Solution to Project Euler Problem 1 in C

Here's the simple solution of the first problem of Project Euler which said:

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 code is as follows and took just 13 lines:

#include<stdio.h>
main()
{
    int i,j=0;
    for(i=3; i<1000; i++)
    {
        if(i%3==0 || i%5==0)
        {
            j=j+i;
        }
    }
    printf("The sum of all multiples is %d", j);
}

Execution time was  0.01055 seconds

The answer to this question is 233168.

For any queries, you are free to comment.