Question

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

Answer 1

Given A28, 3B9 and 62C are divisible by $k$,

then $A28=n_{1}k=100A+20+8$ ...(i)
$3B9=n_{2}k=300+10B+9$ ....(ii)
$62C=n_{3}k=600+20+C$ ...(iii)
where $n_1,n_2$ and $n_3$ are integers.
Let $\Delta =\left|\begin{array}{ccc}A& 3& 6\\ 8& 9& C\\ 2& B& 2\end{array}\right|$
Applying $R_2\to 100R_1+10R_3+R_2$
$\therefore \Delta =\left|\begin{array}{ccc}A& 3& 6\\ 100A+20+8& 300+10B+9& 600+20+C\\ 2& B& 2\end{array}\right|$

$\therefore \Delta =\left|\begin{array}{ccc}A& 3& 6\\ n_{1}k& n_{2}k& n_{3}k\\ 2& B& 2\end{array}\right|$ ...{using (i), (ii) & (iii)}
$=k\left|\begin{array}{ccc}A& 3& 6\\ n_1& n_2& n_3\\ 2& B& 2\end{array}\right|$

Hence, $\Delta$ is divisible by $k$.