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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...

Евгений
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3 Answers3

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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, $n \times n$).

Jean Marie
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$\begin{vmatrix}7&3&7\\8&7&1\\9&3&8 \end{vmatrix}=\begin{vmatrix}7&3&707\\8&7&801\\9&3&908 \end{vmatrix}$ 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.

MasB
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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.

Piquito
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