2005 iTest Problems/Problem 37

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Since we want to find the number of zeros at the end of $209!$, it is the same as finding the largest value of $n$ such that $10^n$ is a divisor of $209!.$ Since $10=2\cdot5$ and there are more factors of $2$ than $5,$ finding the number of zeros at the end of $209!$ is the same as finding the largest value of $m$ such that $5^m$ that is a divisor of $209!.$ We then can use the floor function to find the factors of $5$ in $209!.$