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If $x$ and $y$ are digits such that $17 !=3556xy428096000,$ then $x+y$ equals
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Verified Answer
The correct answer is:
15
Given, $171=3556 \times 4428096000$
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 $|10+x-y \mid$ must be multiple of 11 or $0 .$ The only possibility is
$$
\begin{array}{l}
|x-y|=1 \\
\therefore \quad x+y=15
\end{array}
$$
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 $|10+x-y \mid$ must be multiple of 11 or $0 .$ The only possibility is
$$
\begin{array}{l}
|x-y|=1 \\
\therefore \quad x+y=15
\end{array}
$$
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