Difference between revisions of "2022 AMC 12A Problems/Problem 21"

Latest revision as of 13:00, 5 November 2024

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4 (Describe the Roots)
6 Solution 5 (Simple Elimination)
- 6.1 Solution 5a (Elimination but slightly different)
- 6.2 Solution 5b (elimination but even faster and cheesier)
7 Video Solution by Math-X (Smart and Simple)
8 Video Solution
9 Video Solution by ThePuzzlr
10 See Also

Problem

Let $P(x) = x^{2022} + x^{1011} + 1.$ Which of the following polynomials is a factor of $P(x)$ ?

$\textbf{(A)} \, x^2 -x + 1 \qquad\textbf{(B)} \, x^2 + x + 1 \qquad\textbf{(C)} \, x^4 + 1 \qquad\textbf{(D)} \, x^6 - x^3 + 1 \qquad\textbf{(E)} \, x^6 + x^3 + 1$

Solution 1

$P(x) = x^{2022} + x^{1011} + 1$ is equal to $\frac{x^{3033}-1}{x^{1011}-1}$ by difference of powers.

Therefore, the answer is a polynomial that divides $x^{3033}-1$ but not $x^{1011}-1$ .

Note that any polynomial $x^m-1$ divides $x^n-1$ if and only if $m$ is a factor of $n$ .

The prime factorizations of $1011$ and $3033$ are $3*337$ and $3^2*337$ , respectively.

Hence, $x^9-1$ is a divisor of $x^{3033}-1$ but not $x^{1011}-1$ .

By difference of powers, $x^9-1=(x^3-1)(x^6+x^3+1)$ . Therefore, the answer is $\boxed{E}$ .

Solution 2

We simply test roots for each, as $2022,1011$ are multiples of three, we need to make sure the roots are in the form of $e^{i\frac{k\pi}{9}}$ , so we only have to look at $D,E$ .

If we look at choice $E$ , $x=e^{i\frac{\pm2\pi}{9}}$ which works perfectly, the answer is just $E$

~bluesoul

Solution 3

Let $x^{1011} = u$ , now we can rewrite our polynomial as $u^2+u+1$ . Using the quadratic formula to solve for the roots of this polynomial, we have $x^{1011} = \frac{-1\pm i\sqrt{3}}{2}$ Looking at our answer choices, we want to find a polynomial whose roots satisfy this expression. Since the expression $x^6+x^3+1$ is in a similar form to our original polynomial, except with $x^3$ in place of $x^{1011}$ , this would be a good place to start. Solving for the roots of $x^3$ in a similar fashion, $x^3= \frac{-1\pm i\sqrt{3}}{2}$ for the solution we are testing. Now notice that we can rewrite the roots of $x^3$ as $x^3 = \operatorname{cis}{\frac{2\pi}{3}}, \operatorname{cis}{\frac{4\pi}{3}}$ Both of which are third roots of unity. We want to now check if this value of $x^3$ satisfies $x^{1011} = \frac{-1\pm i\sqrt{3}}{2}$ . Notice that $x^{1011} = (x^{3})^{112\cdot3}\cdot x^3$ , and since both values of $x^3$ are roots of unity, we can simplify the expression we want satisfied to the expression to $x^{1011}=x^3$ . Since both values of $x^3$ are also values of $x^{1011}$ , the roots for our $x^6+x^3+1$ are also roots of $x^{2022}+x^{1011}+1$ , meaning that $x^6+x^3+1 | x^{2022}+x^{1011}+1$ so Therefore, the answer is $\boxed{E}$ .

- DavidHovey

Solution 4 (Describe the Roots)

We know that a monic polynomial $q$ divides a monic polynomial $p$ if and only if all the roots of $q$ are roots of $p.$ Since $P(x)=x^{2022}+x^{1011}+1=\frac{x^{3033}-1}{x^{1011}-1}$ , the roots of $P$ are the $3033$ rd roots of unity that aren't $1011$ th roots of unity.

Now, note that:

1: The roots of polynomial $A$ are the primitive $6$ th roots of unity.

2: The roots of polynomial $B$ are the primitive cube roots of unity.

3: The roots of polynomial $C$ are the primitive $8$ th roots of unity.

4: The roots of polynomial $D$ are the primitive $18$ th roots of unity.

5: The roots of polynomial $E$ are the primitive $9$ th roots of unity.

