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

Revision as of 01:37, 2 October 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 (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

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: $\begin{align*} z^3 &= 1 \\ z^3 &= e^{i 0} \\ z &= e^{i 0}, e^{i \frac{2\pi}{3}}, e^{i -\frac{2\pi}{3}}. \end{align*}$ 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 ;-))

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 4: / Line 4: @@
 <math>\textbf{(A) }-z \qquad \textbf{(B) }-1 \qquad \textbf{(C) }2021\qquad \textbf{(D) }z+1 \qquad \textbf{(E) }2z+1</math>
-==Solution 1==
+==Solution 1 (Difference of Cubes)==
-===Solution 1.1===
+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>
+Note that
+<cmath>\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*}</cmath>
+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>
+~MRENTHUSIASM
+==Solution 2 (Finds Q(z) Using Patterns)==
+Note that the equation above is in the form of polynomial division, with <math>z^{2021}+1</math> being the dividend, <math>z^2+z+1</math> being the divisor, and <math>Q(x)</math> and <math>R(x)</math> being the quotient and remainder respectively. Since the degree of the dividend is <math>2021</math> and the degree of the divisor is <math>2</math>, that means the degree of the quotient is <math>2021-2 = 2019</math>. Note that <math>R(x)</math> can't influence the degree of the right hand side of this equation since its degree is either <math>1</math> or <math>0</math>. Since the coefficients of the leading term in the dividend and the divisor are both <math>1</math>, that means the coefficient of the leading term of the quotient is also <math>1</math>. Thus, the leading term of the quotient is <math>z^{2019}</math>. Multiplying <math>z^{2019}</math> by the divisor gives <math>z^{2021}+z^{2020}+z^{2019}</math>. We have our <math>z^{2021}</math> term but we have these unnecessary terms like <math>z^{2020}</math>. We can get rid of these terms by adding <math>-z^{2018}</math> to the quotient to cancel out these terms, but this then gives us <math>z^{2021}-z^{2018}</math>. Our first instinct will probably be to add <math>z^{2017}</math>, but we can't do this as although this will eliminate the <math>-z^{2018}</math> term, it will produce a <math>z^{2019}</math> term. Since no other term of the form <math>z^n</math> where <math>n</math> is an integer less than <math>2017</math> will produce a <math>z^{2019}</math> term when multiplied by the divisor, we can't add <math>z^{2017}</math> to the quotient. Instead, we can add <math>z^{2016}</math> to the coefficient to get rid of the <math>-z^{2018}</math> term. Continuing this pattern, we get the quotient as <cmath>z^{2019}-z^{2018}+z^{2016}-z^{2015}+\cdots-z^2+1.</cmath>
+The last term when multiplied with the divisor gives <math>z^2+z+1</math>. This will get rid of the <math>-z^2</math> term but will produce the expression <math>z+1</math>, giving us the dividend as <math>z^{2021}+z+1</math>. Note that the dividend we want is of the form <math>z^{2021}+1</math>. Therefore, our remainder will have to be <math>-z</math> in order to get rid of the <math>z</math> term in the expression and give us <math>z^{2021}+1</math>, which is what we want. Therefore, the remainder is <math>\boxed{\textbf{(A) }-z}.</math>
+~ rohan.sp ~rocketsri
+==Solution 3 (Modular Arithmetic in Polynomials)==
 Note that
 <cmath>z^3-1\equiv 0\pmod{z^2+z+1}</cmath>
@@ Line 16: / Line 37: @@
 The answer is <math>\boxed{\textbf{(A) }-z}.</math>
-===Solution 1.2 (More Thorough Version of Solution 1.1)===
+==Solution 4 (Complex Numbers)==
