+489 What is the greatest common divisor of \(2^{1001}-1\) and \(2^{1012}-1\)?
+6248 The theorem being used here is that
\(gcd(a^n -1,~a^m-1) = a^{gcd(m,n)}-1\)
in this case \(n=1001,~m=1012\)
\(1001 = 7 \cdot 11 \cdot 13 \\ 1012 = 2 \cdot 11 \cdot 23 \\ \text{so clearly }gcd(1001,~1012)=11\\ 2^{11}-1 = 2047 \text{ as guest correctly computed}\)
The proof of this theorem can be found online
I think the GCD of ((2^1001) - 1), ((2^1012) - 1) =2^(1012 - 1001) - 1 =2^11 - 1 =2,047
+6248 The theorem being used here is that
\(gcd(a^n -1,~a^m-1) = a^{gcd(m,n)}-1\)
in this case \(n=1001,~m=1012\)
\(1001 = 7 \cdot 11 \cdot 13 \\ 1012 = 2 \cdot 11 \cdot 23 \\ \text{so clearly }gcd(1001,~1012)=11\\ 2^{11}-1 = 2047 \text{ as guest correctly computed}\)
The proof of this theorem can be found online
