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

(→‎Solution 2 (Somewhat of a long method))
(→‎Solution 7 (Factorization))
 
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Let <math>Q(z)</math> and <math>R(z)</math> be the unique polynomials such that<cmath>z^{2021}+1=(z^2+z+1)Q(z)+R(z)</cmath>and the degree of <math>R</math> is less than <math>2.</math> What is <math>R(z)?</math>
 
Let <math>Q(z)</math> and <math>R(z)</math> be the unique polynomials such that<cmath>z^{2021}+1=(z^2+z+1)Q(z)+R(z)</cmath>and the degree of <math>R</math> is less than <math>2.</math> What is <math>R(z)?</math>
  
−
<math>\textbf{(A) }-z \qquad \textbf{(B) }-1 \qquad \textbf{(C) }2021\qquad \textbf{(D) }z+1 \qquad \textbf{(E) }2z+1</math>
+
<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)==
 +
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> but <math>s\neq1.</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> 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
 
Note that
 
<cmath>z^3-1\equiv 0\pmod{z^2+z+1}</cmath>
 
<cmath>z^3-1\equiv 0\pmod{z^2+z+1}</cmath>
Line 13: Line 35:
 
So <math>F(z) = z^2+1</math>, and
 
So <math>F(z) = z^2+1</math>, and
 
<cmath>R(z)\equiv F(z) \equiv -z\pmod{z^2+z+1}</cmath>
 
<cmath>R(z)\equiv F(z) \equiv -z\pmod{z^2+z+1}</cmath>
−
The answer is <math>\boxed{\textbf{(A) }-z}.</math>
+
The answer is <math>\boxed{\textbf{(A) }{-}z}.</math>
  
−
==Solution 2 (Complex numbers)==
+
==Solution 4 (Complex Numbers)==
−
One thing to note is that <math>R(z)</math> takes the form of <math>Az + B</math> for some constants A and B.
+
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>
+
Note that the roots of <math>z^2 + z + 1</math> are part of the solutions of <math>z^3 -1 = 0.</math>
 
They can be easily solved with roots of unity:
 
They can be easily solved with roots of unity:
−
<cmath>z^3 = 1</cmath>
+
<cmath>\begin{align*}
−
<cmath>z^3 = e^{i 0}</cmath>
+
z^3 &= 1 \\
−
<cmath>z = e^{i 0}, e^{i \frac{2\pi}{3}}, e^{i -\frac{2\pi}{3}}</cmath> <cmath>\newline</cmath>
+
z^3 &= e^{i 0} \\
−
Obviously the right two solutions are the roots of <math>z^2 + z + 1 = 0</math>
+
z &= e^{i 0}, e^{i \frac{2\pi}{3}}, e^{i\left(-\frac{2\pi}{3}\right)}.
−
We substitute <math>e^{i \frac{2\pi}{3}}</math> into the original equation, and <math>z^2 + z + 1</math> becomes 0. Using De Moivre's theorem, we get:
+
\end{align*}</cmath>
−
<cmath>e^{i\frac{4042\pi}{3}} + 1 = A \cdot e^{i \frac{2\pi}{3}} + B</cmath>
+
Obviously the right two solutions are the roots of <math>z^2 + z + 1 = 0.</math>
−
<cmath>e^{i\frac{4\pi}{3}} + 1 = A \cdot e^{i \frac{2\pi}{3}} + B</cmath>
+
We substitute <math>e^{i \frac{2\pi}{3}}</math> into the original equation, and <math>z^2 + z + 1</math> becomes <math>0.</math> Using De Moivre's Theorem, we get
 +
<cmath>\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*}</cmath>
 
Expanding into rectangular complex number form:
 
