[Number Theory]
Problem: Compute the sum of all positive two-digit factors of $2^{32}-1$.

Solution: Using the difference of squares, $2^{32}-1=\left(2^{16}-1\right)\left(2^{16}+1\right)$. The second factor, $2^{16}+1$, is the Fermat prime 65537 , so continue with the first factor:

$$
\begin{aligned}
2^{16}-1 & =\left(2^{8}+1\right)\left(2^{8}-1\right) \\
2^{8}-1 & =\left(2^{4}+1\right)\left(2^{4}-1\right) \\
2^{4}-1 & =15=3 \cdot 5
\end{aligned}
$$



Because the problem does not specify that the two-digit factors must be prime, the possible two-digit factors are $17,3 \cdot 17=51,5 \cdot 17=85$ and $3 \cdot 5=15$, for a sum of $17+51+85+15=\mathbf{1 6 8}$.

Final answer: 168