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

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== Problem ==
 
== Problem ==
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Let <math> a_1 , a_2 , ... </math> be a sequence for which
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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>?
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<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>.  
 
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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>
  
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== Solution ==
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== Solution 1 ==
 
Looking at the first few terms of the sequence:
 
Looking at the first few terms of the sequence:
  
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Since <math> 2006 \equiv 2\bmod{6}</math>,
 
Since <math> 2006 \equiv 2\bmod{6}</math>,
  
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<math> a_{2006} = a_2 = 3 \Rightarrow E </math>
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<math> a_{2006} = a_2 = \boxed{\textbf{(E) }3}</math>
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== Solution 2 ==
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<math> 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}} </math> , so <math> a_n = a_{n-6} </math> and because <math> 2006 = 2 + 334 \times 6 </math> , so <math> a_{2006} = a_2 = \boxed{\textbf{(E) }3}</math>
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~thatmathsguy
  
 
== See Also ==
 
== See Also ==
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*[[2006 AMC 10B Problems]]
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{{AMC10 box|year=2006|ab=B|num-b=17|num-a=19}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 00:33, 29 May 2023

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 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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