Question

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

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

Answer 1

The correct option is A $15$

Given, $17!=3556xy428096000$

Since, $17!$ Is divisible by $9$, so sum of the digits $(48+x+y)$ must be divisible by $9$.

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

Also, $17!$ Is divisible by $11$, so $|10+x-y|$ must be multiply of $11$ or $0$. The only possibility is $|x-y|=1$

$\therefore x+y=15$