\gcd(2^{12} - 1, 2^{18} - 1) = 2^6 - 1 = 64 - 1 = 63
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- \gcd(2^a - 1, 2^b - 1) = 2^{\gcd(a,b)} - 1
- Here, $ a = 12 $, $ b = 18 $, so $ \gcd(12, 18) = 6 $.
- Thus,
- Therefore, the greatest common divisor is $ \boxed{63} $.
- Question: What is the largest integer that must divide the product of any four consecutive positive integers?
- Solution: Let the four consecutive integers be $ n, n+1, n+2, n+3 $.
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