However, since $6$ , $8$ , and $18$ don't divide $3033$ , the roots of polynomial $A$ are not all $3033$ rd roots of unity, and the same is true for polynomials $C$ and $D$ , eliminating choices $A$ , $C$ and $D.$ Also, since $3$ divides $1011$ , the roots of polynomial $B$ are all $1011$ th roots of unity, eliminating choice $B.$ That leaves choice $\boxed{E}$ , and we can confirm that this is correct by noticing that $9$ divides $3033$ but not $1011.$ From that, we can see that the roots of polynomial $E$ are $3033$ rd roots of unity but not $1011$ th roots of unity, so they are all roots of $P.$ Therefore, $E$ divides $P.$

~pianoboy

Solution 5 (Simple Elimination)

Put the value $x = -1$ . This gives $P(-1) = (-1)^{2022} + (-1)^{1011} + 1 = 1$ . This automatically eliminates choices $A$ , $C$ and $D$ since they do not form a factor of $1$ at $x = -1$ .

Solution $5.1$ : Now, put $x = \omega$ (omega) at choice $B$ . We get that, $\omega^2 + \omega + 1 = 0$ Thus choice B has $(x - \omega)$ as a factor. If choice $B$ were to be a factor of $P(x)$ , $(x - \omega)$ would also have to be a factor of $P(x)$ , which is clearly not the case, as $P(\omega) \neq 0$ . This eliminates choice $B$ , leaving us with answer $\boxed{E}$ .

Solution $5.2$ (Mods): Plug in $2$ for $x$ at choice $B$ . $2^2+2*2+1=7$ . The cycle of mods for $2$ and $7$ is $2$ , $4$ , $1$ . Both $2022$ and $1011$ are divisible by $3$ , so we have $1$ + $1$ + $1$ = $3$ . $3$ is not divisible by $7$ , so we eliminate choice $B$ . This leaves us with $\boxed{(E) x^{6}+x^{3}+1}$ .

~SouradipClash_03

Solution $5.2$ ~ Bread 10

Solution 5a (Elimination but slightly different)

Like Solution 5, let $x=-1$ which eliminates the choices of $A$ , $C$ , and $D$ as they do not divide $P(-1)=1$ as they form $3, 2, 3$ respectively by letting $x=-1$ .

This leaves us with only $2$ choices, $B$ and $E$ . Notice that letting $x=0$ or $x=1$ still make these answer choices work and the other values will leave large numbers for us to check which is not feasible in a 75 minutes math competition.

However, we notice answer choice $B$ is quadratic so if answer choice $B$ divides the given polynomial, then the roots of the quadratic must also be roots of the polynomial.

Through quadratic formula, we find the roots of this quadratic as $\dfrac{-1+i\sqrt{3}}{2}$ and $\dfrac{-1-i\sqrt{3}}{2}$ .

We notice that these roots can be written nicely in polar form $\cos(\dfrac{2\pi}{3})+i\sin(\dfrac{2\pi}{3})$ or $\cos(\dfrac{4\pi}{3})+i\sin(\dfrac{4\pi}{3})$ .

We plug either one of these (preferably the root on the left as it is smaller) and see that the polynomial doesn't equal $0$ suggesting that $B$ is not the correct answer choice.

As we only have one answer choice left, we choose $\boxed{E}$

~Batmanstark

Solution 5b (elimination but even faster and cheesier)

We first substitute $-1$ and eliminate $A$ , $C$ , and $D$ . We are now left with $B$ and $E$ . When $x=3$ , $B=13$ ; we then calculate $3^{2022}+3^{1011}+1$ mod $13$ . Note that $2022$ and $1011$ are both multiples of $3$ , and $3^3=27\equiv1\pmod{13}$ . Thus, $3^{2022}\equiv3^{1011}\equiv3^3\equiv1\pmod{13}$ , so it turns out that the given expression is $1+1+1=3$ mod $13$ . We need it to be $0$ , and since $3\neq0$ , then $B$ is not the correct answer; $\boxed{\textbf{(E) }x^6+x^3+1}$ is correct.

~Technodoggo

Video Solution by Math-X (Smart and Simple)

