But $d$ must be an integer. So $m + n$ must divide 2025 evenly, and be at least 2. We seek the largest divisor $d$ of 2025 such that \( \frac{2025}{d} \geq 2 $ and $m + n = \frac{2025}{d}$ is minimized among valid coprime pairs. The smallest possible $m + n \geq 2$ is 3 (e.g., $m = 1, n = 2$), which is coprime. Then:

February 22, 2026

$But $d$ must be an integer. So $m + n$ must divide 2025 evenly, and be at least 2. We seek the largest divisor $d$ of 2025 such that \( \frac{2025}{d} \geq 2 $ and $m + n = \frac{2025}{d}$ is minimized among valid coprime pairs. The smallest possible $m + n \geq 2$ is 3 (e.g., $m = 1, n = 2$), which is coprime. Then:$

a + b = d(m + n) = 2025
Maximizing $d$ requires minimizing $m + n$, with $m, n \geq 1$ and $\gcd(m, n) = 1$. The minimum value of $m + n$ is 2, achieved when $m = n = 1$, which are coprime. Then:
d = \frac{2025}{2} = 1012.5
d = \frac{2025}{3} = 675
Check: $a = 675$, $b = 1350$, $\gcd(675, 1350) = 675$. Can we do better? Try $m + n = 2$: $d = 2025/2 = 1012.5$, not integer. Next smallest $m + n = 3$ gives $d = 675$. Any larger $m + n$ gives smaller $d$. So the maximum possible $\gcd(a, b)$ is $\boxed{675}$.
Question: A paleobotanist studying ancient plant symmetry observes that a fossilized fern has a repeating pattern every $n$ fronds, where $n$ is the smallest three-digit number divisible by both 12 and 18. What is $n$?

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