Difference between revisions of "2025 AIME II Problems/Problem 4"
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Multiplying these together gets us the original product, which is <math>\frac{31}{13} \cdot 3 = \frac{93}{13}</math>.\\ | Multiplying these together gets us the original product, which is <math>\frac{31}{13} \cdot 3 = \frac{93}{13}</math>.\\ | ||
Thus <math>m+n=\boxed{106}</math>. | Thus <math>m+n=\boxed{106}</math>. | ||
| + | |||
| + | == Solution 3 == | ||
| + | |||
| + | Using logarithmic identities and the change of base formula, the product can be rewritten as <cmath>\prod_{k=4}^{63}\frac{k^2-1}{k^2-4}\frac{\log(k+1)}{\log(k)}</cmath>. Then we can separate this into two series. | ||
| + | The latter series is a telescoping series, and it can be pretty easily evaluated to be <math>\frac{\log(64)}{\log(4)}=3</math>. The former can be factored as <math>\frac{(k-1)(k+1)}{(k-2)(k+2)}</math>, and writing out the first terms could tell us that this is a telescoping series as well. Cancelling out the terms would yield <math>\frac{5}{2}\cdot\frac{62}{65}=\frac{31}{13}</math>. | ||
| + | Multiplying the two will give us <math>\frac{93}{13}</math>, which tells us that the answer is <math>\boxed{106}</math>. | ||
==See also== | ==See also== | ||
Revision as of 22:15, 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
Solution 2
We can move the exponents to the front of the logarithms like this:
\begin{align*}
\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{15\log_4 (5)}{12\log_5 (5)} \cdot \frac{24\log_5 (5)}{21\log_6 (5)}\cdot \frac{35\log_6 (5)}{32\log_7 (5)} \cdots
\end{align*}
Now we multiply the logs and fractions seperately.\\
Let's do it for the logs first:
\begin{align*}
\frac{\log_4 (5)}{\log_5 (5)} \cdot \frac{\log_5 (5)}{\log_6 (5)}\cdot \frac{\log_6 (5)}{\log_7 (5)} \cdots \frac{\log_{63} (5)}{\log_{64} (5)} = \frac{\log_4 (5)}{\log_{64} (5)} = 3
\end{align*}
Now fractions:
\begin{align*}
\frac{15}{12} \cdot \frac{24}{21} \cdot \frac{35}{32} \cdots = \frac{3\cdot 5}{2\cdot 6} \cdot \frac{4\cdot 6}{3\cdot 7} \cdot \frac{5\cdot 7}{4\cdot 8} \cdots \frac{62\cdot 64}{61\cdot 65} = \frac{5}{2} \cdot \frac{62}{65} = \frac{31}{13}
\end{align*}
Multiplying these together gets us the original product, which is 
.\\
Thus 
.
Solution 3
Using logarithmic identities and the change of base formula, the product can be rewritten as ![\[\prod_{k=4}^{63}\frac{k^2-1}{k^2-4}\frac{\log(k+1)}{\log(k)}\]](http://latex.artofproblemsolving.com/6/2/4/624b094770bd79bcf4a4e66d85031039a2a23fc4.png)
. Then we can separate this into two series.
The latter series is a telescoping series, and it can be pretty easily evaluated to be 
. The former can be factored as 
, and writing out the first terms could tell us that this is a telescoping series as well. Cancelling out the terms would yield 
.
Multiplying the two will give us 
, which tells us that the answer is 
.
See also
| 2025 AIME II (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
