Question

Let the digit number $A28,3B9,62C$ where $A,B,C$ integers between $0$ and $9$, are divisible by a fixed integer $k$ then $\left[\begin{array}{ccc}A& 3& 6\\ 8& 9& C\\ 2& B& 2\end{array}\right]$ is divisible by

• A. $k+1$
• B. $k$
• C. $k-1$
• D. None of these

Answer 1

The correct option is B $k$

Since $A28,3B9$ and $62C$ are divisible by $k$
Therefore,
$A28=xk=100A+20+83B2=yk=300+10B+962C=zk=600+20+C$
Where $x,y,z$ are integers
Now let
$\Delta =\left|\begin{array}{ccc}A& 3& 6\\ 8& 9& C\\ 2& B& 2\end{array}\right|$

Applying $R_2\to R_2+100R_1+10R_3$

$\Delta =\left|\begin{array}{ccc}A& 3& 6\\ 100A+20+8& 300+10B+9& 600+20C\\ 2& B& 2\end{array}\right|=\left|\begin{array}{ccc}A& 3& 6\\ xk& yk& zk\\ 2& b& 2\end{array}\right|=k\left|\begin{array}{ccc}A& 3& 6\\ x& y& z\\ 2& B& 2\end{array}\right|$ is divisible by $k$.
Hence, option 'B' is correct.