Sure! For this one you'll need to use nested loops. Your first loop will count from 1 to 10,000. You'll then have to find a way to figure out if there are any numbers below you current number  (from your first loop, without counting that number) that evenly divide into it. You'll then need to verify if they sum up to the original number, and if so print them. I believe Keenan gave us an outline on how the loop structure could be.