2022 AMC 12A Problems/Problem 21

Revision as of 20:54, 13 November 2022 by Bluesoul (talk | contribs) (Solution 2)

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

See Also

2022 AMC 12A (ProblemsAnswer KeyResources)
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

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