Finding the Sum of all positive numbers less than N

Given three positive integers NA and B (A < B < N), find the sum of all positive integers less than N, which are divisible by either A or B.
For example, when N = 20, A = 4 and B = 7, the possible values are 4, 7, 8, 12, 14, and 16. Their sum is 61.

Method 1: (Brute Force)

  1. Starting from i = 1, loop till N-1
  2. For every i, check ether A divides i or B divides i and add it to sum , if it does
  3. Print the sum
Implementation:


#include <stdio.h>
int main()
{
    int N,A,B,sum=0,i;
    scanf("%d %d %d",&N,&A,&B);
    for(i=1;i<N;i++)
    {
            if( (i%A)==0 || (i%B)==0 )
            sum=sum+i;
    }
    printf("%d",sum);
    return 0;
}


Input Format:

N, A and B.

Output Format:

 Sum

Sample Input
20 4 7

Sample Output
61