On determinants and common divisors

Question

Let $n\in\mathbb N$ and let $a_1,\ldots,a_n$ be natural numbers smaller than $10^n$. Write each $a_k$ in base $10$ and add $0$'s to the left of each decimal expansion, if needed, so that each $a_k$ is written with $n$ digits. Consider the matrix such that the entries of the $k$^th row are the digits of $a_k$ (written in the same order). Let $d\in\mathbb N$ be such that $d$ divides each $a_k$. Prove that $d\mid\det A$.

For instance, suppose that $n=4$ and that your numbers are $3876$, $2784$, $684$, and $8388$, each of which is a multiple of $12$. Then$$A=\begin{bmatrix}3 & 8 & 7 & 6 \\ 2 & 7 & 8 & 4 \\ 0 & 6 & 8 & 4 \\ 8 & 3 & 8 & 8\end{bmatrix}$$and $\det A=-360$, which is, in fact, a multiple of $12$.

I learned about this problem yesterday. I found it quite cute and I decided to share it with all of you.

Note: I know how to prove it.

Answer 1

Let $t$ be the column vector with entries $1,10,10^2,\dots,10^{n-1}$.

Let $b$ be the column vector with entries $a_1,\ldots,a_n$.

Then $At = b$ and so $(\det A) t = \operatorname{adj}(A) A t = \operatorname{adj}(A) b$.

The first entry in both vectors says that $\det A$ is an integer linear combination of $a_1,\ldots,a_n$. Therefore, every common divisor of $a_1,\ldots,a_n$ also divides $\det A$.

This proof works unchanged for any base, not just $10$.

Answer 2

Here's my answer to my problem. Take the matrix $A$ and add the following columns to the last one:

• the first column times $10^{n-1}$;
• the second column times $10^{n-2}$;
• …
• the next to last column times $10$.

The new matrix $A^\star$ will have the same determinant as $A$, but $A^\star$ is a matrix with integer entries such that the entries of its last column are the original numbers $a_1,\ldots,a_n$. Since they are all multiples of $d$, $d\mid\det A^\star=\det A$.