Question

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

Answer

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 $\frac{(k^2-1)/\log k}{(k^2-4)/\log(k+1)} = \frac{(k-1)(k+1)}{(k-2)(k+2)} \cdot \frac{\log(k+1)}{\log k}.$

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

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

Solution:

Problem 4 in Other Years