# 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?

HINT $$17!$$ is divisible by $$9$$. What is an easy test for divisibility by 9, involving the digits of a number?
The alternating sum of digits must be divisible by $$11$$, i.e., $$11\mid 18-x$$. It follows that $$x=7$$.
• If the answer was $0$ or $9$, the other method (divisibility by $9$) would not be enough. This method guarantees unambiguity by itself. On the other hand alternating sum requires little more attention and discipline. – Kamil Maciorowski Mar 1 at 4:47