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

• 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) + ....```

## Java

 `// Java program to count``// trailing 0s in n!``import` `java.io.*;` `class` `GFG {``    ``// Function to return trailing``    ``// 0s in factorial of n``    ``static` `int` `findTrailingZeros(``int` `n)``    ``{``        ``// Initialize result``        ``int` `count = ``0``;` `        ``// Keep dividing n by powers``        ``// of 5 and update count``        ``for` `(``int` `i = ``5``; n / i >= ``1``; i *= ``5``)``            ``count += n / i;` `        ``return` `count;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``100``;``        ``System.out.println(``"Count of trailing 0s in "``                          ``+ n + ``"! is "` `+ findTrailingZeros(n));``    ``}``}` `// This code is contributed by Pramod Kumar`

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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