<s>[INST] $17!=3556xy428096000,$then $(x+y)$ equals?(without using a calculator) $17!=3556xy428096000,$then $(x+y)$ equals?
a)$15$ b)$6$ c)$12$ d)$13$
With help of calculator $(x+y)$ can be easily calculated as $15$.But without a calculator,I can only conclude that the sum of digits $(3+5+5+6+x+y+4+2+8+9+6)$ is a multiple of $3$, & excluding $(x+y)$ the sum of the digits $(3+5+5+6+4+2+8+9+6) =48$.Thus excluding option d) the ans can be any of a),b) or c).
Please help me to find the exact answer.
Thank you 
 [/INST] Actually, the number was copied incorrectly (somehow) and it was driving me crazy trying to figure out what I was doing wrong.
Since $17 > 13$, $17!$ must be divisible by $7 \cdot 11 \cdot 13 = 1001$.  That means if we take the digits preceding the trailing zeros, and place them in groups of three, like so—$355, 6\text{XY}, 428, 096$—the alternating sums must be the same modulo $1001$.  We have, then,
$$
355+428 = 6\text{XY}+96
$$
and then $\text{XY} = 87$.
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