Python Program to Count trailing zeroes in factorial of a number

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) + ....

Python3

# Python3 program to
# count trailing 0s
# in n !

# Function to return
# trailing 0s in
# factorial of n
def findTrailingZeros(n):

    # Initialize result
    count = 0

    # Keep dividing n by
    # powers of 5 and
    # update Count
    i = 5
    while (n / i>= 1):
        count += int(n / i)
        i *= 5

    return int(count)

# Driver program
n = 100
print("Count of trailing 0s "+
    "in 100 ! is", findTrailingZeros(n))

# This code is contributed by Smitha Dinesh Semwal

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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