Count trailing zeroes in factorial of a number | GeeksforGeeks

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

Trailing 0s in n! = Count of 5s in prime factors of n!
                  = floor(n/5) + floor(n/25) + floor(n/125) + ....

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;

}

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