2011 AMC 12B Problem 15

Factoring, $$\begin{aligned} 2^{24}-1 &=(2^{12}-1)(2^{12}+1) \\ &=(2^6-1)(2^6+1) \\ &\quad {}\cdot(2^4+1)(2^8-2^4+1), \end{aligned}$$ which equals $63\cdot65\cdot17\cdot241$ $=3^2\cdot5\cdot7\cdot13\cdot17\cdot241.$ Since $241$ is a three-digit prime, the two-digit factors come from $3^2\cdot5\cdot7\cdot13\cdot17.$ They are $$\begin{gathered} 13,15,17,21,35,39,45,51, \\ 63,65,85,91, \end{gathered}$$ for a total of $12.$ Thus, the correct answer is D.