Python counting the trailing zero of factorial

Working on an example but it does not work. There must be a function which counts trailing zero in n! the factorial formula. where, n is the number for whose factorial we want to find the number of trailing zeros. Used some imports but it did not work.

My code:

def is_positive_integer(x):
    try:
        x = float(x)
    except ValueError:
        return False
    else:
        if x.is_integer() and x > 0:
            return True
        else:
            return False


def trailing_zeros(num):
    if is_positive_integer(num):
        # The above function call has done all the sanity checks for us
        # so we can just convert this into an integer here
        num = int(num)

        k = math.floor(math.log(num, 5))
        zeros = 0
        for i in range(1, k + 1):
            zeros = zeros + math.floor(num/math.pow(5, i))
        return zeros
    else:
        print("Factorial of a non-positive non-integer is undefined")

Ex:

    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

Output must be:

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

Answer

This code can do the job, your code is very complex.

def Zeros(n):
    count = 0

    i = 5
    while n / i >= 1:
        count += int(n / i)
        i *= 5

    return int(count)


n = 100
print("Trailing zeros " +
      "in 100! is", Zeros(n))

Output:

Trailing zeros in 100! is 24