### Instruction: What is the greatest common divisor of $2^{1001}-1$ and $2^{1012}-1$?### Response: To find the greatest common divisor of two numbers, I can use the Euclidean algorithm, which repeatedly applies the division algorithm to the pair of numbers until I get a remainder of zero.
The division algorithm says that for any two positive integers $a$ and $b$, there exist unique integers $q$ and $r$ such that $a = bq + r$ and $0 \leq r < b$.
The Euclidean algorithm then says that the greatest common divisor of $a$ and $b$ is the same as the greatest common divisor of $b$ and $r$.
So I can start by applying the division algorithm to $2^{1012}-1$ and $2^{1001}-1$.
To find the quotient $q$, I can notice that $2^{1012}-1 = 2^{11}(2^{1001}-1) + 2^{11}-1$, so $q = 2^{11}$ and $r = 2^{11}-1$.
This means that the greatest common divisor of $2^{1012}-1$ and $2^{1001}-1$ is the same as the greatest common divisor of $2^{1001}-1$ and $2^{11}-1$.
I can repeat this process with $2^{1001}-1$ and $2^{11}-1$.
To find the quotient $q$, I can notice that $2^{1001}-1 = 2^{990}(2^{11}-1) + 2^{990}-1$, so $q = 2^{990}$ and $r = 2^{990}-1$.
This means that the greatest common divisor of $2^{1001}-1$ and $2^{11}-1$ is the same as the greatest common divisor of $2^{11}-1$ and $2^{990}-1$.
I can continue this way, reducing the exponents by multiples of 11 until I reach a pair of numbers that are easy to handle.
Eventually, I will get to $2^{22}-1$ and $2^{11}-1$.
To find the quotient $q$, I can notice that $2^{22}-1 = 2^{11}(2^{11}-1) + 0$, so $q = 2^{11}$ and $r = 0$.
This means that the greatest common divisor of $2^{22}-1$ and $2^{11}-1$ is just $2^{11}-1$.
Since this is the last nonzero remainder in the Euclidean algorithm, it is also the greatest common divisor of the original pair of numbers.
Therefore, the greatest common divisor of $2^{1001}-1$ and $2^{1012}-1$ is $2^{11}-1$.