Factorial

Problem

Idea : Given an integer n , you have to find out number of zeros at the end of n! .

There are one zero at the end of a number x . What is its meaning ? That means , x can be expressed as multiple of 10(5*2) . Similarly , There are two zero at the end of a number x . What is its meaning ? That means , x can be expressed as multiple of 10(5*2)*10(5*2) . If we want to count the number of zero , we should count the number of 5*2 appears in the number . The number of zero is equal to the number of 5*2 .

To do so , we first look up at 5 and 2 . How many 5 and 2 is in the number .

As example , if n=10

Number of 5 = 10/5 = 2
Number of 2 = 10/2 = 5

We see that , the number of 2 > the number of 5 .

So , cheaking 5 is enough .

In an integer n , number of 5 is n/5 .

Some numbers have two , three or more munber of 5 as 25 , 50 , 75 , 100

So , we can cheak in n the total occurance of 5 by -

Single 5 = n/5
Double 5 = n/25
Triple 5 = n/75 and so on......

The total number of 5 = The total number of 5*2 = The total number of zero .


 My Solution