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2021 AMC 12B Problems/Problem 20: Difference between revisions

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==Problem 20==
==Problem==
Let <math>Q(z)</math> and <math>R(z)</math> be the unique polynomials such that<cmath>z^{2021}+1=(z^2+z+1)Q(z)+R(z)</cmath>and the degree of <math>R</math> is less than <math>2.</math> What is <math>R(z)?</math>
Let <math>Q(z)</math> and <math>R(z)</math> be the unique polynomials such that<cmath>z^{2021}+1=(z^2+z+1)Q(z)+R(z)</cmath>and the degree of <math>R</math> is less than <math>2.</math> What is <math>R(z)?</math>


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<cmath>R(z)\equiv F(z) \equiv -z\pmod{z^2+z+1}</cmath>
<cmath>R(z)\equiv F(z) \equiv -z\pmod{z^2+z+1}</cmath>
The answer is <math>\boxed{\textbf{(A) }-z}.</math>
The answer is <math>\boxed{\textbf{(A) }-z}.</math>
==See Also==
{{AMC12 box|year=2021|ab=B|num-b=19|num-a=21}}
{{MAA Notice}}

Revision as of 19:36, 11 February 2021

Problem

Let $Q(z)$ and $R(z)$ be the unique polynomials such that\[z^{2021}+1=(z^2+z+1)Q(z)+R(z)\]and the degree of $R$ is less than $2.$ What is $R(z)?$

$\textbf{(A) }-z \qquad \textbf{(B) }-1 \qquad \textbf{(C) }2021\qquad \textbf{(D) }z+1 \qquad \textbf{(E) }2z+1$

Solution

Note that \[z^3-1\equiv 0\pmod{z^2+z+1}\] so if $F(z)$ is the remainder when dividing by $z^3-1$, \[F(z)\equiv R(z)\pmod{z^2+z+1}.\] Now, \[z^{2021}+1= (z^3-1)(z^{2018} + z^{2015} + \cdots + z^2) + z^2+1\] So $F(z) = z^2+1$, and \[R(z)\equiv F(z) \equiv -z\pmod{z^2+z+1}\] The answer is $\boxed{\textbf{(A) }-z}.$

See Also

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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