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

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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) + ....

C#




// C# program to count
// trailing 0s in n!
using System;
 
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()
    {
        int n = 100;
        Console.WriteLine("Count of trailing 0s in "
                + n + "! is " + findTrailingZeros(n));
    }
}
 
// This code is contributed by vt_m


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!


Last Updated : 14 Jun, 2022
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