Python Program to Count trailing zeroes in factorial of a number

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

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