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$17!$ 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?

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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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  • $\begingroup$ The sum od its digits Has to be divisible by 9 $\endgroup$ – a_man_with_no_name Feb 28 at 19:21
  • $\begingroup$ How could i miss it! $\endgroup$ – a_man_with_no_name Feb 28 at 19:21
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The alternating sum of digits must be divisible by $11$, i.e., $11\mid 18-x$. It follows that $x=7$.

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  • 1
    $\begingroup$ 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. $\endgroup$ – Kamil Maciorowski Mar 1 at 4:47

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