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AIME 2025b Problem 4

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The product∏k=463log⁡k(5k2−1)log⁡k+1(5k2−4)=log⁡4(515)log⁡5(512)⋅log⁡5(524)log⁡6(521)⋅log⁡6(535)log⁡7(532)⋯log⁡63(53968)log⁡64(53965)\prod^{63}_{k=4} \frac{\log_k (5^{k^2 - 1})}{\log_{k + 1} (5^{k^2 - 4})} = \frac{\log_4 (5^{15})}{\log_5 (5^{12})} \cdot \frac{\log_5 (5^{24})}{\log_6 (5^{21})}\cdot \frac{\log_6 (5^{35})}{\log_7 (5^{32})} \cdots \frac{\log_{63} (5^{3968})}{\log_{64} (5^{3965})}is equal to mn,\tfrac mn, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Dataset text AIME 1983 to 2026. Quoted with source.

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