For all integers n greater than 1, define a_n=(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

2002 AMC 12B Problem 22 — Solution

2002 AMC 12B Problem 22

Below is the professionally curated solution for Problem 22 of the 2002 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12B solutions, or check the answer key.

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Concepts:logarithm

Difficulty rating: 1630

22.

For all integers $n$ greater than $1,$ define $a_n=\dfrac{1}{\log_n 2002}.$ 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

$-2$

$-1$

$\dfrac{1}{2002}$

$\dfrac{1}{1001}$

$\dfrac{1}{2}$

Solution:

By change of base, $a_n=\dfrac{1}{\log_n 2002}=\log_{2002} n.$ So $b-c=\log_{2002}\frac{2\cdot3\cdot4\cdot5}{10\cdot11\cdot12\cdot13\cdot14}.$

The fraction equals $\dfrac{120}{240240}=\dfrac{1}{2002},$ so $b-c=\log_{2002}\dfrac{1}{2002}=-1.$

Thus, the correct answer is B.

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