How many zeroes does $$\frac{50!}{2^95^5}$$ end in?
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Figure out the highest powers of 2 and 5 that divide 50!, and divide by $2^9*5^5$. – user2357112 Apr 08 '14 at 00:09
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1Hint: Count how many times $5$ appears as a factor of $50!$ and how many times $2$ appears as factor of $50!$. – Git Gud Apr 08 '14 at 00:10
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Hint: Use the Lemma from this answer, that says the number of factors of a prime $p$ that divide $n!$ is $$ \frac{n-\sigma_p(n)}{p-1} $$ where $\sigma_p(n)$ is the sum of the digits in the base-$p$ representation of $n$.
$50=110010_\text{two}$.
$50=200_\text{five}$.
