Highly selective Backlog Qs for JEE MainMathematicsPermutation and Combination
The number of zeros at the end of 100! is
Options
- A21
- B22
- C23
- D24
Correct answer
D. 24
Step-by-step solution
E ₅(100 !)= [ 100 5 ]+ [ 100 25 ]+ [ 100 125 ]=20+4+0=24