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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

18.

Let a1,a_1, a2,a_2, …\ldots be a sequence for which a1=2,a_1=2, a2=3,a_2=3, and an=an−1an−2a_n=\dfrac{a_{n-1}}{a_{n-2}} for each positive integer n≥3.n\ge 3. What is a2006?a_{2006}?
12\dfrac{1}{2}
23\dfrac{2}{3}
32\dfrac{3}{2}
22
33
Answer: E
Concepts:recursionpattern recognition
Small hint:
Compute the first several terms to detect a repeating cycle
Big hint:
The sequence repeats every 66 terms; reduce 20062006 modulo 66
Solution:
The terms are 2, 3, 32, 12, 13, 23,2,\,3,\,\tfrac32,\,\tfrac12,\,\tfrac13,\,\tfrac23, then 2, 3,…,2,\,3,\ldots, a cycle of length 6.6.
Since 2006=6⋅334+2,2006=6\cdot334+2, we have a2006=a2=3.a_{2006}=a_2=3.
Thus, the correct answer is E.
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