Question

If $x$ and $y$ are digits such that $17!=3556xy428096000$, then $x+y$ equal

• A. $15$
• B. $6$
• C. $12$
• D. $13$

Answer 1

The correct option is A $15$

$17!=3556xy428096000$

Since $17!$ is divisible by $9$ so sum of the digits $48+x+y$ must be divisible by $9$. (Using divisibility test of $9$)

So, $x+y$ can either be $15$ or $6$.

Also $17!$ is divisible by $11$ so $|10+x-y|$ must be multiple of $11$ or zero. (Using divisibility test of $11$)

The only possibility is $|x-y|=1$

So, $x+y=15$