2005 iTest Problems/Problem 37

Revision as of 01:16, 24 July 2024 by Kzs (talk | contribs) (Created page with "Since we want to find the number of zeros at the end of <math>209!</math>, it is the same as finding the largest value of <math>n</math> such that <math>10^n</math> is a divis...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $209!.$