Question

If three-digit numbers $A28, 3B9$ and $62C$ , where $A, B$ and $C$ are integers between $0$ and $9$ , are divisible by a fixed integer $k$ , then the determinant $\begin{vmatrix}A&3&6\\ 8&9&C\\ 2&B&2\end{vmatrix}$ is

• A. divisible by $k$
• B. divisible by $k^2$
• C. divisible by $2k$
• D. None of these

Answer 1

The Correct Option is A

Given, $A28$, $3B9$ and $62C$ are divisible by $k$
$\therefore A28=k$
$\Rightarrow 100A+20+8=k : 3B9 = k$
$\Rightarrow 300+10B+9=k$ and $62C=k$
$\Rightarrow 600+20+C=k$
Let $\Delta=\left|\begin{matrix}A&3&6\\ 8&9&C\\ 2&B&2\end{matrix}\right|$
$=\frac{1}{100}\times\frac{1}{10} \left|\begin{matrix}100 A&300&600\\ 8&9&C\\ 20&10 B&20\end{matrix}\right|$
Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$
$=\frac{1}{1000}$
$\left|\begin{matrix}100A+20+8&300+10B+9&600+20+C\\ 8&9&C\\ 20&10B&20\end{matrix}\right|$
$=\frac{1}{1000} \left|\begin{matrix}k&k&k\\ 8&9&C\\ 20&10B&20\end{matrix}\right|$
$=\frac{k}{100} \left|\begin{matrix}1&1&1\\ 8&9&C\\ 20&10B&20\end{matrix}\right|$
Hence, it is always divisible by $k$