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- So the largest divisor is $2024$ itself, but for $d = 2024$, we would need $a = 2024$, $b = 0$, but $b$ must be positive â invalid.
- Try the next largest divisor: $1012 = 2024/2$. Can we have $a = 1012$, $b = 1012$? Then $a + b = 2024$, both even, and $\gcd(1012,1012) = 1012$.
- So $d = 1012$ is achievable.
- Hence, the maximum $\gcd$ is $1012$.
- Therefore, the largest possible value is $oxed{1012}$.
- Question: How many of the 100 smallest positive integers are congruent to $3 \pmod{7}$?
