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Home » But ENIAC: 10 × 2.5 = 25 - 500apps

But ENIAC: 10 × 2.5 = 25 - 500apps

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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📌 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

📌 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

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