AMC12 2021 B

AMC12 2021 B · Q20

AMC12 2021 B · Q20. It mainly tests Polynomials, Manipulating equations.

Let $Q(z)$ and $R(z)$ be the unique polynomials such that\[z^{2021}+1=(z^2+z+1)Q(z)+R(z)\]and the degree of $R$ is less than $2.$ What is $R(z)?$

设 $Q(z)$ 和 $R(z)$ 是唯一多项式，使得\[z^{2021}+1=(z^2+z+1)Q(z)+R(z)\]且 $R$ 的度数小于 $2$。$R(z)$ 是什么？

(A) {-}z {-}z

(B) {-}1 {-}1

(D) z+1 z+1

(E) 2z+1 2z+1

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let $z=s$ be a root of $z^2+z+1$ so that $s^2+s+1=0.$ It follows that \[(s-1)\left(s^2+s+1\right)=s^3-1=0,\] from which $s^3=1,$ but $s\neq1.$ Note that \begin{align*} s^{2021}+1 &= s^{3\cdot673+2}+1 \\ &= (s^3)^{673}\cdot s^2+1 \\ &= s^2+1 \\ &= \left(s^2+s+1\right)-s \\ &= -s. \end{align*} Since $z^{2021}+1=-z$ for each root $z=s$ of $z^2+z+1,$ the remainder when $z^{2021}+1$ is divided by $z^2+z+1$ is $R(z)=\boxed{\textbf{(A) }{-}z}.$

设 $z=s$ 是 $z^2+z+1$ 的根，即 $s^2+s+1=0$。则 \[(s-1)\left(s^2+s+1\right)=s^3-1=0,\] 由此 $s^3=1$，但 $s\neq1$。注意到 \begin{align*} s^{2021}+1 &= s^{3\cdot673+2}+1 \\ &= (s^3)^{673}\cdot s^2+1 \\ &= s^2+1 \\ &= \left(s^2+s+1\right)-s \\ &= -s. \end{align*} 由于对于 $z^2+z+1$ 的每个根 $z=s$，有 $z^{2021}+1=-z$，因此 $z^{2021}+1$ 除以 $z^2+z+1$ 的余数是 $R(z)=\boxed{\textbf{(A) }{-}z}$。

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