//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.

#include <stdio.h>

int sumOfMultiple( int max, int multiple ) {
   int sum = 0;
   int i = 0;
   
   for( i = 0; i < max; i += multiple ) {
      sum += i;
   }
   
   return sum;
}

int main( int argc, char* argv[] ) {
   int threes      = sumOfMultiple(1000, 3);
   int fives      = sumOfMultiple(1000, 5);
   int fifteens   = sumOfMultiple(1000, 15);
   int sum         = threes + fives - fifteens;

   printf("Sum of threes: %d\n", threes);
   printf("Sum of fives: %d\n", fives);
   printf("Total sum: %d\n", sum);
}

for( i = 0; i < max; i++) {
    if( i % 3 == 0 || i % 5 == 0 ) {
        sum += i;
    }
}

int score = 0;
for(int i = 0; i<=1000; i++)
{
        if(i%3 ==0 || i%5==0)
        {
            score+=i;
            Console.WriteLine(i);
        }
}
Console.WriteLine("Total is {0}", score);