Find the sum of all the multiples of 3 or 5 below 1000

Question

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. I have the following code but the answer do

Answer 1

#include
#include
int main()
{
    int x,y,n;
    int sum=0;
    printf("enter the valeus of x,y and z\n");
    scanf("%d%d%d",&x,&y,&n);
    printf("entered   valeus of x=%d,y=%d and z=%d\n",x,y,n);
    sum=x*((n/x)*((n/x)+1)/2)+y*((n/y)*((n/y)+1)/2)-x*y*(n/(x*y))*((n/(x*y))+1)/2;
    printf("sum is %d\n",sum);
    return 0;
}
// give x,y and n  as 3 5 and 1000