Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2025 AIME II Problems/Problem 4"

(→‎Solution 1)
(→‎Solution 2)
Line 42: Line 42:
 
\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}
 
\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*}
 
\end{align*}
−
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>.
 +
 +
~ Edited by [https://artofproblemsolving.com/wiki/index.php/User:Aoum aoum]
  
 
== Solution 3 ==
 
== Solution 3 ==

Revision as of 16:19, 14 February 2025

Problem

The product\[\prod^{63}_{k=4} \frac{\log_k (5^{k^2 - 1})}{\log_{k + 1} (5^{k^2 - 4})} = \frac{\log_4 (5^{15})}{\log_5 (5^{12})} \cdot \frac{\log_5 (5^{24})}{\log_6 (5^{21})}\cdot \frac{\log_6 (5^{35})}{\log_7 (5^{32})} \cdots \frac{\log_{63} (5^{3968})}{\log_{64} (5^{3965})}\]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:

\begin{align*} &= \dfrac{15}{12} \cdot \dfrac{24}{21} \cdot \dfrac{35}{32} \cdot \dots \cdot \dfrac{3968}{3965} \cdot \dfrac{\log_4{5}}{\log_{64}{5}} \\ &= \log_4{64} \cdot \dfrac{(4+1)(4-1)(5+1)(5-1)\cdot \dots \cdot (63+1)(63-1)}{(4+2)(4-2)(5+2)(5-2)\cdot \dots \cdot (63+2)(63-2)} \\ &= 3 \cdot \dfrac{5 \cdot 3 \cdot 6 \cdot 4 \cdot \dots \cdot 64 \cdot 62}{6 \cdot 2 \cdot 7 \cdot 3 \cdot \dots \cdot 65 \cdot 61} \\ &= 3 \cdot \dfrac{5 \cdot 62}{65 \cdot 2} \\ &= 3 \cdot \dfrac{5 \cdot 2 \cdot 31}{5 \cdot 13 \cdot 2} \\ &= 3 \cdot \dfrac{31}{13} \\ &= \dfrac{93}{13} \end{align*}


Desired answer: $93 + 13 = \boxed{106}$

(Feel free to correct any $\LaTeX$ and formatting.)

~ Mitsuihisashi14

~ $\LaTeX$ by Tacos_are_yummy_1

~ Additional edits by aoum

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 $\frac{31}{13} \cdot 3 = \frac{93}{13}$. Thus $m+n=\boxed{106}$.

~ Edited by aoum

Solution 3

Using logarithmic identities and the change of base formula, the product can be rewritten as \[\prod_{k=4}^{63}\frac{k^2-1}{k^2-4}\frac{\log(k+1)}{\log(k)}\]. Then we can separate this into two series. The latter series is a telescoping series, and it can be pretty easily evaluated to be $\frac{\log(64)}{\log(4)}=3$. The former can be factored as $\frac{(k-1)(k+1)}{(k-2)(k+2)}$, and writing out the first terms could tell us that this is a telescoping series as well. Cancelling out the terms would yield $\frac{5}{2}\cdot\frac{62}{65}=\frac{31}{13}$. Multiplying the two will give us $\frac{93}{13}$, which tells us that the answer is $\boxed{106}$.

Solution 4 (thorough)

The product is equal to $\prod^{63}_{k=4} \frac{(k-1)(k+1)\log_k 5}{(k-2)(k+2)\log_{k + 1} 5}$ from difference of squares and properties of logarithms. We can now expand:

\[\prod^{63}_{k=4} \frac{(k-1)(k+1)\log_k 5}{(k-2)(k+2)\log_{k + 1} 5}=\prod^{63}_{k=4}\frac{\log_k5}{\log_{k+1}5}\cdot\frac{3\cdot5\cdot4\cdot6\cdot5\cdot7\cdots62\cdot64}{2\cdot6\cdot3\cdot7\cdot4\cdot8\cdots61\cdot65}\] \[=\frac{\log_4 5\cdot\log_5 5\cdots\log_{63}5}{\log_{5}5\cdot\log_6 5\cdots\log_{64} 5}\cdot\frac{3\cdot4\cdot(5^2\cdot6^2\cdots62^2)\cdot63\cdot64}{2\cdot3\cdot4\cdot5\cdot(6^2\cdot7^2\cdots61^2)\cdot62\cdot63\cdot64\cdot65}\] \[=\frac{\log_4 5}{\log_{64} 5}\cdot\frac{3\cdot4\cdot5^2\cdot(6^2\cdots61^2)\cdot62^2\cdot63\cdot64}{2\cdot3\cdot4\cdot5\cdot(6^2\cdots61^2)\cdot62\cdot63\cdot64\cdot65}\] \[=\log_{64}4\cdot\frac{3\cdot4\cdot5^2\cdot62^2\cdot63\cdot64}{2\cdot3\cdot4\cdot5\cdot62\cdot63\cdot64\cdot65}\] \[=3\cdot\frac{5\cdot62}{2\cdot65}\] \[=3\cdot\frac{31}{13}\] \[=\frac{93}{13}\] Thus the answer is $93+13=\boxed{106}$. ~eevee9406

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2025_AIME_II_Problems/Problem_4&oldid=242567"