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 ;-))

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

https://youtu.be/nnjr17q7fS0

Video Solution (Long Division, Not Brutal)

https://youtu.be/kxPDeQRGLEg ~hippopotamus1

~ 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS