2025 AIME II Problems/Problem 4

Revision as of 22:05, 13 February 2025 by Mitsuihisashi14 (talk | contribs) (Solution)

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 1

We can rewrite the equation as:

= 15/12 * 24/21 * 35/32 * ... * 3968/3965 * \log_4 5 / \log_64 5 = \log_4 64 * (4+1)(4-1)(5+1)(5-1)* ... * (63+1)(63-1)/(4+2)(4-2)(5+2)(5-2)* ... * (63+2)(63-2) = 3 * 5 * 3 * 6 * 4 * ... * 64 * 62 / 6 * 2 * 7 * 3 * ... * 65 * 61 = 3 * 5 * 62 / 65 * 2 = 3 * 5 * 2 * 31 / 5 * 13 * 2 = 3 * 31 / 13 = 93/13 Desired answer: 93 + 13 = 106

(Feel free to correct any latexes and formattings) ~Mitsuihisashi14