https://youtu.be/7yAh4MtJ8a8?si=OkLQzwiqd9AnIS5K&t=6374 ~Math-X

Video Solution

includes review of factoring polynomials

https://youtu.be/UIxePPu0Zus

~MathProblemSolvingSkills.com

Video Solution by ThePuzzlr

https://youtu.be/YRcaIrwA2AU

~ MathIsChess

@@ Line 1: / Line 1: @@
-==Solution==
+==Problem==
+Let <cmath>P(x) = x^{2022} + x^{1011} + 1.</cmath> Which of the following polynomials is a factor of <math>P(x)</math>?
+<math>\textbf{(A)} \, x^2 -x + 1 \qquad\textbf{(B)} \, x^2 + x + 1 \qquad\textbf{(C)} \, x^4 + 1 \qquad\textbf{(D)} \, x^6 - x^3 + 1  \qquad\textbf{(E)} \, x^6 + x^3 + 1 </math>
+==Solution 1==
 <math>P(x) = x^{2022} + x^{1011} + 1</math> is equal to <math>\frac{x^{3033}-1}{x^{1011}-1}</math> by difference of powers.
@@ Line 11: / Line 16: @@
 By difference of powers, <math>x^9-1=(x^3-1)(x^6+x^3+1)</math>.
-Therefore, the answer is E.
+Therefore, the answer is <math>\boxed{E}</math>.
+==Solution 2==
+We simply test roots for each, as <math>2022,1011</math> are multiples of three, we need to make sure the roots are in the form of <math>e^{i\frac{k\pi}{9}}</math>, so we only have to look at <math>D,E</math>.
+If we look at choice <math>E</math>, <math>x=e^{i\frac{\pm2\pi}{9}}</math> which works perfectly, the answer is just <math>E</math>
+~bluesoul
+==Solution 3==
+Let <math>x^{1011} = u</math>, now we can rewrite our polynomial as <math>u^2+u+1</math>. Using the quadratic formula to solve for the roots of this polynomial, we have <cmath>x^{1011} = \frac{-1\pm i\sqrt{3}}{2}</cmath> Looking at our answer choices, we want to find a polynomial whose roots satisfy this expression. Since the expression <math>x^6+x^3+1</math> is in a similar form to our original polynomial, except with <math>x^3</math> in place of <math>x^{1011}</math>, this would be a good place to start. Solving for the roots of <math>x^3</math> in a similar fashion, <cmath>x^3= \frac{-1\pm i\sqrt{3}}{2}</cmath> for the solution we are testing. Now notice that we can rewrite the roots of <math>x^3</math> as <cmath>x^3 = \operatorname{cis}{\frac{2\pi}{3}}, \operatorname{cis}{\frac{4\pi}{3}}</cmath> Both of which are third roots of unity. We want to now check if this value of <math>x^3</math> satisfies <math>x^{1011} = \frac{-1\pm i\sqrt{3}}{2}</math>. Notice that <math>x^{1011} = (x^{3})^{112\cdot3}\cdot x^3</math>, and since both values of <math>x^3</math> are roots of unity, we can simplify the expression we want satisfied to the expression to <math>x^{1011}=x^3</math>. Since both values of <math>x^3</math> are also values of <math>x^{1011}</math>, the roots for our <math>x^6+x^3+1</math> are also roots of <math>x^{2022}+x^{1011}+1</math>, meaning that <cmath>x^6+x^3+1 | x^{2022}+x^{1011}+1</cmath> so Therefore, the answer is <math>\boxed{E}</math>.
+- DavidHovey
+==Solution 4 (Describe the Roots)==
+We know that a monic polynomial <math>q</math> divides a monic polynomial <math>p</math> if and only if all the roots of <math>q</math> are roots of <math>p.</math> Since <cmath>P(x)=x^{2022}+x^{1011}+1=\frac{x^{3033}-1}{x^{1011}-1}</cmath>, the roots of <math>P</math> are the <math>3033</math>rd roots of unity that aren't <math>1011</math>th roots of unity.
+Now, note that:
+: The roots of polynomial <math>A</math> are the primitive <math>6</math>th roots of unity.
+: The roots of polynomial <math>B</math> are the primitive cube roots of unity.
+: The roots of polynomial <math>C</math> are the primitive <math>8</math>th roots of unity.
+: The roots of polynomial <math>D</math> are the primitive <math>18</math>th roots of unity.
+: The roots of polynomial <math>E</math> are the primitive <math>9</math>th roots of unity.
+However, since <math>6</math>, <math>8</math>, and <math>18</math> don't divide <math>3033</math>, the roots of polynomial <math>A</math> are not all <math>3033</math>rd roots of unity, and the same is true for polynomials <math>C</math> and <math>D</math>, eliminating choices <math>A</math>, <math>C</math> and <math>D.</math> Also, since <math>3</math> divides <math>1011</math>, the roots of polynomial <math>B</math> are all <math>1011</math>th roots of unity, eliminating choice <math>B.</math> That leaves choice <math>\boxed{E}</math>, and we can confirm that this is correct by noticing that <math>9</math> divides <math>3033</math> but not <math>1011.</math> From that, we can see that the roots of polynomial <math>E</math> are <math>3033</math>rd roots of unity but not <math>1011</math>th roots of unity, so they are all roots of <math>P.</math> Therefore, <math>E</math> divides <math>P.</math>
