-ProjectEuler #1 + Solution

Problem Description:

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. Try it yourself before you look at the solution:

#include 

int main()
{
 unsigned int sum = 0;
 for(int i = 0; i < 1000; i++) //go through until 1000
  if(i % 3 == 0 || i % 5 == 0) //if it is divisble by 3 or 5
   sum += i; //add the number to sum.
 printf("Sum of natural numbers below 1000, that are divisible by 3 or 5 is %d", sum); //print the sum
 getchar();
 return 0;
}

Solution source: ProjectEuler #1 Output:

Sum of natural numbers below 1000, that are divisible by 3 or 5 is 233168

I'll start uploading more ProjectEulers; if you have any suggestions, comments, or questions, post 'em. ...Sri Krishna.