Project Euler Problem 1

Hi Friends,

I started solving Project Euler problems recently and found them to be really interesting. So I though of discussing my approach and solutions with the world. I will code my answers in C. So here we start.

Problem 1: If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.

The problem is pretty easy and straight forward. Just check all the numbers from 3 to 999 if they are divisible by 3 or 5 and add them if they are.
Here is my code:

1:  #include <stdio.h>  
2:  int main(void)  
3:  {  
4:       int sum=0,i;  
5:       for(i=3;i<1000;i++){  
6:            if(i%3==0 || i%5==0)  
7:                 sum+=i;  
8:       }  
9:       printf("%d",sum);  
10:       return 0;       
11:  }  

Do comment if you have any alternate solution.