Question 420125
We could assign k to be an integer such that {{{n/(20-n) = k^2}}}. Solve for n and you will get {{{n = k^2(20-n) = 20k^2 - nk^2}}} --> {{{n(1+k^2) = 20k^2}}} --> {{{n = 20k^2/(1+k^2) = 20 - (20/(1+k^2))}}}. Here, the positive factors of 20 are 1,2,5,10,20, so k^2 can equal 0,1,4,9, or 19. This implies that k = {0,1,2,3} and n = 0, 10, 16, or 18.


Another way just as effective is to brute-force the problem. Most math teachers recommend against this, but I will sometimes brute-force a problem if I know for sure the method will work, if I know I'm counting all possible cases, and if the method does not take too much time. Here, we know that n can only be between 0 and 20; otherwise, {{{n/(20-n)}}} would be negative.