2025 AIME II Problems/Problem 4: Difference between revisions
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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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})}{\log_6 (5^{21})}\cdot \frac{\log_6 (5^{35})}{\log_7 (5^{32})} \cdots \frac{\log_{63} (5^{3968})}{\log_{64} (5^{3965})}</cmath>is equal to <math>\tfrac mn,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n.</math> | 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})}{\log_6 (5^{21})}\cdot \frac{\log_6 (5^{35})}{\log_7 (5^{32})} \cdots \frac{\log_{63} (5^{3968})}{\log_{64} (5^{3965})}</cmath>is equal to <math>\tfrac mn,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n.</math> | ||
== Solution == | == 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 | |||
Revision as of 22:05, 13 February 2025
Problem
The product![\[\prod^{63}_{k=4} \frac{\log_k (5^{k^2 - 1})}{\log_{k + 1} (5^{k^2 - 4})} = \frac{\log_4 (5^{15})}{\log_5 (5^{12})} \cdot \frac{\log_5 (5^{24})}{\log_6 (5^{21})}\cdot \frac{\log_6 (5^{35})}{\log_7 (5^{32})} \cdots \frac{\log_{63} (5^{3968})}{\log_{64} (5^{3965})}\]](http://latex.artofproblemsolving.com/1/8/a/18ad3a97e3df2da4f41cbdf62f07db5272a9948b.png)
is equal to 
where 
and 
are relatively prime positive integers. Find 
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
