For how many positive whole numbers $n$ is $n^2+6$ a multiple of $n+6$?

Rewrite $n^2+6$ as $(n+6)(n-6)+42$. The first part is already a multiple of $n+6$, so $n+6$ must divide 42. The divisors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42, but $n$ is positive, so $n+6$ is at least 7 — only 7, 14, 21 and 42 survive, giving $n=1$, 8, 15 and 36. Each checks out: 7 divides 7, 14 divides 70, 21 divides 231 and 42 divides 1302. The trap 8 is the number of divisors of 42 and forgets that the four small ones force $n$ to be zero or negative. Answering 6 or 2 stops the search early after finding only the easy values.