projecteuler.net [Problem 1]

Problem:

 

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.

 

 

Answer:

 

 public static double count(int max,int count){
  int a = 0;
  double sum =0;
   for(int i=2;i<max;i++){
    if(count*i >= max) break;
    a = i;
   }
   out(a);
   for(int i=1;i<=a;i++){
    sum+=i;
   }
   out(sum);
   return sum*count;
 }
 public static void main(String[] args){
  out(count(1000,3)+count(1000,5)-count(1000,15));
 }