2006 AMC 10B Problem 18

Attempt Problem 18 of the 2006 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10B solutions, or check the answer key.

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18.

Let $a_1,a_2,\ldots$ be a sequence for which $a_1=2,$ $a_2=3,$ and $a_n=\dfrac{a_{n-1}}{a_{n-2}}$ for each positive integer $n\ge 3.$ What is $a_{2006}?$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{2}$

$2$

$3$

Answer: E

Concepts:recursion pattern recognition

Difficulty rating: 1280

Solution:

The terms are $2,\,3,\,\tfrac32,\,\tfrac12,\,\tfrac13,\,\tfrac23,$ then $2,\,3,\ldots,$ a cycle of length $6.$

Since $2006=6\cdot334+2,$ we have $a_{2006}=a_2=3.$

Thus, the correct answer is E.

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