AMC10 2022 B

AMC10 2022 B · Q21

AMC10 2022 B · Q21. It mainly tests Polynomials, Algebra misc.

Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$, the remainder is $x+2$, and when $P(x)$ is divided by the polynomial $x^2+1$, the remainder is $2x+1$. There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?

设 $P(x)$ 是一个具有有理系数的多项式，使得当 $P(x)$ 被多项式 $x^2 + x + 1$ 除时，余数是 $x+2$，当 $P(x)$ 被多项式 $x^2+1$ 除时，余数是 $2x+1$。存在一个唯一的最低次数的多项式具有这两个性质。该多项式的系数平方和是多少？

(A) 10 10

(B) 13 13

(D) 20 20

(E) 23 23

Answer

Correct choice: (E)

正确答案：（E）

Solution

Given that all the answer choices and coefficients are integers, we hope that $P(x)$ has positive integer coefficients. Throughout this solution, we will express all polynomials in base $x$. E.g. $x^2 + x + 1 = 111_{x}$. We are given: \[111a + 12 = 101b + 21 = P(x).\] We add $111$ and $101$ to each side and balance respectively: \[111(a - 1) + 123 = 101(b - 1) + 122 = P(x).\] We make the unit's digits equal: \[111(a - 1) + 123 = 101(b - 2) + 223 = P(x).\] We now notice that: \[111(a - 11) + 1233 = 101(b - 12) + 1233 = P(x).\] Therefore $a = 11_{x} = x + 1$, $b = 12_{x} = x + 2$, and $P(x) = 1233_{x} = x^3 + 2x^2 + 3x + 3$. $3$ is the minimal degree of $P(x)$ since there is no way to influence the $x$‘s digit in $101b + 21$ when $b$ is an integer. The desired sum is $1^2 + 2^2 +3^2+ 3^2 = \boxed{\textbf{(E)} \ 23}$ P.S. The four computational steps can be deduced through quick experimentation.

给定所有答案选项和系数都是整数，我们希望 $P(x)$ 具有正整数系数。在整个解法中，我们将所有多项式用基数 $x$ 表示。例如 $x^2 + x + 1 = 111_{x}$。我们有： \[111a + 12 = 101b + 21 = P(x).\] 我们将 $111$ 和 $101$ 分别加到等式两边并平衡： \[111(a - 1) + 123 = 101(b - 1) + 122 = P(x).\] 我们使个位数相等： \[111(a - 1) + 123 = 101(b - 2) + 223 = P(x).\] 我们现在注意到： \[111(a - 11) + 1233 = 101(b - 12) + 1233 = P(x).\] 因此 $a = 11_{x} = x + 1$，$b = 12_{x} = x + 2$，且 $P(x) = 1233_{x} = x^3 + 2x^2 + 3x + 3$。$3$ 是 $P(x)$ 的最小次数，因为当 $b$ 是整数时，没有办法影响 $101b + 21$ 中的 $x$ 的位数。所需的和是 $1^2 + 2^2 +3^2+ 3^2 = \boxed{\textbf{(E)} \ 23}$ P.S. 这四个计算步骤可以通过快速实验推导出来。

Topics