-Instead of dealing with a nasty <math>z^2+z+1</math>, we can instead deal with the nice <math>z^3 - 1</math>, as <math>z^2+z+1</math> is a factor of <math>z^3-1</math>. Then, we try to see what <math>\frac{z^{2021} + 1}{z^3 - 1}</math> is. Of course, we will need a <math>z^{2018}</math>, getting <math>z^{2021} - z^{2018}</math>. Then, we need to get rid of the <math>z^{2018}</math> term, so we add a <math>z^{2015}</math>, to get <math>z^{2021} - z^{2015}</math>. This pattern continues, until we add a <math>z^2</math> to get rid of <math>z^5</math>, and end up with <math>z^{2021} - z^2</math>. We can't add anything more to get rid of the <math>z^2</math>, so our factor is <math>z^{2018} + z^{2015} + z^{2012} + \cdots + z^2</math>. Then, to get rid of the <math>z^2</math>, we must have a remainder of <math>+z^2</math>, and to get the <math>+1</math> we have to also have a <math>+1</math> in the remainder. So, our product is <cmath>z^{2021}+1= (z^3-1)(z^{2018} + z^{2015} + \cdots + z^2) + z^2+1.</cmath> Then, our remainder is <math>z^2+1</math>. The remainder when dividing by <math>z^3-1</math> must be the same when dividing by <math>z^2+z+1</math>, modulo <math>z^2+z+1</math>. So, we have that <math>z^2 + 1 \equiv R(z) \pmod{z^2+z+1}</math>, or <math>R(z) \equiv -z\pmod{z^2+z+1}</math>. This corresponds to answer choice <math>\boxed{\textbf{(A) }-z}</math>.
-~rocketsri
-==Solution 2 (Complex Numbers)==
 One thing to note is that <math>R(z)</math> takes the form of <math>Az + B</math> for some constants <math>A</math> and <math>B.</math>
 Note that the roots of <math>z^2 + z + 1</math> are part of the solutions of <math>z^3 -1 = 0.</math>
@@ Line 43: / Line 59: @@
 ~Jamess2022 (burntTacos ;-))
-==Solution 3 (Directly Finding the Quotient by Using Patterns)==
-Note that the equation above is in the form of polynomial division, with <math>z^{2021}+1</math> being the dividend, <math>z^2+z+1</math> being the divisor, and <math>Q(x)</math> and <math>R(x)</math> being the quotient and remainder respectively. Since the degree of the dividend is <math>2021</math> and the degree of the divisor is <math>2</math>, that means the degree of the quotient is <math>2021-2 = 2019</math>. Note that <math>R(x)</math> can't influence the degree of the right hand side of this equation since its degree is either <math>1</math> or <math>0</math>. Since the coefficients of the leading term in the dividend and the divisor are both <math>1</math>, that means the coefficient of the leading term of the quotient is also <math>1</math>. Thus, the leading term of the quotient is <math>z^{2019}</math>. Multiplying <math>z^{2019}</math> by the divisor gives <math>z^{2021}+z^{2020}+z^{2019}</math>. We have our <math>z^{2021}</math> term but we have these unnecessary terms like <math>z^{2020}</math>. We can get rid of these terms by adding <math>-z^{2018}</math> to the quotient to cancel out these terms, but this then gives us <math>z^{2021}-z^{2018}</math>. Our first instinct will probably be to add <math>z^{2017}</math>, but we can't do this as although this will eliminate the <math>-z^{2018}</math> term, it will produce a <math>z^{2019}</math> term. Since no other term of the form <math>z^n</math> where <math>n</math> is an integer less than <math>2017</math> will produce a <math>z^{2019}</math> term when multiplied by the divisor, we can't add <math>z^{2017}</math> to the quotient. Instead, we can add <math>z^{2016}</math> to the coefficient to get rid of the <math>-z^{2018}</math> term. Continuing this pattern, we get the quotient as <cmath>z^{2019}-z^{2018}+z^{2016}-z^{2015}+\cdots-z^2+1.</cmath>
-The last term when multiplied with the divisor gives <math>z^2+z+1</math>. This will get rid of the <math>-z^2</math> term but will produce the expression <math>z+1</math>, giving us the dividend as <math>z^{2021}+z+1</math>. Note that the dividend we want is of the form <math>z^{2021}+1</math>. Therefore, our remainder will have to be <math>-z</math> in order to get rid of the <math>z</math> term in the expression and give us <math>z^{2021}+1</math>, which is what we want. Therefore, the remainder is <math>\boxed{\textbf{(A) }-z}.</math>
-~ rohan.sp
-==Solution 4 (Difference of Cubes)==
-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>
-Note that
-<cmath>\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*}</cmath>
-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>
-~MRENTHUSIASM
 == Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-Solving) ==

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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