Multiples_of_3_or_5.c

#include <stdio.h>
#include <stdlib.h>
#define LIMIT 1000
/*
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.
*/

int main(){
    int sum=0,i;

    for(i=1;i<LIMIT;i++){
        if((i%3==0)||(i%5==0)){
            sum += i;
        }
    }
    printf("Sum: %d",sum);
    return 0;
}