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2006 AMC 10B Problem 18

Problem 18 of 25IntermediateAlgebraProblem-Solving Techniques

Let a1,a_1, a2,a_2, …\ldots be a sequence for which a1=2,a_1=2, a2=3,a_2=3, and an=an−1an−2a_n=\dfrac{a_{n-1}}{a_{n-2}} for each positive integer n≥3.n\ge 3. What is a2006?a_{2006}?

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Solution

The terms are 2, 3, 32, 12, 13, 23,2,\,3,\,\tfrac32,\,\tfrac12,\,\tfrac13,\,\tfrac23, then 2, 3,…,2,\,3,\ldots, a cycle of length 6.6. Since 2006=6⋅334+2,2006=6\cdot334+2, we have a2006=a2=3.a_{2006}=a_2=3. Thus, the correct answer is E.
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