Question

Three digits numbers $7x,36y$ and $12z$ where $x,y,z$ are integers from $0$ to $9,$ are divisible by a fixed constant $k.$ Then the determinant $\left|\begin{array}{ccc}x& 3& 1\\ 7& 6& z\\ 1& y& 2\end{array}\right|$ $+48$ must be divisible by

• A. $k$
• B. $k^2$
• C. $k^3$
• D. $k^4$

Answer 1

The correct option is A $k$

Since $7x,36y,12z$ are divisible by $k$

Let us Assume $x=ka,y=kb,z=kc$

So the det $⟹\left|\begin{array}{ccc}ka& 3& 1\\ 7& 6& kc\\ 1& kb& 2\end{array}\right|$

$⟹12ka-k^{3}abc+3kc+7kb-48$

$⟹FromQuestion,12ka-k^{3}abc+3kc+7kb-48+48$

Hence A