If 3 integers are divisible by 67 prove that determinant of matrix of their digits is also devisible without direct calculation So we got three numbers:
\begin{align}
737 \\
871 \\
938 \\
\end{align}
are divisible by 67.
Easy to check. Our numbers consist of:
\begin{align}
N=\mathrm{X}\!\cdot\!\mathrm{10}+\mathrm{Y}\\
737=\mathrm{73}\!\cdot\!\mathrm{10}+\mathrm{Y}
\end{align}
So signature of divisibility is:
\begin{align}
if \ \ \mathrm{X}-20\!\cdot\!\mathrm{Y}\ is \ divisible \ by \ 67 \ then \ N \ is \ also \ divisible \\
73-\mathrm{20}\!\cdot\!\mathrm{7}=-67 \ is \ divisible  \ by \ 67
\end{align}
The same with other two integers.
And in this case we need to prove that determinant:
\begin{vmatrix}7&3&7\\8&7&1\\9&3&8  \end{vmatrix}
is also divisible by 67 without calculating determinant itself!
So what is the main idea here? I see, that all digits here are co-factors of integers itselves, so maybe we now should use properties of Laplace formula...but I can't catch how...
 A: I will attempt the difficult exercise to avoid the use determinant's properties, stricto sensu.
Let us write the property of divisibility in this way:
$$\underbrace{\begin{pmatrix}7&3&7\\8&7&1\\9&3&8  \end{pmatrix}}_{M} \begin{pmatrix}100\\10\\1 \end{pmatrix}=67\begin{pmatrix}a\\b\\c \end{pmatrix}\tag{1}$$
for certain integers $a,b,c$.
Besides, using Cramer's formulas,
$$M^{-1}=\dfrac{1}{\det M}N$$ where $N=adj(M)$ has integer coefficients (transpose of the matrix of cofactors).
Multiplying LHS and RHS of (1) by $M^{-1}$, we have:
$$\begin{pmatrix}100\\10\\1 \end{pmatrix}=\dfrac{67}{\det M} N \begin{pmatrix}a\\b\\c \end{pmatrix}$$
Considering the last line, we have, for a certain integer $K$:
$$1=\dfrac{67}{\det M}K \ \ \iff \ \ \det{M}=67 K$$
As $67$ is prime, $\det(M)$ is a divisor of $67$.
Remark: As we haven't used neither $10$ nor $100$, it means that  this property is valid for any numeration basis (and, besides, any matrix size, $n \times n$).
A: $\begin{vmatrix}7&3&7\\8&7&1\\9&3&8  \end{vmatrix}=\begin{vmatrix}7&3&707\\8&7&801\\9&3&908  \end{vmatrix}$ by adding 100 times the first column to the last. Now add 10 times the second column to the last and conclude using a determinant property.
A: HINT.-Generalizing, let $a_1b_1c_1,a_2b_2c_2$ and $a_3b_3c_3$ be three integers divisible by a prime $p$ distinct of $2$ and $5$. Show that $\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}$ is divisible by $p$.
We have
$$\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=\frac{1}{10^3}\begin{vmatrix}10^2a_1&10b_1&c_1\\10^2a_2&10b_2&c_2\\10^2a_3&10b_3&c_3\end{vmatrix}=\frac{1}{10^3}\begin{vmatrix}10^2a_1+10b_1+c_1&10b_1&c_1\\10^2a_2+10b_2+c_2&10b_2&c_2\\10^2a_3+10b_3+c_3&10b_3&c_3\end{vmatrix}\equiv0\pmod p$$ because the terms of the first row are all divisible by $p$.
REMARK.- Longer generalization to $n$ integers and the corresponding determinant of order $n$, is almost obvious.
