jmc

By the Euclidean algorithm, $$\begin{array}{rl}& \text{gcd}(2^{1012}-1,2^{1001}-1)\\ & =\text{gcd}(2^{1012}-1-2^{11}(2^{1001}-1),2^{1001}-1)\\ & =\text{gcd}(2^{11}-1,2^{1001}-1)\end{array}$$ Using the divisibility rule for $11$, we know that $11$ divides into $1001$. Writing $2^{1001}$ as $(2^{11}{)}^{91}$ and $1$ as $1^{91}$, we use the difference of odd powers factorization to find that $$2^{1001}-1=(2^{11}{)}^{91}-1^{91}=(2^{11}-1)((2^{11}{)}^{90}+(2^{11}{)}^{89}+\cdots (2^{11}{)}^1+1).$$ Thus $2^{1001}-1$ is divisible by $2^{11}-1$, so the greatest common divisor is $2^{11}-1=\boxed{2047}$.