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

(Problem)
Line 1: Line 1:
 
==Problem==
 
==Problem==
Let <cmath>P(x) = x^{2022} + x^{1011} + 1</cmath>. Which of the following polynomials is a factor of <math>P(x)</math>?
+
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>
 
<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>

Revision as of 12:51, 12 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 E.

Video Solution by ThePuzzlr

https://youtu.be/YRcaIrwA2AU

~ MathIsChess