0

How many zeros are at the end of the number $236!$? Are there any special formulas or ways to solve this that would work for the factorial of any number?

Thanks, |-(<>)-| Tie Fighter

Tie Fighter
  • 43
  • 3
  • 1
    Yes, there are formulas. I'll just say here that you need to count how many factors of 5 divide $236!.$ – B. Goddard Jan 20 '17 at 19:47
  • $\lfloor 236 /5 \rfloor = 47$ multiples of $5$; $\lfloor47/5\rfloor = 9$ multiples of $25$; $\lfloor9/5\rfloor = 1$ multiple of $125$ giving $47+9+1 = 57$ multiples of $5$ contributing to the value of $236!$, hence $57$ zeroes at the end. – Joffan Jan 20 '17 at 21:03

0 Answers0