Additionally, since both $a$ and $b$ are even, $d$ must be even (because any common divisor of two even numbers is even).
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- Therefore, the remainder is $oxed{0}$.
- Question: What is the largest possible value of $\gcd(a,b)$ if the sum of two positive integers $a$ and $b$ is $2024$ and both are even?
- 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 $2024$.
- To maximize $d$, we take the largest even divisor of $2024$ such that $a$ and $b$ are positive integers.
- Factor $2024$:
- = 2^3 imes 11 imes 23.
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