Question

乘积 $\begin{gathered} \prod_{k=4}^{63} \frac{\log_k (5^{k^2 - 1})}{\log_{k+1} (5^{k^2 - 4})} \\ = \frac{\log_4 (5^{15})}{\log_5 (5^{12})} \\ \quad {}\cdot \frac{\log_5 (5^{24})}{\log_6 (5^{21})} \\ \quad {}\cdot \frac{\log_6 (5^{35})}{\log_7 (5^{32})} \\ \quad \cdots \frac{\log_{63} (5^{3968})}{\log_{64} (5^{3965})} \end{gathered}$ 等于 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

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 $m + n.$

答案

答案：106

知识点：对数裂项相消

难度评级：2300

小提示：

由换底公式，每个因子等于 $\frac{k^2-1}{k^2-4} \cdot \frac{\log(k+1)}{\log k}$

By change of base, each factor equals $\frac{k^2-1}{k^2-4} \cdot \frac{\log(k+1)}{\log k}$

大提示：

分解 $k^2 - 1 = (k-1)(k+1)$ 和 $k^2 - 4 = (k-2)(k+2)$ ；所有部分都会裂项相消

Factor $k^2 - 1 = (k-1)(k+1)$ and $k^2 - 4 = (k-2)(k+2);$ everything telescopes

解答：

由换底公式， $\log_k (5^{k^2-1}) = \frac{(k^2 - 1)\log 5}{\log k}$ ，所以乘积的每个因子等于 $\begin{gathered} \frac{\frac{k^2-1}{\log k}}{\frac{k^2-4}{\log(k+1)}} \\ = \frac{(k-1)(k+1)}{(k-2)(k+2)} \\ \quad {}\cdot \frac{\log(k+1)}{\log k} \end{gathered}\text{。}$

三个部分在 $k = 4, \ldots, 63$ 上都会裂项相消： $\prod_{k=4}^{63} \frac{k-1}{k-2} = \frac{62}{2} = 31\text{，}$ $\prod_{k=4}^{63} \frac{k+1}{k+2} = \frac{5}{65} = \frac{1}{13}\text{，}$ $\begin{gathered} \prod_{k=4}^{63} \frac{\log(k+1)}{\log k} \\ = \frac{\log 64}{\log 4} = 3 \end{gathered}\text{。}$

该乘积为 $31 \cdot \frac{1}{13} \cdot 3 = \frac{93}{13}$ ，已经是最简分数，所以 $m + n = 93 + 13 = 106$ 。

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 $m + n = 93 + 13 = 106.$

2025 AIME II 第 4 题

答案：106

难度评级：2300

小提示：

大提示：

解答：

其他年份的第 4 题