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

Revision as of 21:54, 13 November 2022

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

$P(x) = x^{2022} + x^{1011} + 1$ 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

Video Solution by ThePuzzlr

https://youtu.be/YRcaIrwA2AU

~ MathIsChess

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
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 20: / Line 20: @@
 ==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>.
+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>
+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
 == Video Solution by ThePuzzlr ==

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