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?

Solution 1:

HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?

Solution 2:

The alternating sum of digits must be divisible by $11$, i.e., $11\mid 18-x$. It follows that $x=7$.