2021 AMC 12B Problems/Problem 20

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 $\boxed{\textbf{(A) }-z}.$

Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-solving)

https://youtu.be/nnjr17q7fS0

~ pi_is_3.14

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.

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2021 AMC 12B Problems/Problem 20

Contents

Problem

Solution

Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-solving)

See Also