23. [12] Let $d$ be a randomly chosen divisor of 2016 . Find the expected value of
$$
\frac{d^{2}}{d^{2}+2016}
$$
Answer:
$\frac{1}{2}$

Let $a b=2016$. Then
$$
\frac{a^{2}}{a^{2}+2016}+\frac{b^{2}}{b^{2}+2016}=\frac{a^{2}}{a^{2}+2016}+\frac{\left(\frac{2016}{a}\right)^{2}}{\left(\frac{2016}{a}\right)^{2}+2016}=\frac{a^{2}}{a^{2}+2016}+\frac{2016}{a^{2}+2016}=1
$$

Thus, every divisor $d$ pairs up with $\frac{2016}{d}$ to get 1 , so our desired expected value is $\frac{1}{2}$.