# 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 1 (Difference of Cubes)

Let $z=s$ be a root of $z^2+z+1$ so that $s^2+s+1=0.$ It follows that $$(s-1)\left(s^2+s+1\right)=s^3-1=0,$$ from which $s^3=1,$ but $s\neq1.$

Note that \begin{align*} s^{2021}+1 &= s^{3\cdot673+2}+1 \\ &= (s^3)^{673}\cdot s^2+1 \\ &= s^2+1 \\ &= \left(s^2+s+1\right)-s \\ &= -s. \end{align*} Since $z^{2021}+1=-z$ for each root $z=s$ of $z^2+z+1,$ the remainder when $z^{2021}+1$ is divided by $z^2+z+1$ is $R(z)=\boxed{\textbf{(A) }{-}z}.$

~MRENTHUSIASM

## Solution 2 (Finds Q(z) Using Patterns)

Note that the equation above is in the form of polynomial division, with $z^{2021}+1$ being the dividend, $z^2+z+1$ being the divisor, and $Q(x)$ and $R(x)$ being the quotient and remainder respectively. Since the degree of the dividend is $2021$ and the degree of the divisor is $2$, that means the degree of the quotient is $2021-2 = 2019$. Note that $R(x)$ can't influence the degree of the right hand side of this equation since its degree is either $1$ or $0$. Since the coefficients of the leading term in the dividend and the divisor are both $1$, that means the coefficient of the leading term of the quotient is also $1$. Thus, the leading term of the quotient is $z^{2019}$. Multiplying $z^{2019}$ by the divisor gives $z^{2021}+z^{2020}+z^{2019}$. We have our $z^{2021}$ term but we have these unnecessary terms like $z^{2020}$. We can get rid of these terms by adding $-z^{2018}$ to the quotient to cancel out these terms, but this then gives us $z^{2021}-z^{2018}$. Our first instinct will probably be to add $z^{2017}$, but we can't do this as although this will eliminate the $-z^{2018}$ term, it will produce a $z^{2019}$ term. Since no other term of the form $z^n$ where $n$ is an integer less than $2017$ will produce a $z^{2019}$ term when multiplied by the divisor, we can't add $z^{2017}$ to the quotient. Instead, we can add $z^{2016}$ to the coefficient to get rid of the $-z^{2018}$ term. Continuing this pattern, we get the quotient as $$z^{2019}-z^{2018}+z^{2016}-z^{2015}+\cdots-z^2+1.$$ The last term when multiplied with the divisor gives $z^2+z+1$. This will get rid of the $-z^2$ term but will produce the expression $z+1$, giving us the dividend as $z^{2021}+z+1$. Note that the dividend we want is of the form $z^{2021}+1$. Therefore, our remainder will have to be $-z$ in order to get rid of the $z$ term in the expression and give us $z^{2021}+1$, which is what we want. Therefore, the remainder is $\boxed{\textbf{(A) }{-}z}.$

~ rohan.sp ~rocketsri

## Solution 3 (Modular Arithmetic in Polynomials)

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 4 (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: \begin{align*} z^3 &= 1 \\ z^3 &= e^{i 0} \\ z &= e^{i 0}, e^{i \frac{2\pi}{3}}, e^{i\left(-\frac{2\pi}{3}\right)}. \end{align*} 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 \begin{align*} 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. \end{align*} Expanding into rectangular complex number form: $$\frac{1}{2} - \frac{\sqrt{3}}{2} i = \left(-\frac{1}{2}A + B\right) + \frac{\sqrt{3}}{2} A i.$$ Comparing the real and imaginary parts, we get $(A,B) = (-1,0).$ So, the answer is $\boxed{\textbf{(A) }{-}z}.$

~Jamess2022 (burntTacos ;-))

## Solution 5 (Complex Numbers but Easier)

We have motivation to get rid of the $Q(z)$ term by subtituting in either $e^{\frac{2\pi i}{3}}$ or $e^{\frac{4\pi i}{3}}$ for $z$, as this sets $z^2+z+1$ equal to $0$. Doing so, \begin{align*} \left(e^{\frac{2\pi i}{3}}\right)^{2021} + 1 &= R\left(e^{\frac{2\pi i}{3}}\right) \\ e^{\frac{4042\pi i}{3}} + 1 &= R\left(e^{\frac{2\pi i}{3}}\right) \\ e^{\frac{4\pi i}{3}} + 1 &= R\left(e^{\frac{2\pi i}{3}}\right) \\ \frac{1}{2}-\frac{\sqrt{3}}{2}i &= R\left(-\frac{1}{2} + \frac{\sqrt{3}}{2} i \right). \end{align*} This immediately eliminates choices $\textbf{(B)}$ and $\textbf{(C)},$ as they are constants. Upon a quick check, $R(z) = 2z+1$ and $z + 1$ both don't work, as the sign of $\frac{\sqrt{3}}{2}$ has to be negative. Checking $R(z)$ (or realizing that it's the only one that actually works), we see that $\boxed{\textbf{(A) }{-}z}$ is the right answer.

-Benedict T (countmath1)

## Solution 6 (More Complex Numbers)

Since $z^{2021}+1=(z^2+z+1)Q(z)+R(z),$ we can plug in either $\text{cis} (240^{\circ})$ or $\text{cis} (240^{\circ})$. Assuming $R(z)=ax+b,$ we have $a(\text{cis}(240^{\circ}))+b=\frac{1}{2}+\frac{i\sqrt{3}}{2}$ and $a(\text{cis}(120^{\circ}))+b=\frac{1}{2}-\frac{i\sqrt{3}}{2}.$ Solving gives us $a=-1$ and $b=0$, thus $R(z)=\boxed{\textbf{(A) }{-}z}.$

~SirAppel

## Solution 7 (Factorization)

Notice that $(z^{2018} + z^{2015} + z^{2012} + \cdots + z^5 + z^2)(z^2 + z + 1) = z^{2020} + z^{2019} + \cdots + z^{4} + z^3 + z^2$, so \begin{align*} z^{2021} + 1 = z^{2021} - 1 + 2 &= (z-1)( z^{2020} + z^{2019} + \cdots + z^{2} + z + 1 ) + 2 \\ &= (z-1)[(z^{2018} + z^{2015} + z^{2012} + \cdots + z^5 + z^2)(z^2 + z + 1) + z + 1] + 2 \\ &= (z-1)(z^{2018} + z^{2015} + z^{2012} + \cdots + z^5 + z^2)(z^2 + z + 1) + (z-1)(z+1) + 2 \\ &= (z-1)(z^{2018} + z^{2015} + z^{2012} + \cdots + z^5 + z^2)(z^2 + z + 1) + z^2 + 1 \\ &= (z-1)(z^{2018} + z^{2015} + z^{2012} + \cdots + z^5 + z^2)(z^2 + z + 1) + z^2 + z + 1 - z \\ &= (z^2 + z + 1)[(z^{2018} + z^{2015} + z^{2012} + \cdots + z^5 + z^2)(z-1) + 1] - z. \end{align*} Therefore, we have $$Q(z) = (z^{2018} + z^{2015} + z^{2012} + \cdots + z^5 + z^2)(z-1) + 1,$$ from which $R(z) = \boxed{\textbf{(A) }{-}z}$.

## Video Solution

~MathProblemSolvingSkills.com

~ pi_is_3.14

~hippopotamus1