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2025 AIME II Problems/Problem 4

Revision as of 21:33, 13 February 2025 by Dondee123 (talk | contribs) (Created page with "== Problem == The product<cmath>\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})}...")
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Problem

The product\[\prod^{63}_{k=4} \frac{\log_k (5^{k^2 - 1})}{\log_{k + 1} (5^{k^2 - 4})} = \frac{\log_4 (5^{15})}{\log_5 (5^{12})} \cdot \frac{\log_5 (5^{24})}{\log_6 (5^{21})}\cdot \frac{\log_6 (5^{35})}{\log_7 (5^{32})} \cdots \frac{\log_{63} (5^{3968})}{\log_{64} (5^{3965})}\]is equal to $\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution

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