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

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(Solution)
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Since <math> 2006 \equiv 2\bmod{6}</math>,
 
Since <math> 2006 \equiv 2\bmod{6}</math>,
  
<math> a_{2006} = a_2 = 3 \Rightarrow E </math>
+
<math> a_{2006} = a_2 = \boxed{\textbf{(E) }3}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 13:51, 7 March 2022

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

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

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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