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Difference between revisions of "2021 AMC 12B Problems/Problem 20"
(→‎Solution 4 (Division Analysis))
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z^{2021}+1&=\left(z^3-1\right)Q'(z)+\left(z^2+1\right) \\
 
z^{2021}+1&=\left(z^3-1\right)Q'(z)+\left(z^2+1\right) \\
 
&=(z-1)\left(z^2+z+1\right)Q'(z)+\left(z^2+1\right) \\
 
&=(z-1)\left(z^2+z+1\right)Q'(z)+\left(z^2+1\right) \\
−
&=\left(z^2+z+1\right)\left[(z-1)Q'(z)\right]+\left(z^2+1\right).
+
&=\left(z^2+z+1\right)\left[(z-1)Q'(z)\right]+\left(z^2+1\right), \ \ \ \ \ \ \ \ \ \ \ \ (**)
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
which almost resembles to the original equation <cmath>z^{2021}+1=(z^2+z+1)Q(z)+R(z).</cmath> Since we require that <math>\text{deg}\left(R(z)\right)<2,</math> the expression <math>z^2+1</math> goes into the quotient for one more time. Rewriting <math>(**)</math> produces
 +
<cmath>\begin{align*}
 +
z^{2021}+1&=\left(z^2+z+1\right)\left[(z-1)Q'(z)+1\right]+\left(\left(z^2+1\right)-\left(z^2+z+1\right)\right) \\
 +
&=\left(z^2+z+1\right)\left[(z-1)Q'(z)+1\right]+(-z),
 +
\end{align*}</cmath>
 +
from which <math>R(z)=\boxed{\textbf{(A) }-z}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 22:31, 20 February 2021

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

Solution 1.1

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 1.2 (More Thorough Version of Solution 1.1)

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;-;)

Solution 3 (similar to Solution 1.2)

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 \[z^{2019}-z^{2018}+z^{2016}-z^{2015}+....-z^2+1.\] 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 $\boxed{\textbf{(A) }-z \qquad}$

~ rohan.sp

Solution 4 (Division Analysis)

By the difference of cubes / short geometric series, \[\left(z-1\right)\left(z^2+z+1\right)=z^3-1.\] We rewrite $\left(z^{2021}+1\right)\div\left(z^3-1\right)$ by the polynomial division algorithm: \[z^{2021}+1=\left(z^3-1\right)Q'(z)+R'(z), \ \ \ \ \ \ \ \ \ \ \ \ (*)\] where $Q'(z)$ and $R'(z)$ are unique polynomials such that $\text{deg}\left(R'(z)\right)<3.$ Taking $(*)$ in modulo $z^3-1$ (in which $z^3=1$), we have \[z^{2021}+1=z^{3\cdot673+2}+1=z^{3\cdot673}z^2+1=\left(z^3\right)^{673}z^2+1=z^2+1=R'(z).\] Substituting $R'(z)=z^2+1$ back into $(*)$ gives \begin{align*} z^{2021}+1&=\left(z^3-1\right)Q'(z)+\left(z^2+1\right) \\ &=(z-1)\left(z^2+z+1\right)Q'(z)+\left(z^2+1\right) \\ &=\left(z^2+z+1\right)\left[(z-1)Q'(z)\right]+\left(z^2+1\right), \ \ \ \ \ \ \ \ \ \ \ \ (**) \end{align*} which almost resembles to the original equation \[z^{2021}+1=(z^2+z+1)Q(z)+R(z).\] Since we require that $\text{deg}\left(R(z)\right)<2,$ the expression $z^2+1$ goes into the quotient for one more time. Rewriting $(**)$ produces \begin{align*} z^{2021}+1&=\left(z^2+z+1\right)\left[(z-1)Q'(z)+1\right]+\left(\left(z^2+1\right)-\left(z^2+z+1\right)\right) \\ &=\left(z^2+z+1\right)\left[(z-1)Q'(z)+1\right]+(-z), \end{align*} from which $R(z)=\boxed{\textbf{(A) }-z}.$

~MRENTHUSIASM

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

https://youtu.be/nnjr17q7fS0

Video Solution using long division(not brutal)

https://youtu.be/kxPDeQRGLEg ~hippopotamus1

~ pi_is_3.14

See Also

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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