#include<stdio.h> int main(void){
int n1, n2,n3;
n1=333*(3+999)/2;
n2=199*(5+995)/2;
n3=66*(15+990)/2;
printf("%d\n",n1+n2-n3);
n1=getchar();
return 0;
}
Multiples of 3 and 5
Problem 1
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.