My math teacher asked me this problem for homework and I am unsure how to solve it.
Which numbers contain a number of factors equivalent to the value of one of their divisors?
I found that 8 works, because it has 4 factors and 4 divides 8, but I did that by manually computing the number of factors of each number. I am unsure how to develop a general process of proof to find all such numbers.
You're asked to solve $\sigma_0(n)|n$, with $\sigma_0$ the divisor function. Luckily, $\sigma_0$ is a multiplicative function. Writing an arbitrary $n\in\Bbb N$ as $\prod_ip_i^{a_i}$, with the $p_i$ finitely many distinct primes and the $a_i$ positive integers, $\sigma_0(n)=\prod_i(a_i+1)$, so $n$ qualifies iff each $a_i+1$ is a product of powers of the $p_i$. Since once you choose which prime factors you want the legal choices of the $a_i$ are thereby specified, that's about as elegant an all-solutions characterization as you'll get.