Home Competitions AIME 2025b Problem 4
The product∏ k = 4 63 log k ( 5 k 2 − 1 ) log k + 1 ( 5 k 2 − 4 ) = log 4 ( 5 15 ) log 5 ( 5 12 ) ⋅ log 5 ( 5 24 ) log 6 ( 5 21 ) ⋅ log 6 ( 5 35 ) log 7 ( 5 32 ) ⋯ log 63 ( 5 3968 ) log 64 ( 5 3965 ) \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})} k = 4 ∏ 63 log k + 1 ( 5 k 2 − 4 ) log k ( 5 k 2 − 1 ) = log 5 ( 5 12 ) log 4 ( 5 15 ) ⋅ log 6 ( 5 21 ) log 5 ( 5 24 ) ⋅ log 7 ( 5 32 ) log 6 ( 5 35 ) ⋯ log 64 ( 5 3965 ) log 63 ( 5 3968 ) is equal to m n , \tfrac mn, n m , where m m m and n n n are relatively prime positive integers. Find m + n . m + n. m + n .
Dataset text AIME 1983 to 2026 . Quoted with source.
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