Difference between revisions of "2021 AMC 12A Problems/Problem 8"

(Solution)
(Solution)
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<math>\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
 
<math>\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
D\textsubscript{0}&D\textsubscript{1}&D_2&D_3&D_4&D_5&D_6&D_7&D_8&D_9\\\hline
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D0&D1&D2&D3&D4&D5&D6&D7&D8&D9\\\hline
 
0&0&1&1&1&2&3&4&6&9\\\hline
 
0&0&1&1&1&2&3&4&6&9\\\hline
 
E&E&O&O&O&E&O&E&E&O
 
E&E&O&O&O&E&O&E&E&O

Revision as of 16:00, 11 February 2021

Problem

A sequence of numbers is defined by $D_0=0,D_0=1,D_2=1$ and $D_n=D_{n-1}+D_{n-3}$ for $n\ge 3$. What are the parities (evenness or oddness) of the triple of numbers $(D_{2021},D_{2022},D_{2023})$, where $E$ denotes even and $O$ denotes odd?

$\textbf{(A) }(O,E,O) \qquad \textbf{(B) }(E,E,O) \qquad \textbf{(C) }(E,O,E) \qquad \textbf{(D) }(O,O,E) \qquad \textbf{(E) }(O,O,O)$

Solution

Making a small chart, we have

$\begin{tabular}{c|c|c|c|c|c|c|c|c|c} D0&D1&D2&D3&D4&D5&D6&D7&D8&D9\\\hline 0&0&1&1&1&2&3&4&6&9\\\hline E&E&O&O&O&E&O&E&E&O \end{tabular}$

This starts repeating every 7 terms, so $D_{2021}=D_5=E$, $D_{2022}=D_6=O$, and $D_{2023}=D_7=E$. Thus, the answer is $\boxed{\textbf{(C) }(E, O, E)}$ ~JHawk0224

Video Solution by Hawk Math

https://www.youtube.com/watch?v=P5al76DxyHY

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 12 Problems and Solutions

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