question
Given an integer, write a function to find the number of trailing zeros of its factorial.

ex. factorialZeroCount(10)
    10! = 3,628,800
    returns: 2

ex 2. factorialZeroCount(27)
    27! = 10,888,869,450,418,352,160,768,000,000
    returns: 6
Which two single-digit numbers multiplied with each other will result in a zero.
Trailing zeros are formed when a multiple of 5 is multiplied with a multiple of 2. The number of 2's in the factorial product is abundant and thus we will care only about the number of multiples of fives.

For example, 27 has five 5's: 5, 10, 15, 20, 25
In addition, 27 contains one multiple of 5: 25
In total, five 5's + one 25 = six zeros

factorialZeroCount(int n)
    int powerFive = 1;
    int count = 0;

    while (powerFive <= n)
       powerFive *= 5;
       count += floor(n/powerFive);

    return count;


See here more information on trailing zeros.