+~pianoboy
+==Solution 5 (Simple Elimination)==
+Put the value <math>x = -1</math>. This gives <math>P(-1) = (-1)^{2022} + (-1)^{1011} + 1 = 1</math>.
+This automatically eliminates choices <math>A</math> , <math>C</math> and <math>D</math> since they do not form a factor of <math>1</math> at <math>x = -1</math>.
+Solution <math>5.1</math>:
+Now, put <math>x = \omega</math> (omega) at choice <math>B</math>. We get that,
+<cmath>\omega^2 + \omega + 1 = 0</cmath>
+Thus choice B has <math>(x - \omega)</math> as a factor. If choice <math>B</math> were to be a factor of <math>P(x)</math>, <math>(x - \omega)</math> would also have to be a factor of <math>P(x)</math>, which is clearly not the case, as <math>P(\omega) \neq 0</math>.
+This eliminates choice <math>B</math>, leaving us with answer <math>\boxed{E}</math>.
+Solution <math>5.2</math>(Mods):
+Plug in <math>2</math> for <math>x</math> at choice <math>B</math>. <math>2^2+2*2+1=7</math>.
+The cycle of mods for <math>2</math> and <math>7</math> is <math>2</math>, <math>4</math>, <math>1</math>. Both <math>2022</math> and <math>1011</math> are divisible by <math>3</math>, so we have <math>1</math> + <math>1</math> + <math>1</math> = <math>3</math>. <math>3</math> is not divisible by <math>7</math>, so we eliminate choice <math>B</math>. This leaves us with <math>\boxed{(E) x^{6}+x^{3}+1}</math>.
+~SouradipClash_03
+Solution <math>5.2</math> ~ Bread 10
+===Solution 5a (Elimination but slightly different)===
+Like Solution 5, let <math>x=-1</math> which eliminates the choices of <math>A</math>, <math>C</math>, and <math>D</math> as they do not divide <math>P(-1)=1</math> as they form <math>3, 2, 3</math> respectively by letting <math>x=-1</math>.
+This leaves us with only <math>2</math> choices, <math>B</math> and <math>E</math>. Notice that letting <math>x=0</math> or <math>x=1</math> still make these answer choices work and the other values will leave large numbers for us to check which is not feasible in a 75 minutes math competition.
+However, we notice answer choice <math>B</math> is quadratic so if answer choice <math>B</math> divides the given polynomial, then the roots of the quadratic must also be roots of the polynomial.
+Through quadratic formula, we find the roots of this quadratic as <math>\dfrac{-1+i\sqrt{3}}{2}</math> and <math>\dfrac{-1-i\sqrt{3}}{2}</math>.
+We notice that these roots can be written nicely in polar form <math>\cos(\dfrac{2\pi}{3})+i\sin(\dfrac{2\pi}{3})</math> or <math>\cos(\dfrac{4\pi}{3})+i\sin(\dfrac{4\pi}{3})</math>.
+We plug either one of these (preferably the root on the left as it is smaller) and see that the polynomial doesn't equal <math>0</math> suggesting that <math>B</math> is not the correct answer choice.
+As we only have one answer choice left, we choose <math>\boxed{E}</math>
+~Batmanstark
+===Solution 5b (elimination but even faster and cheesier)===
+We first substitute <math>-1</math> and eliminate <math>A</math>, <math>C</math>, and <math>D</math>. We are now left with <math>B</math> and <math>E</math>. When <math>x=3</math>, <math>B=13</math>; we then calculate <math>3^{2022}+3^{1011}+1</math> mod <math>13</math>. Note that <math>2022</math> and <math>1011</math> are both multiples of <math>3</math>, and <math>3^3=27\equiv1\pmod{13}</math>. Thus, <math>3^{2022}\equiv3^{1011}\equiv3^3\equiv1\pmod{13}</math>, so it turns out that the given expression is <math>1+1+1=3</math> mod <math>13</math>. We need it to be <math>0</math>, and since <math>3\neq0</math>, then <math>B</math> is not the correct answer; <math>\boxed{\textbf{(E) }x^6+x^3+1}</math> is correct.
+~Technodoggo
+==Video Solution by Math-X (Smart and Simple)==
+https://youtu.be/7yAh4MtJ8a8?si=OkLQzwiqd9AnIS5K&t=6374 ~Math-X
+== Video Solution==
+includes review of factoring polynomials
+https://youtu.be/UIxePPu0Zus
+~MathProblemSolvingSkills.com
+== Video Solution by ThePuzzlr ==
+https://youtu.be/YRcaIrwA2AU
+~ MathIsChess
+== See Also ==
+{{AMC12 box|year=2022|ab=A|num-b=20|num-a=22}}
+[[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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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