\ 3\cdot 10^{4} & 8 \cdot 10^3 & 7\cdot 10^2 & 9\cdot 10 & 8 \\ 3\cdot 10^{4} & 4\cdot 10^3 & 1\cdot 10^2 & 6\cdot 10 & 2 \\ 4\cdot 10^{4} & 0 \cdot 10^3 & 2 \cdot 10^2 & 2 \cdot 10 & 3 \\ 7 \cdot 10^{4} & 9\cdot 10^3 & 1\cdot 10^2 & 5\cdot 10 & 4 \end{pmatrix}$$ $$ = \det \begin{pmatrix} 21375 & 1\cdot 10^3 & 3\cdot 10^2 & 7\cdot 10 & 5 \\ 38798 & 8 \cdot 10^3 & 7\cdot 10^2 & 9\cdot 10 & 8 \\ 34162 & 4\cdot 10^3 & 1\cdot 10^2 & 6\cdot 10 & 2 \\ 40223 & 0 \cdot 10^3 & 2 \cdot 10^2 & 2 \cdot 10 & 3 \\ 79154 & 9\cdot 10^3 & 1\cdot 10^2 & 5\cdot 10 & 4 \end{pmatrix}$$ which is evidently divisibly by $19$ when calculated via expansion by minors along the first column.