Question

Suppose three digit numbers $A28,3B9$ and $62C$ where $A,B$ and $C$ are integers between $0$ and $9,$ are divisible by a fixed integer $k$. Prove that determinant $\left|\begin{array}{ccc}A& 3& 6\\ 8& 9& C\\ 2& B& 2\end{array}\right|$ is also divisible by $k$.

Answer 1

Since $A28$, $3B9$ and $62C$ are divisible by $k$, we can write
$A28=100A+20+8=n_{1}k$ (1)
$3B9=300+10B+9=n_{2}k$ (2)
$62C=600+20+C=n_{3}k$ (3)
Now $\Delta =\left|\begin{array}{ccc}A& 3& 6\\ 8& 9& C\\ 2& B& 2\end{array}\right|$
$=\left|\stackrel{A}{\underset{2}{100A+20+8}}\stackrel{3}{\underset{B}{300+10B+9}}\stackrel{6}{\underset{2}{600+20+C}}\right|$
[By operating $R_2+(100R_1+10R_3)$]
$=\left|\stackrel{A}{\underset{2}{n_{1}k}}\stackrel{3}{\underset{B}{n_{2}k}}\stackrel{6}{\underset{2}{n_{3}k}}\right|$, by $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$
$=k\left|\stackrel{A}{\underset{2}{n_1}}\stackrel{3}{\underset{B}{n_2}}\stackrel{6}{\underset{2}{n_3}}\right|=k$ (an integer).
Hence $\Delta$ is also divisible by $k$.