Art of Problem Solving
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2005 iTest Problems/Problem 37

Problem

How many zeroes appear at the end of $209$ factorial?

Solution 1

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 $209!$.This is done by Legendre's formula. Since $5^4>209$, we only need to compute up to 3. Using the formula, we find that $\sum_{i=1}^{3}\lfloor \frac{209}{5^n} \rfloor=\boxed{50}$.

See Also

2005 iTest (Problems, Answer Key)
Preceded by:
Problem 36
Followed by:
Problem 38
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