Anders is solving a math problem, and he encounters the expression $\sqrt{15 !}$. He attempts to simplify this radical by expressing it as $a \sqrt{b}$ where $a$ and $b$ are positive integers. The sum of all possible distinct values of $a b$ can be expressed in the form $q \cdot 15$ ! for some rational number $q$. Find $q$.
Note that $15 !=2^{11} \cdot 3^{6} \cdot 5^{3} \cdot 7^{2} \cdot 11^{1} \cdot 13^{1}$. The possible $a$ are thus precisely the factors of $2^{5} \cdot 3^{3} \cdot 5^{1} \cdot 7^{1}=$ 30240. Since $\frac{a b}{15 !}=\frac{a b}{a^{2} b}=\frac{1}{a}$, we have \[ \begin{aligned} q & =\frac{1}{15 !} \sum_{\substack{a, b: \\ a \sqrt{b}=\sqrt{15 !}}} a b \\ & =\sum_{a \mid 30420} \frac{a b}{15 !} \\ & =\sum_{a \mid 30420} \frac{1}{a} \\ & =\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)\left(1+\frac{1}{5}\right)\left(1+\frac{1}{7}\right) \\ & =\left(\frac{63}{32}\right)\left(\frac{40}{27}\right)\left(\frac{6}{5}\right)\left(\frac{8}{7}\right) \\ & =4 \end{aligned} \] $\fbox{4}$.