Difference between revisions of "2006 AMC 10B Problems/Problem 18"

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

Latest revision as of 23:38, 10 January 2018

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

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

$a_{2006} = a_2 = 3 \Rightarrow E$

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 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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