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 $R(z)?$

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

Video Solution using long division(not brutal)

https://youtu.be/kxPDeQRGLEg ~hippopotamus1

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 1b (More Thorough Version of 1)

Instead of dealing with a nasty $z^2+z+1$ , we can instead deal with the nice $z^3 - 1$ , as $z^2+z+1$ is a factor of $z^3-1$ . Then, we try to see what $\frac{z^{2021} + 1}{z^3 - 1}$ is. Of course, we will need a $z^{2018}$ , getting $z^{2021} - z^{2018}$ . Then, we've gotta get rid of the $z^{2018}$ term, so we add a $z^{2015}$ , to get $z^{2021} - z^{2015}$ . This pattern continues, until we add a $z^2$ to get rid of $z^5$ , and end up with $z^{2021} - z^2$ . We can't add anything more to get rid of the $z^2$ , so our factor is $z^{2018} + z^{2015} + z^{2012} + \cdots + z^2$ . Then, to get rid of the $z^2$ , we must have a remainder of $+z^2$ , and to get the $+1$ we have to also have a $+1$ in the remainder. So, our product is $z^{2021}+1= (z^3-1)(z^{2018} + z^{2015} + \cdots + z^2) + z^2+1.$ Then, our remainder is $z^2+1$ . The remainder when dividing by $z^3-1$ must be the same when dividing by $z^2+z+1$ , modulo $z^2+z+1$ . So, we have that $z^2 + 1 \equiv R(z) \pmod{z^2+z+1}$ , or $R(z) \equiv -z\pmod{z^2+z+1}$ . This corresponds to answer choice $\boxed{\textbf{(A)} ~ -z}$ . ~rocketsri

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

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

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