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

Revision as of 15:34, 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

\[\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
D_0&D_1&D_2&D_3&D_4&D_5&D_6&D_7&D_8&D_9\\\hline
0&0&1&1&1&2&3&4&6&9\\\hline
E&E&O&O&O&E&O&E&E&O
\end{tabular}\] (Error compiling LaTeX. Unknown error_msg)

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

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

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

@@ Line 11: / Line 11: @@
 E&E&O&O&O&E&O&E&E&O
 \end{tabular}</cmath>
-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 <math>\boxed{\textbf{(C)} \text{(E, O, E)}}</math>
+This starts repeating every 7 terms, so <math>D_{2021}=D_5=E</math>, <math>D_{2022}=D_6=O</math>, and <math>D_{2023}=D_7=E</math>. Thus, the answer is <math>\boxed{\textbf{(C)} (E, O, E)}</math>
 ~JHawk0224

Page

Toolbox

Search

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

Revision as of 15:34, 11 February 2021

Problem

Solution

See also