# 2002 AMC 12B Problems/Problem 22

## Problem

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 $\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1 \qquad\mathrm{(C)}\ \frac{1}{2002} \qquad\mathrm{(D)}\ \frac{1}{1001} \qquad\mathrm{(E)}\ \frac 12$

## 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*}

## Solution 2

Note that $\frac{1}{\log_a b}=\log_b a$. Thus $a_n=\log_{2002} n$. Also notice that if we have a log sum, we multiply, and if we have a log product, we divide. Using these properties, we get that the result is the following: $$\log_{2002}\left(\frac{2*3*4*5}{10*11*12*13*14}=\frac{1}{11*13*14}=\frac{1}{2002}\right)=\boxed{\textbf{(B)}-1}$$

~yofro

## Solution 3

Note that $a_2 = \frac{1}{\log_2 2002}$. $1$ is also equal to $\log_2 2$. So $a_2 = \frac{\log_2 2}{\log_2 2002}$. By the change of bases formula, $a_2 = \log_{2002} 2$. Following the same reasoning, $a_3 = \log_{2002} 3$, $a_4 = \log_{2002} 4$ and so on. $$b = \log_{2002} 2 + \log_{2002} 3 + .....+ \log_{2002} 5 = \log_{2002} 5! = \log_{2002} 120$$ Now solving for $c$, we see that it equals $\log_{2002} (10\cdot 11 \cdot 12 \cdot 13 \cdot 14)$ $$b-c = \log_{2002} 120 - \log_{2002} 240240 \rightarrow \log_{2002} \frac{1}{2002} = \boxed{-1}$$

~YBSuburbanTea

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 