Difference between revisions of "2017 AMC 10A Problems/Problem 13"

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Define a sequence recursively by <math>F_{0}=0,~F_{1}=1,</math> and <math>F_{n}=</math> the remainder when <math>F_{n-1}+F_{n-2}</math> is divided by <math>3,</math> for all <math>n\geq 2.</math> Thus the sequence starts <math>0,1,1,2,0,2,\ldots</math> What is <math>F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?</math>
 
Define a sequence recursively by <math>F_{0}=0,~F_{1}=1,</math> and <math>F_{n}=</math> the remainder when <math>F_{n-1}+F_{n-2}</math> is divided by <math>3,</math> for all <math>n\geq 2.</math> Thus the sequence starts <math>0,1,1,2,0,2,\ldots</math> What is <math>F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?</math>
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<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math>

Revision as of 16:51, 8 February 2017

Define a sequence recursively by $F_{0}=0,~F_{1}=1,$ and $F_{n}=$ the remainder when $F_{n-1}+F_{n-2}$ is divided by $3,$ for all $n\geq 2.$ Thus the sequence starts $0,1,1,2,0,2,\ldots$ What is $F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?$

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$