$17!=3556xy428096000,$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
First use the fact that the factorial is divisible by $9$, and therefore the digit sum is divisible by $9$.
Then use the fact that the number with decimal representation $abcdefg\dots$ is divisible by $11$ if and only if $a-b+c-d+e-f+\cdots$ is divisible by $11$.
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 $\text{XY} = 87$.
Testing divisibility by 9 gives $$ 48 + x + y \equiv 0 \pmod {9}. $$ Testing divisibility by 11 gives $$ 12 - x + y \equiv 0 \pmod {11} $$ So $x - y \equiv 1 \pmod {11}$ and $x + y \equiv 6 \pmod{9}$. Since $x, y$ are both digits, $x = 1 + y$, and $x + y$ is odd.
