2006 AMC 10B Problem 18

Below is the professionally curated solution for Problem 18 of the 2006 AMC 10B, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:recursion pattern recognition

Difficulty rating: 1280

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 $a_{2006}?$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{2}$

$2$

$3$

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.

← Problem 17 Full Exam Problem 19 →

Problem 18 in Other Years

2000 AMC 10 · 2001 AMC 10 · 2002 AMC 10A · 2002 AMC 10B · 2003 AMC 10A · 2003 AMC 10B · 2004 AMC 10A · 2004 AMC 10B · 2005 AMC 10A · 2005 AMC 10B · 2006 AMC 10A · 2007 AMC 10A · 2007 AMC 10B · 2008 AMC 10A · 2008 AMC 10B · 2009 AMC 10A · 2009 AMC 10B · 2010 AMC 10A · 2010 AMC 10B · 2011 AMC 10A · 2011 AMC 10B · 2012 AMC 10A · 2012 AMC 10B · 2013 AMC 10A · 2013 AMC 10B · 2014 AMC 10A · 2014 AMC 10B · 2015 AMC 10A · 2015 AMC 10B · 2016 AMC 10A · 2016 AMC 10B · 2017 AMC 10A · 2017 AMC 10B · 2018 AMC 10A · 2018 AMC 10B · 2019 AMC 10A · 2019 AMC 10B · 2020 AMC 10A · 2020 AMC 10B · 2021 AMC 10A Spring · 2021 AMC 10B Spring · 2021 AMC 10A Fall · 2021 AMC 10B Fall · 2022 AMC 10A · 2022 AMC 10B · 2023 AMC 10A · 2023 AMC 10B · 2024 AMC 10A · 2024 AMC 10B · 2025 AMC 10A · 2025 AMC 10B