2025 AIME II Problem 4

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

By change of base, each factor equals

\frac{k^2-1}{k^2-4} \cdot \frac{\log(k+1)}{\log k}

Factor

k^2 - 1 = (k-1)(k+1)

and

k^2 - 4 = (k-2)(k+2);

everything telescopes

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{\frac{k^2-1}{\log k}}{\frac{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.

Problem 4 in other years