2017 AMC 10A Problems/Problem 13: Difference between revisions
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==Problem== | |||
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> | ||
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math> | <math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math> | ||
==Solution== | |||
==See Also== | |||
{{AMC10 box|year=2017|ab=A|num-b=12|num-a=14}} | |||
{{MAA Notice}} | |||
Revision as of 15:53, 8 February 2017
Problem
Define a sequence recursively by 
and 
the remainder when 
is divided by 
for all 
Thus the sequence starts 
What is 
Solution
See Also
| 2017 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
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