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- Solution: Let $a + b = 2024$, and suppose $d = \gcd(a,b)$. Since $d \mid a$ and $d \mid b$, then $d \mid (a + b) = 2024$. So $d$ is a divisor of $2024$.
- Additionally, since both $a$ and $b$ are even, $d$ must be even (because any common divisor of two even numbers is even).
- To maximize $d$, we take the largest even divisor of $2024$ such that $a$ and $b$ are positive integers.
- = 2^3 imes 11 imes 23.
- 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 $\gcd(1012,1012) = 1012$.
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