Interview Question for Software Development Engineer at Microsoft:

not only the zeros in the end.

This is in fact a complicated math problem. What you want to do is to find how many trailing ceros does 100! Have with out having to Calculate the number. We know that the only way to get a trailing cero is to multiply a number by 10, or what would be the same, to miltiply it by 5 and 2. So what you want to do here is very easy, you need to find how many 2 and fives are there from 2-100. At last you would just have to check if you have more 2s or more 5s because you want to return only the smaller number. Your code would look something like this:

two = five = 0;
For(i=0;i<101;i++)
      while(i%2==0)
            two++;
            i=i/2;
       while(i%5==0)
            five++;
            i=i/5;

if(two<five)
      return two;
else
      return five;

Any questions feel free to ask.

here's the code for my previous solution, it's in c++.

#include <iostream>

using namespace std;

int main(){

    int two=0, five=0, i;
    int num;
    for(i=2;i<101;i++)
    {
        num=i;
        while(num%2==0)
        {
            two++;
            num=num/2;
        }
        while(num%5==0)
        {
            five++;
            num=num/5;
        }
    }

    if(two<five)
        cout << "100! has " << two << " trailing ceros.\n";
    else
        cout << "100! has " << five << " trailing ceros.\n";

    return 0;
}

The question asks for the number of 0s in the decimal representation of 100!, not the number of trailing 0s, which is quite easy to find.