Difference between revisions of "2025 AIME II Problems/Problem 4"
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= 15/12 * 24/21 * 35/32 * ... * 3968/3965 * \log_4 5 / \log_64 5 | = 15/12 * 24/21 * 35/32 * ... * 3968/3965 * \log_4 5 / \log_64 5 | ||
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= \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) | = \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) | ||
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= 3 * 5 * 3 * 6 * 4 * ... * 64 * 62 / 6 * 2 * 7 * 3 * ... * 65 * 61 | = 3 * 5 * 3 * 6 * 4 * ... * 64 * 62 / 6 * 2 * 7 * 3 * ... * 65 * 61 | ||
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= 3 * 5 * 62 / 65 * 2 | = 3 * 5 * 62 / 65 * 2 | ||
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= 3 * 5 * 2 * 31 / 5 * 13 * 2 | = 3 * 5 * 2 * 31 / 5 * 13 * 2 | ||
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= 3 * 31 / 13 | = 3 * 31 / 13 | ||
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= 93/13 | = 93/13 | ||
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Desired answer: 93 + 13 = 106 | Desired answer: 93 + 13 = 106 | ||
(Feel free to correct any latexes and formats) | (Feel free to correct any latexes and formats) | ||
~Mitsuihisashi14 | ~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 formats) ~Mitsuihisashi14
