imc

First, we can write the expression in a more primitive form which will allow us to start factoring. $$10^{1002}-4^{501}=2^{1002}\cdot 5^{1002}-2^{1002}$$ Now, we can factor out $2^{1002}$. This leaves us with $5^{1002}-1$. Call this number $N$. Thus, our final answer will be $2^{1002+k}$, where $k$ is the largest power of $2$ that divides $N$. Now we can consider $N(\mathrm{mod}16)$, since $k\le 4$ by the answer choices. Note that $$\begin{array}{rl}{5}^1& \equiv 5(\mathrm{mod}16)\\ 5^2& \equiv 9(\mathrm{mod}16)\\ 5^3& \equiv 13(\mathrm{mod}16)\\ 5^4& \equiv 1(\mathrm{mod}16)\\ 5^5& \equiv 5(\mathrm{mod}16)\\ & ⋮\end{array}$$ The powers of $5$ cycle in $\mathrm{mod}16$ with a period of $4$. Thus, $$5^{1002}\equiv 5^2\equiv 9(\mathrm{mod}16)⟹5^{1002}-1\equiv 8(\mathrm{mod}16)$$ This means that $N$ is divisible by $8=2^3$ but not $16=2^4$, so $k=3$ and our answer is $2^{1002+3}=\boxed{2^{1005}}$.