Difference between revisions of "2021 AMC 12B Problems/Problem 20"

Revision as of 01:12, 2 October 2021

1 Problem
2 Solution 1
- 2.1 Solution 1.1
- 2.2 Solution 1.2 (More Thorough Version of Solution 1.1)
3 Solution 2 (Complex Numbers)
4 Solution 3 (Directly Finding the Quotient by Using Patterns)
5 Solution 4 (Difference of Cubes)
6 Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-Solving)
7 Video Solution (Long Division, Not Brutal)
8 See Also

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 need to 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: $\begin{align*} z^3 &= 1 \\ z^3 &= e^{i 0} \\ z &= e^{i 0}, e^{i \frac{2\pi}{3}}, e^{i -\frac{2\pi}{3}}. \end{align*}$ 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: $A = -1, B = 0.$ The answer is $\boxed{\textbf{(A) }-z}.$

~Jamess2022 (burntTacos ;-))

Solution 3 (Directly Finding the Quotient by 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

Solution 4 (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 $s^3=1.$

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 $z^2+z+1,$ we conclude that the remainder when $z^{2021}+1$ is divided by $z^2+z+1$ is $R(z)=\boxed{\textbf{(A) }-z}.$

~MRENTHUSIASM

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

@@ Line 51: / Line 51: @@
 ~ rohan.sp
-==Solution 4 (Division Analysis Without Finding Q(z))==
+==Solution 4 (Difference of Cubes)==
-By the difference of cubes or the short geometric series, we get <cmath>\left(z-1\right)\left(z^2+z+1\right)=z^3-1.</cmath>
+Let <math>z=s</math> be a root of <math>z^2+z+1</math> so that <math>s^2+s+1=0.</math> It follows that <cmath>(s-1)\left(s^2+s+1\right)=s^3-1=0,</cmath> from which <math>s^3=1.</math>
-We rewrite <math>\left(z^{2021}+1\right)\div\left(z^3-1\right)</math> by the polynomial division algorithm:
-<cmath>z^{2021}+1=\left(z^3-1\right)Q'(z)+R'(z), \hspace{33mm}(*)</cmath>
+Note that
-where <math>Q'(z)</math> and <math>R'(z)</math> are unique polynomials such that <math>\mathrm{deg}\left(R'(z)\right)<3.</math> Taking <math>(*)</math> modulo <math>z^3-1</math> (in which <math>z^3\equiv1</math>), we have <cmath>z^{2021}+1=z^{3\cdot673+2}+1=\left(z^3\right)^{673}z^2+1\equiv z^2+1=R'(z).</cmath>
-Substituting <math>R'(z)=z^2+1</math> back into <math>(*)</math> gives
 <cmath>\begin{align*}
-z^{2021}+1&=\left(z^3-1\right)Q'(z)+\left(z^2+1\right) \\
+s^{2021}+1 &= s^{3\cdot673+2}+1 \\
-&=(z-1)\left(z^2+z+1\right)Q'(z)+\left(z^2+1\right) \\
+&= (s^3)^{673}\cdot s^2+1 \\
-&=\left(z^2+z+1\right)\left[(z-1)Q'(z)\right]+\left(z^2+1\right), \hspace{7.25mm} (**)
+&= s^2+1 \\
+&= \left(s^2+s+1\right)-s \\
+&= -s.
 \end{align*}</cmath>
-which almost resembles the original equation <cmath>z^{2021}+1=\left(z^2+z+1\right)Q(z)+R(z).</cmath> Since we require that <math>\mathrm{deg}\left(R(z)\right)<2,</math> the divisor <math>z^2+z+1</math> goes into the remaining <math>z^2+1</math> for one more time. Rewriting <math>(**)</math> produces
+Since <math>z^{2021}+1=-z</math> for each root <math>z=s</math> of <math>z^2+z+1,</math> we conclude that the remainder when <math>z^{2021}+1</math> is divided by <math>z^2+z+1</math> is <math>R(z)=\boxed{\textbf{(A) }-z}.</math>
-<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)\underbrace{\left[(z-1)Q'(z)+1\right]}_{Q(z)}+\underbrace{[-z]}_{R(z)},
-\end{align*}</cmath>
-from which <math>R(z)=\boxed{\textbf{(A) }-z}.</math>
 ~MRENTHUSIASM

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
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