# Python Program to Count trailing zeroes in factorial of a number

• Difficulty Level : Medium
• Last Updated : 14 Jun, 2022

Given an integer n, write a function that returns count of trailing zeroes in n!.  Examples :

```Input: n = 5
Output: 1
Factorial of 5 is 120 which has one trailing 0.

Input: n = 20
Output: 4
Factorial of 20 is 2432902008176640000 which has
4 trailing zeroes.

Input: n = 100
Output: 24```
```Trailing 0s in n! = Count of 5s in prime factors of n!
= floor(n/5) + floor(n/25) + floor(n/125) + ....```

## Python3

 `# Python3 program to``# count trailing 0s``# in n !` `# Function to return``# trailing 0s in``# factorial of n``def` `findTrailingZeros(n):``    ` `    ``# Initialize result``    ``count ``=` `0` `    ``# Keep dividing n by``    ``# powers of 5 and``    ``# update Count``    ``i ``=` `5``    ``while` `(n ``/` `i>``=` `1``):``        ``count ``+``=` `int``(n ``/` `i)``        ``i ``*``=` `5` `    ``return` `int``(count)` `# Driver program``n ``=` `100``print``(``"Count of trailing 0s "``+``    ``"in 100 ! is"``, findTrailingZeros(n))` `# This code is contributed by Smitha Dinesh Semwal`

Output:

`Count of trailing 0s in 100 ! is 24`

Time Complexity:  O(log5n)

Auxiliary Space: O(1)

Please refer complete article on Count trailing zeroes in factorial of a number for more details!

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