Write $$\begin{aligned} 3^{512}-1 & =\left(3^{256}+1\right)\left(3^{256}-1\right)=\left(3^{256}+1\right)\left(3^{128}+1\right)\left(3^{128}-1\right) \\ & =\cdots=\left(3^{256}+1\right)\left(3^{128}+1\right) \cdots(3+1)(3-1) \end{aligned}$$ Now each factor $3^{2^{k}}+1, k \geq 1$, is divisible by just one factor of 2 , since $3^{2^{k}}+1=$ $\left(3^{2}\right)^{2^{k-1}}+1 \equiv 1^{2^{k-1}}+1=2(\bmod 4)$. Thus we get 8 factors of 2 here, and the remaining terms $(3+1)(3-1)=8$ give us 3 more factors of 2 , for a total of 11.