Expanding into rectangular complex number form:
−
<cmath>\frac{1}{2} - \frac{\sqrt{3}}{2} i = (-\frac{1}{2}A + B) + \frac{\sqrt{3}}{2} i A</cmath>
+
<cmath>\frac{1}{2} - \frac{\sqrt{3}}{2} i = \left(-\frac{1}{2}A + B\right) + \frac{\sqrt{3}}{2} A i.</cmath>
−
Comparing the real and imaginary parts, we get:
+
Comparing the real and imaginary parts, we get <math>(A,B) = (-1,0).</math> So, the answer is <math>\boxed{\textbf{(A) }{-}z}.</math>
−
<cmath>A = -1, B = 0</cmath>
+
 
−
The answer is <math>\boxed{\textbf{(A) }-z}</math>. ~Jamess2022(burntTacos;-;)
+
~Jamess2022 (burntTacos ;-))
 +
 
 +
==Solution 5 (Complex Numbers but Easier)==
 +
 
 +
We have motivation to get rid of the <math>Q(z)</math> term by subtituting in either <math>e^{\frac{2\pi i}{3}}</math> or <math>e^{\frac{4\pi i}{3}}</math> for <math>z</math>, as this sets <math>z^2+z+1</math> equal to <math>0</math>. Doing so,
 +
<cmath>\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*}</cmath>
 +
This immediately eliminates choices <math>\textbf{(B)}</math> and <math>\textbf{(C)},</math> as they are constants. Upon a quick check, <math>R(z) = 2z+1</math> and <math>z + 1</math> both don't work, as the sign of <math>\frac{\sqrt{3}}{2}</math> has to be negative. Checking <math>R(z)</math> (or realizing that it's the only one that actually works), we see that <math>\boxed{\textbf{(A) }{-}z}</math> is the right answer.
 +
 
 +
-Benedict T (countmath1)
 +
 
 +
==Solution 6 (More Complex Numbers)==
 +
Since <math>z^{2021}+1=(z^2+z+1)Q(z)+R(z),</math> we can plug in either <math>\text{cis} (240^{\circ})</math> or <math>\text{cis} (240^{\circ})</math>. Assuming <math>R(z)=ax+b,</math> we have <math>a(\text{cis}(240^{\circ}))+b=\frac{1}{2}+\frac{i\sqrt{3}}{2}</math> and <math>a(\text{cis}(120^{\circ}))+b=\frac{1}{2}-\frac{i\sqrt{3}}{2}.</math> Solving gives us <math>a=-1</math> and <math>b=0</math>, thus <math>R(z)=\boxed{\textbf{(A) }{-}z}.</math>
 +
 
 +
~SirAppel
 +
 
 +
==Solution 7 (Factorization)==
 +
 
 +
Notice that <math>(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</math>, so
 +
<cmath>\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*}</cmath>
 +
Therefore, we have <cmath>Q(z) = (z^{2018} + z^{2015} + z^{2012} + \cdots + z^5 + z^2)(z-1) + 1,</cmath> from which <math>R(z) = \boxed{\textbf{(A) }{-}z}</math>.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
−
== Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-solving) ==
+
== Video Solution==
 +
https://youtu.be/3GYJE9aK83k
 +
 
 +
~MathProblemSolvingSkills.com
 +
 
 +
== Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-Solving) ==
 
https://youtu.be/nnjr17q7fS0
 
https://youtu.be/nnjr17q7fS0
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
==Video Solution (Long Division, Not Brutal)==
 +
https://youtu.be/kxPDeQRGLEg
 +
 +
~hippopotamus1
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021|ab=B|num-b=19|num-a=21}}
 
{{AMC12 box|year=2021|ab=B|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:43, 24 October 2023

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 $R(z) = \boxed{\textbf{(A) }{-}z}$.

~isabelchen

Video Solution

https://youtu.be/3GYJE9aK83k

~MathProblemSolvingSkills.com

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

https://youtu.be/nnjr17q7fS0

~ pi_is_3.14

Video Solution (Long Division, Not Brutal)

https://youtu.be/kxPDeQRGLEg

~hippopotamus1

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

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

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