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 ndef 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 = 100print("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
Please refer complete article on Count trailing zeroes in factorial of a number for more details!
