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

(→‎Solution)
(→‎Solution)
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==Solution==
 
==Solution==
 
Making a small chart, we have
 
Making a small chart, we have
−
<math>\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
+
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
 
D_0&D_1&D_2&D_3&D_4&D_5&D_6&D_7&D_8&D_9\\\hline
 
D_0&D_1&D_2&D_3&D_4&D_5&D_6&D_7&D_8&D_9\\\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
−
\end{tabular}</math>
+
\end{tabular}
 
This starts repeating every 7 terms, so <math>D_{2021}=D_5=</math> E, <math>D_{2022}=D_6=</math> O, and <math>D_{2023}=D_7=</math> E. Thus, the answer is \boxed{C) \text{(E, O, E)}}
 
This starts repeating every 7 terms, so <math>D_{2021}=D_5=</math> E, <math>D_{2022}=D_6=</math> O, and <math>D_{2023}=D_7=</math> E. Thus, the answer is \boxed{C) \text{(E, O, E)}}
 
~JHawk0224
 
~JHawk0224

Revision as of 15:32, 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} D_0&D_1&D_2&D_3&D_4&D_5&D_6&D_7&D_8&D_9\\\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{C) \text{(E, O, E)}} ~JHawk0224

See also

2021 AMC 12A (Problems • Answer Key • Resources)
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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