2025 AIME II Problems/Problem 4

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:

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

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: $\begin{array}{r} \frac{\log_{4} (5^{15})}{\log_{5} (5^{12})} \cdot \frac{\log_{5} (5^{24})}{\log_{6} (5^{21})} \cdot \frac{\log_{6} (5^{35})}{\log_{7} (5^{32})} \dots = \frac{15 \log_{4} (5)}{12 \log_{5} (5)} \cdot \frac{24 \log_{5} (5)}{21 \log_{6} (5)} \cdot \frac{35 \log_{6} (5)}{32 \log_{7} (5)} \dots \end{array}$ Now we multiply the logs and fractions separately. Let's do it for the logs first: $\begin{array}{r} \frac{\log_{4} (5)}{\log_{5} (5)} \cdot \frac{\log_{5} (5)}{\log_{6} (5)} \cdot \frac{\log_{6} (5)}{\log_{7} (5)} \dots \frac{\log_{63} (5)}{\log_{64} (5)} = \frac{\log_{4} (5)}{\log_{64} (5)} = 3 \end{array}$ Now fractions: $\begin{array}{r} \frac{15}{12} \cdot \frac{24}{21} \cdot \frac{35}{32} \dots = \frac{3 \cdot 5}{2 \cdot 6} \cdot \frac{4 \cdot 6}{3 \cdot 7} \cdot \frac{5 \cdot 7}{4 \cdot 8} \dots \frac{62 \cdot 64}{61 \cdot 65} = \frac{5}{2} \cdot \frac{62}{65} = \frac{31}{13} \end{array}$ 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:

$\begin{aligned} \prod_{k = 4}^{63} \frac{(k - 1) (k + 1) \log_{k} 5}{(k - 2) (k + 2) \log_{k + 1} 5} & = \prod_{k = 4}^{63} \frac{\log_{k} 5}{\log_{k + 1} 5} \cdot \frac{3 \cdot 5 \cdot 4 \cdot 6 \cdot 5 \cdot 7 \dots 62 \cdot 64}{2 \cdot 6 \cdot 3 \cdot 7 \cdot 4 \cdot 8 \dots 61 \cdot 65} \\ = \frac{\log_{4} 5 \cdot \log_{5} 5 \dots \log_{63} 5}{\log_{5} 5 \cdot \log_{6} 5 \dots \log_{64} 5} \cdot \frac{3 \cdot 4 \cdot (5^{2} \cdot 6^{2} \dots 62^{2}) \cdot 63 \cdot 64}{2 \cdot 3 \cdot 4 \cdot 5 \cdot (6^{2} \cdot 7^{2} \dots 61^{2}) \cdot 62 \cdot 63 \cdot 64 \cdot 65} \\ = \frac{\log_{4} 5}{\log_{64} 5} \cdot \frac{3 \cdot 4 \cdot 5^{2} \cdot (6^{2} \dots 61^{2}) \cdot 62^{2} \cdot 63 \cdot 64}{2 \cdot 3 \cdot 4 \cdot 5 \cdot (6^{2} \dots 61^{2}) \cdot 62 \cdot 63 \cdot 64 \cdot 65} \\ = \log_{64} 4 \cdot \frac{3 \cdot 4 \cdot 5 \cdot 62}{2 \cdot 65} \\ = 3 \cdot \frac{31}{13} \\ = \frac{93}{13} \end{aligned}$

Thus the answer is $93+13=\boxed{106}$ .

~ eevee9406

~ Edited by aoum

Video Solution

https://youtu.be/Mt_vVxWCo3k

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

Page

Toolbox

Search