2002 AMC 12B Problem 22

Problem 22 of 25HarderAlgebra

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

Show solution

Solution

By change of base,

a_n=\dfrac{1}{\log_n 2002}=\log_{2002} n.

\begin{gathered} b-c \\ {}=\log_{2002}\frac{2\cdot3\cdot4\cdot5}{10\cdot11\cdot12\cdot13\cdot14}. \end{gathered}

The fraction equals

\dfrac{120}{240240}=\dfrac{1}{2002},

b-c=\log_{2002}\dfrac{1}{2002}=-1.

Thus, the correct answer is B.