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2017 AMC 10A Problem 13

Problem 13 of 25IntermediateAlgebraNumber TheoryProblem-Solving Techniques

Define a sequence recursively by F0=0,F_{0}=0,  F1=1,~F_{1}=1, and Fn=F_{n}= the remainder when Fn−1+Fn−2F_{n-1}+F_{n-2} is divided by 3,3, for all n≥2.n\geq 2. Thus the sequence starts 0,0, 1,1, 1,1, 2,2, 0,0, 2,2, ….\ldots. What is F2017+F2018+F2019+F2020+F_{2017}+F_{2018}+F_{2019}+F_{2020}+F2021+F2022+F2023+F2024?F_{2021}+F_{2022}+F_{2023}+F_{2024}?

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Solution

Let us list out the first few values to see if we can find a pattern in this sequence. 0,1,1,2,0,2,2,1,0,1,⋯ 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, \cdots The pair (F8,F9)=(0,1)=(F0,F1).(F_8,F_9)=(0,1)=(F_0,F_1). Because each later term is determined by the preceding two terms, the sequence therefore repeats every 88 terms. The desired answer is the sum of 88 consecutive numbers, which is fixed. This sum is 0+1+1+2+0+2 0 + 1 + 1 + 2 + 0 + 2+2+1=9. + 2 + 1 = 9. Thus, D is the correct answer.
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Tagged: Fibonacci · modular arithmetic · pattern recognition

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