Euler Problem 1-Multiples of 3 and 5 (C Language)

Using for loop and checking if-else conditions

 /* 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. */

#include <stdio.h>

int main()

{

    int sum = 0;

    for (int i = 0; i < 1000; i++)

    {

        if ((i % 3 == 0) || (i % 5 == 0))

        {

            sum = sum + i;

        }

    }

    printf("Sum : %d", sum);

    return 0;

}