2021 AMC 12B Problems/Problem 20

Revision as of 11:39, 12 February 2021 by Jamess2022 (talk | contribs) (Solution 2 (Somewhat of a long method))

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 1

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

Solution 2 (Complex numbers)

One thing to note is that $R(z)$ takes the form of $Az + B$ for some constants A and B. Note that the roots of $z^2 + z + 1$ are part of the solutions of $z^3 -1 = 0$ They can be easily solved with roots of unity: \[z^3 = 1\] \[z^3 = e^{i 0}\] \[z = e^{i 0}, e^{i \frac{2\pi}{3}}, e^{i -\frac{2\pi}{3}}\] \[\newline\] Obviously the right two solutions are the roots of $z^2 + z + 1 = 0$ We substitute $e^{i \frac{2\pi}{3}}$ into the original equation, and $z^2 + z + 1$ becomes 0. Using De Moivre's theorem, we get: \[e^{i\frac{4042\pi}{3}} + 1 = A \cdot e^{i \frac{2\pi}{3}} + B\] \[e^{i\frac{4\pi}{3}} + 1 = A \cdot e^{i \frac{2\pi}{3}} + B\] Expanding into rectangular complex number form: \[\frac{1}{2} - \frac{\sqrt{3}}{2} i = (-\frac{1}{2}A + B) + \frac{\sqrt{3}}{2} i A\] Comparing the real and imaginary parts, we get: \[A = -1, B = 0\] The answer is $\boxed{\textbf{(A) }-z}$. ~Jamess2022(burntTacos;-;)

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

https://youtu.be/nnjr17q7fS0

~ pi_is_3.14

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
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

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