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

Solution 1

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

$a_n = \frac{1}{\log_n 2002} = \frac{\log_n n}{log_n 2002} = \log_(2002) n$ So

\begin{align*}
b- c &= \left(log_2002 2 + log_2002 3 + log_2002 4 + log_2002 5 - log_2002 10 - log_2002 11 - log_2002 12 - log_2002 13 - log_2002 14))\\
&= \left(log_2002 (\frac{2 \cdot 3 \cdot 4 \cdot 5}{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}))\\ 
&= \left(log_2002 2002^-1) = -1 \Rightarrow \mathrm{(B)}
\end{align*} (Error compiling LaTeX. Unknown error_msg)

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2002 AMC 12B Problems/Problem 22

Contents

Problem

Solution

Solution 1

Solution 2

See also