2025 AIME II Problems/Problem 4

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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 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

Solution 2

We can move the exponents to the front of the logarithms like this: log4(515)log5(512)log5(524)log6(521)log6(535)log7(532)=15log4(5)12log5(5)24log5(5)21log6(5)35log6(5)32log7(5) Now we multiply the logs and fractions seperately.\ Let's do it for the logs first: log4(5)log5(5)log5(5)log6(5)log6(5)log7(5)log63(5)log64(5)=log4(5)log64(5)=3 Now fractions: 151224213532=35264637574862646165=526265=3113 Multiplying these together gets us the original product, which is $\frac{31}{13} \cdot 3 = \frac{93}{13}$.\ Thus $m+n=\boxed{106}$.

See also

2025 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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