Project Euler Problem1

  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.

Solution:-

#include <bits/stdc++.h>
using namespace std;

int main() {

    int sum = 0,n=3;
    while(n < 1000)
    {
        if(n%3==0 || n%5==0)
        {
             sum+=n;
        }
        n++;
    }
   
    cout<<sum;
    return 0;
}