# 2006 AMC 10B Problems/Problem 18

## Problem

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

$\mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{3}{2}\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } 3$

## Solution 1

Looking at the first few terms of the sequence:

$a_1=2 , a_2=3 , a_3=\frac{3}{2}, a_4=\frac{1}{2} , a_5=\frac{1}{3} , a_6=\frac{2}{3} , a_7=2 , a_8=3 , ....$

Clearly, the sequence repeats every 6 terms.

Since $2006 \equiv 2\bmod{6}$,

$a_{2006} = a_2 = \boxed{\textbf{(E) }3}$

## Solution 2

$a_n = \frac{a_{n-1}}{a_{n-2}} = \frac{\frac{a_{n-2}}{a_{n-3}}}{a_{n-2}} = \frac{1}{a_{n-3}}$ , so $a_n = a_{n-6}$ and because $2006 = 2 + 334 \times 6$ , so $a_{2006} = a_2 = \boxed{\textbf{(E) }3}$

~thatmathsguy

## See Also

 2006 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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