# Python program to count number of trailing zeros in Factorial of number N

In this Python program, we are going to learn how to count number of trailing zeros in factorial of N?
Submitted by Ankit Rai, on June 15, 2019

Formula used:

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

Example:

```    Input: N = 23
Output: 4
Factorial of 23 is 25852016738884976640000 which has four trailing 0.

Input: N = 25
Output: 6
Factorial of 25 is 15511210043330985984000000 which has six trailing 0.
```

Code:

```# Define a function for finding
# number of trailing zeros in N!
def find_trailing_zeros(num) :
sum = 0
i = 1

# iterating untill quotient is not zero
while(True) :

# take integer divison
quotient = num // (5 ** i)

if(quotient == 0) :
break

sum += quotient
i += 1

return(sum)

# Driver code
if __name__ == "__main__" :

# assigning a number
num = 10

# function call
print("Number of trailing zeros in factorial of",num,"is :",find_trailing_zeros(num))

num = 20

print("Number of trailing zeros in factorial of",num,"is :",find_trailing_zeros(num))
```

Output

```Number of trailing zeros in factorial of 10 is : 2
Number of trailing zeros in factorial of 20 is : 4
```

Preparation