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📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 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

📌 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:

📌 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 $n$?

📌 Solution: We seek the smallest three-digit number divisible by both 12 and 18, i.e., their least common multiple.

📌 $12 = 2^2 \cdot 3$, $18 = 2 \cdot 3^2$, so:

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