Question

The product $\begin{gathered} \prod_{k=4}^{63} \frac{\log_k (5^{k^2 - 1})}{\log_{k+1} (5^{k^2 - 4})} \\ = \frac{\log_4 (5^{15})}{\log_5 (5^{12})} \\ \quad {}\cdot \frac{\log_5 (5^{24})}{\log_6 (5^{21})} \\ \quad {}\cdot \frac{\log_6 (5^{35})}{\log_7 (5^{32})} \\ \quad \cdots \frac{\log_{63} (5^{3968})}{\log_{64} (5^{3965})} \end{gathered}$ is equal to $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Answer: 106

Concepts:logarithm telescoping

Difficulty rating: 2300

Solution:

By the change-of-base formula, $\log_k (5^{k^2-1}) = \frac{(k^2 - 1)\log 5}{\log k},$ so each factor of the product equals $\begin{gathered} \frac{(k^2-1)/\log k}{(k^2-4)/\log(k+1)} \\ = \frac{(k-1)(k+1)}{(k-2)(k+2)} \\ \quad {}\cdot \frac{\log(k+1)}{\log k}. \end{gathered}$

All three pieces telescope over $k = 4, \ldots, 63:$ $\prod_{k=4}^{63} \frac{k-1}{k-2} = \frac{62}{2} = 31,$ $\prod_{k=4}^{63} \frac{k+1}{k+2} = \frac{5}{65} = \frac{1}{13},$ $\begin{gathered} \prod_{k=4}^{63} \frac{\log(k+1)}{\log k} \\ = \frac{\log 64}{\log 4} = 3. \end{gathered}$

The product is $31 \cdot \frac{1}{13} \cdot 3 = \frac{93}{13},$ which is in lowest terms, so $m + n = 93 + 13 = 106.$

2025 AIME II Problem 4

Answer: 106

Difficulty rating: 2300

Solution:

Problem 4 in Other Years