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

Latest revision as of 22:25, 22 October 2022

Problem

A sequence of numbers is defined by $D_0=0,D_1=0,D_2=1$ 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

We construct the following table: $\begin{array}{c||c|c|c|c|c|c|c|c|c|c|c} &&&&&&&&&&& \\ [-2.5ex] \textbf{Term} &\boldsymbol{D_0}&\boldsymbol{D_1}&\boldsymbol{D_2}&\boldsymbol{D_3}&\boldsymbol{D_4}&\boldsymbol{D_5}&\boldsymbol{D_6}&\boldsymbol{D_7}&\boldsymbol{D_8}&\boldsymbol{D_9}&\boldsymbol{\cdots} \\ \hline \hline &&&&&&&&&&& \\ [-2.25ex] \textbf{Value} & 0&0&1&1&1&2&3&4&6&9&\cdots \\ \hline &&&&&&&&&&& \\ [-2.25ex] \textbf{Parity} & E&E&O&O&O&E&O&E&E&O&\cdots \end{array}$ Note that $(D_7,D_8,D_9)$ and $(D_0,D_1,D_2)$ have the same parities, so the parity is periodic with period $7.$ Since the remainders of $(2021\div7,2022\div7,2023\div7)$ are $(5,6,7),$ we conclude that $(D_{2021},D_{2022},D_{2023})$ and $(D_5,D_6,D_7)$ have the same parities, namely $\boxed{\textbf{(C) }(E,O,E)}.$

~JHawk0224 ~MRENTHUSIASM

Video Solution (Quick and Easy)

https://youtu.be/ecLkESGj-pY

~Education, the Study of Everything

Video Solution by Aaron He (Finding Cycles)

https://www.youtube.com/watch?v=xTGDKBthWsw&t=7m43s

Video Solution by Hawk Math

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

Video Solution by OmegaLearn (Using Parity and Pattern Finding)

https://youtu.be/TSBjbhN_QKY

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/cckGBU2x1zg?t=227

~IceMatrix

@@ Line 5: / Line 5: @@
 ==Solution==
-Making a small chart, we have
+We construct the following table:
+<cmath>\begin{array}{c||c|c|c|c|c|c|c|c|c|c|c}
+&&&&&&&&&&& \\ [-2.5ex]
+\textbf{Term} &\boldsymbol{D_0}&\boldsymbol{D_1}&\boldsymbol{D_2}&\boldsymbol{D_3}&\boldsymbol{D_4}&\boldsymbol{D_5}&\boldsymbol{D_6}&\boldsymbol{D_7}&\boldsymbol{D_8}&\boldsymbol{D_9}&\boldsymbol{\cdots} \\
+\hline \hline
+&&&&&&&&&&& \\ [-2.25ex]
+\textbf{Value} & 0&0&1&1&1&2&3&4&6&9&\cdots \\ \hline
+&&&&&&&&&&& \\ [-2.25ex]
+\textbf{Parity} & E&E&O&O&O&E&O&E&E&O&\cdots
+\end{array}</cmath>
+Note that <math>(D_7,D_8,D_9)</math> and <math>(D_0,D_1,D_2)</math> have the same parities, so the parity is periodic with period <math>7.</math> Since the remainders of <math>(2021\div7,2022\div7,2023\div7)</math> are <math>(5,6,7),</math> we conclude that <math>(D_{2021},D_{2022},D_{2023})</math> and <math>(D_5,D_6,D_7)</math> have the same parities, namely <math>\boxed{\textbf{(C) }(E,O,E)}.</math>
-<math>\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
+~JHawk0224 ~MRENTHUSIASM
-D0&D1&D2&D3&D4&D5&D6&D7&D8&D9\\\hline
-&0&1&1&1&2&3&4&6&9\\\hline
-E&E&O&O&O&E&O&E&E&O
-\end{tabular}</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>
+==Video Solution (Quick and Easy)==
-~JHawk0224
+https://youtu.be/ecLkESGj-pY
-==Video Solution by Aaron He (Finding cycles)==
+~Education, the Study of Everything
+==Video Solution by Aaron He (Finding Cycles)==
 https://www.youtube.com/watch?v=xTGDKBthWsw&t=7m43s
 ==Video Solution by Hawk Math==
 https://www.youtube.com/watch?v=P5al76DxyHY
-== Video Solution (Using parity and pattern finding) ==
+== Video Solution by OmegaLearn (Using Parity and Pattern Finding) ==
 https://youtu.be/TSBjbhN_QKY
 ~ pi_is_3.14
+==Video Solution by TheBeautyofMath==
+https://youtu.be/cckGBU2x1zg?t=227
+~IceMatrix
 ==See also==
 {{AMC12 box|year=2021|ab=A|num-b=7|num-a=9}}
 {{MAA Notice}}

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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