<s>[INST] Find some digits of $17!$ 
$17!$ is equal to $$35568x428096y00$$
  Both $x$ and $y$, are digits. Find $x$ and $y$.

So, $$17!=2^{15}\times 3^6\times 5^3\times 7^2\times 11\times 13\times 17=(2^3\times 5^3)\times 2^{12}\times 3^6\times 7^2\times 11\times 13\times 17$$
If there`s a product of $(2\times 5)^3$
Then this number has $3$ zeros at the end, so $y=0$
How do I find the $x$ now?
 [/INST] HINT $17!$ is divisible by $9$.  What is an easy test for divisibility by 9, involving the digits of a number?
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