Project Euler #1: Multiples of 3 and 5
You don't need iteration at all for this problem.

Consider; the sum of all numbers from 1 to n is equal to n*(n+1)/2.  Also the sum of all numbers less than n that divides d equals d times the sum of all numbers less than n/d.  

So the sum of all numbers less than 1000 that divides 3 is

3*floor(999/3)*(floor(999/3)+1)/2


Likewise the sum of all numbers less than 1000 that divides 5 is

5*floor(999/5)*(floor(999/5)+1)/2


Adding the two numbers would overcount though.  Since the numbers that divides both 3 and 5 would get counted twice.  The numbers that divides both 3 and 5 is precisely the numbers that divides 3*5/gcd(3,5)=15/1=15.

The sum of all numbers less than 1000 that divides 15 is

15*floor(999/15)*(floor(999/15)+1)/2


So the final result is that the sum of all numbers less than 1000 that divides either 3 or 5 equals:

  3 * (floor(999/3)  *  (floor(999/3)+1))/2
+ 5 * (floor(999/5)  *  (floor(999/5)+1))/2
-15 * (floor(999/15) * (floor(999/15)+1))/2