AMC12 2002 B

AMC12 2002 B · Q22

AMC12 2002 B · Q22. It mainly tests Logarithms (rare).

For all integers $n$ greater than $1$, define $a_n = \frac{1}{\log_n 2002}$. Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$. Then $b- c$ equals

对所有大于 $1$ 的整数 $n$，定义 $a_n = \frac{1}{\log_n 2002}$。令 $b = a_2 + a_3 + a_4 + a_5$，$c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$。则 $b- c$ 等于

(A) -2 -2

(B) -1 -1

(D) \frac{1}{1001} \frac{1}{1001}

(E) \frac{1}{2} \frac{1}{2}

Answer

Correct choice: (B)

正确答案：（B）

Solution

By the change of base formula, $a_n = \frac{1}{\frac{\log 2002}{\log n}} = \left(\frac{1}{\log 2002}\right) \log n$. Thus \begin{align*}b- c &= \left(\frac{1}{\log 2002}\right)(\log 2 + \log 3 + \log 4 + \log 5 - \log 10 - \log 11 - \log 12 - \log 13 - \log 14)\\ &= \left(\frac{1}{\log 2002}\right)\left(\log \frac{2 \cdot 3 \cdot 4 \cdot 5}{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}\right)\\ &= \left(\frac{1}{\log 2002}\right) \log 2002^{-1} = -\left(\frac{\log 2002}{\log 2002}\right) = -1 \Rightarrow \mathrm{(B)}\end{align*}

由换底公式，$a_n = \frac{1}{\frac{\log 2002}{\log n}} = \left(\frac{1}{\log 2002}\right) \log n$。因此 \begin{align*}b- c &= \left(\frac{1}{\log 2002}\right)(\log 2 + \log 3 + \log 4 + \log 5 - \log 10 - \log 11 - \log 12 - \log 13 - \log 14)\\ &= \left(\frac{1}{\log 2002}\right)\left(\log \frac{2 \cdot 3 \cdot 4 \cdot 5}{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}\right)\\ &= \left(\frac{1}{\log 2002}\right) \log 2002^{-1} = -\left(\frac{\log 2002}{\log 2002}\right) = -1 \Rightarrow \mathrm{(B)}\end{align*}

Topics

AL.015 Logarithms (rare)