2022 AMC 12A Problem 21

Below is the professionally curated solution for Problem 21 of the 2022 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 12A solutions, or check the answer key.

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Concepts:roots of unity polynomial

Difficulty rating: 2170

21.

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

$x^2-x+1$

$x^2+x+1$

$x^4+1$

$x^6-x^3+1$

$x^6+x^3+1$

Solution:

If $\zeta$ is a root of a divisor, then $P(\zeta)=\zeta^{2022}+\zeta^{1011}+1=0$ is required.

The roots of $x^6+x^3+1$ are the primitive $9$ th roots of unity, so $\zeta^9=1.$ Reducing exponents modulo $9,$ $2022\equiv6$ and $1011\equiv3,$ giving $\zeta^6+\zeta^3+1.$ Here $\omega=\zeta^3$ is a primitive cube root of unity, so this equals $\omega^2+\omega+1=0.$

The other four options fail: substituting their roots yields nonzero values (for instance, the primitive cube roots of unity give $P=3$ ).

Thus, the correct answer is E.

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