Find a value of $n$ that has exactly 32 divisors I know that I could simply multiply the first $32$ primes together but is there some other way to ascertain the answer to this number theory problem?

In general a number with prime factorization $p_1^{a_1}p_2^{a_2}\cdots p_{k}^{a_k}$ has $\prod_{i=1}^{k}(a_i+1)$ divisors: the prime $p_i$ can occur from $0$ to $a_i$ times in each divisor.  So to find all numbers with exactly $32$ divisors, first find all factorizations of $32$.  There are a number of these; each one corresponds to a family of numbers with exactly $32$ divisors:
$$
32=2\cdot 2\cdot 2\cdot 2\cdot 2 \implies p_1p_2p_3p_4p_5
$$
$$
32=4\cdot8\implies p_1^3p_2^7
$$
$$
32=32\implies p_1^{31}
$$
etc.
