Blahhh

$6=2^13^1\\ \text{let }n = \prod \limits_{k=0}^\infty p_k^{n_k} \\ \text{where }p_k \text{ is the }kth \text{ prime}\\ 6n = 2^{n_2+1}\cdot 3^{n_3+1}\cdot \prod \limits_{k=3}^\infty p_k^{n_k}$

$6n \text{ thus has }(n_2+2)(n_3+2)\prod \limits_{k=0}^\infty (n_k+1) = 9 \text{ divisors}$

$\text{This is easily enough accomplished if }n_2=1,~n_3=1,~n_k=0,k>3, \text{ i.e. }\\ 6n=36 = 2^2 3^2\\ \text{There are a total of 2 prime divisors}$