To maximize $ d $, $ m + n $ must be minimized. Since $\gcd(m, n) = 1$, the smallest possible value for $ m + n $ is 2, achieved when $ m = 1 $ and $ n = 1 $. Thus: - AMAZONAWS

📅 March 6, 2026 👤 scraface

Mar 06, 2026

$To maximize $ d $, $ m + n $ must be minimized. Since $\gcd(m, n) = 1$, the smallest possible value for $ m + n $ is 2, achieved when $ m = 1 $ and $ n = 1 $. Thus:$

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