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Home » Den Nenner: $24 \times 120 \times 6 = 17280$ - Dyverse

Den Nenner: $24 \times 120 \times 6 = 17280$ - Dyverse

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 Solution: Let \(d = \gcd(a, b)\), so \(a = d \cdot m\), \(b = d \cdot n\), with \(m\) and \(n\) coprime positive integers. Then:

📌 a + b = d(m + n) = 2024

📌 Thus, \(d\) must be a divisor of 2024, and \(m + n = \frac{2024}{d}\). To maximize \(d\), we minimize \(m + n\), which is minimized when \(m + n = 2\), since \(m, n \geq 1\) and coprime. The only such pair is \(m = 1\), \(n = 1\), which are coprime. Then:

📌 d = \frac{2024}{2} = 1012

📌 This is valid because \(a = 1012\), \(b = 1012\), and \(\gcd(1012, 1012) = 1012\). Therefore, the maximum possible value is \(\boxed{1012}\).

📌 Question: A cartographer uses a coordinate system where two perpendicular roads are aligned with integer-length segments \(a\) and \(b\) such that \(a + b = 2025\) and \(\gcd(a, b)\) is as large as possible. What is the largest possible value of \(\gcd(a, b)\)?

📌 Solution: Let \(d = \gcd(a, b)\), so \(a = d \cdot m\), \(b = d \cdot n\), with \(m\) and \(n\) coprime. Then:

📌 a + b = d(m + n) = 2025

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