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

Revision as of 20:43, 13 July 2006

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 $a_{2006}$ ?

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

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

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

@@ Line 1: / Line 1: @@
 == 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>?
+<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>
 == Solution ==
+Looking at the first few terms of the sequence:
+<math> 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 , .... </math>
+Clearly, the sequence repeats every 6 terms.
+Since <math> 2006 \equiv 2\bmod{6}</math>,
+<math> a_{2006} = a_2 = 3 \Rightarrow E </math>
 == See Also ==
 *[[2006 AMC 10B Problems]]