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

Revision as of 10:22, 24 December 2020

Problem

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 $F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?$

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

Solution

A pattern starts to emerge as the function is continued. The repeating pattern is $0,1,1,2,0,2,2,1\ldots$ The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating sequence, we just need to find the sum of the numbers in the sequence, which is $\boxed{\textbf{(D)}\ 9}$

Video Solution

https://youtu.be/s4vnGlwwHHw

https://youtu.be/Y09BGCES_PA

~savannahsolver

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

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

Revision as of 10:22, 24 December 2020

Contents

Problem

Solution

Video Solution

See Also

@@ Line 11: / Line 11: @@
 ==Video Solution==
 https://youtu.be/s4vnGlwwHHw
+https://youtu.be/Y09BGCES_PA
+~savannahsolver
 ==See Also==