RE04 - Editorial

Author: Manisha Singh
Tester: Charu Kalra
Editorialist: Arun Kumar






Finding the problem is the first task based on the input, output format and exe file provided. Then think the logic and code the same.


The task in this problem is to find number of zeroes in factorial of a number.


A simple method is to first calculate factorial of n, then count trailing 0s in the result (We can count trailing 0s by repeatedly dividing the factorial by 10 till the remainder is 0).

The above method can cause overflow for a slightly bigger numbers as factorial of a number is a big number. The idea is to consider prime factors of a factorial n. A trailing zero is always produced by prime factors 2 and 5. If we can count the number of 5s and 2s, our task is done. Consider the following examples.

n = 5: There is one 5 and 3 2s in prime factors of 5! (2 * 2 * 2 * 3 * 5). So count of trailing 0s is 1.

n = 11: There are two 5s and three 2s in prime factors of 11! (2 8 * 34 * 52 * 7). So count of trailing 0s is 2.

We can easily observe that the number of 2s in prime factors is always more than or equal to the number of 5s. So if we count 5s in prime factors, we are done.


Author’s solution can be found here

Exe file:

Exe file for the probem can be found here