Question

The least values of $x$ and $y$ so that $7x342y$ is divisible by $88$ are ($x,y$ are digits in the number)

• A. $4,4$
• B. $4,3$
• C. $5,6$
• D. $6,7$

Answer 1

The correct option is A $4,4$

For $7x342y$ to be divisible by $88$, it must be divisible by $8$ and $11$.
For $7x342y$ to be divisible by $8$, $42y$ must be a multiple of $8$.
$\therefore y=4$.
For $7x342y$ to be divisible by $11$,

$12-(x+4+y)=0$ or $12-(x+4+y)=11k$

$\Rightarrow 8-(x+y)=0$ or $8-(x+y)=11k$
Since, $8-(x+y)=11k$ is not possible, hence $8=(x+y)$.
We know $y=4$ $\Rightarrow x=4$.