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So a challenge problem states that you have an $n \times n$ matrix, where each entry is an integer between $0$ and $9$, and when each row is read as a base-10 number the number is divisible by a common factor $m$. The problem is to prove that the determinant of the matrix is also divisible by $m$. So far I have two proofs: If $m$ is a power of a prime, then we can consider the field ${\mathbb F}_m$ and the vector $(10^{n-1}, 10^{n-2}, \ldots, 1)$ is a non-zero vector in the kernel of the matrix over ${\mathbb F}_m$ so the determinant is $0$ over the field by general linear algebra. Or, for arbitrary $m$, we can add $10^{n-1}$ times the first column and $10^{n-2}$ times the second column and so forth and add it all to the last column without changing the determinant, and then we get the the last column is all numbers divisible by $m$ so the determinant is divisible by $m$ by expanding the determinant along the last column. However are there any more elementary proofs, say that just use the basic full expansion formula of the determinant, without using so many properties of the determinant?

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    $\begingroup$ I find this request rather strange as long as there is a truly elementary proof using the properties of the determinant and taking into account that no one is using the definition of the determinant to do something with it, except for proving the determinant properties. $\endgroup$ – user26857 Oct 18 '14 at 8:13
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Take the matrix (unity matrix except for last column) $$Q=\pmatrix{1&0&\dots&&10^{n-1}\cr0&1&0&\dots&10^{n-2}\cr\dots&0\cr0&\dots&&&1\cr}$$Its determinant is obviously 1, and the last column of $A\,Q$ is obviously the rows of $A$ interpreted in base 10. If the whole last column of $A\,Q$ can be divided by $m$, so can its determinant, and since the determinant of Q is 1, the determinant of $A\,Q$ is the same as the determinant of $A$.

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  • $\begingroup$ $\det AB=\det A\det B$ is much harder to prove than any elementary property of the determinant! $\endgroup$ – user26857 Oct 22 '14 at 5:22

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