Difference between revisions of "2006 AMC 10B Problems/Problem 18"
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== Problem == | == Problem == | ||
| − | == Solution == | + | 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 1 == | ||
| + | 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 = \boxed{\textbf{(E) }3}</math> | ||
| + | |||
| + | == Solution 2 == | ||
| + | |||
| + | <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> | ||
| + | |||
| + | ~thatmathsguy | ||
| + | |||
== See Also == | == See Also == | ||
| − | + | {{AMC10 box|year=2006|ab=B|num-b=17|num-a=19}} | |
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:33, 29 May 2023
Contents
Problem
Let 
be a sequence for which 
, 
, and 
for each positive integer 
. What is 
?
Solution 1
Looking at the first few terms of the sequence:
Clearly, the sequence repeats every 6 terms.
Since 
,
Solution 2
, so 
and because 
, so
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
