AMC12 2018 B

AMC12 2018 B · Q22

AMC12 2018 B · Q22. It mainly tests Polynomials, Manipulating equations.

Consider polynomials $P(x)$ of degree at most 3, each of whose coefficients is an element of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. How many such polynomials satisfy $P(-1) = -9$?

考虑度数至多为 3 的多项式 $P(x)$，其每个系数均为集合 $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ 中的元素。有多少这样的多项式满足 $P(-1) = -9$？

(A) 110 110

(B) 143 143

(D) 220 220

(E) 286 286

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $P(x)=ax^3+bx^2+cx+d$, where $a$, $b$, $c$, and $d$ are integers between 0 and 9, inclusive. The condition $P(-1)=-9$ is equivalent to $-a+b-c+d=-9$. Adding 18 to both sides gives $(9-a)+b+(9-c)+d=9$ where $0\le 9-a,b,9-c,d\le 9$. By the stars and bars argument, there are $\binom{9+4-1}{4-1}=\binom{12}{3}=220$ nonnegative integer solutions to $x_1+x_2+x_3+x_4=9$. Each of these give rise to one of the desired polynomials.

答案（D）：设 $P(x)=ax^3+bx^2+cx+d$，其中 $a,b,c,d$ 为 0 到 9（含）之间的整数。条件 $P(-1)=-9$ 等价于 $-a+b-c+d=-9$。两边同加 18 得到 $(9-a)+b+(9-c)+d=9$，其中 $0\le 9-a,b,9-c,d\le 9$。由“插板法（stars and bars）”，满足 $x_1+x_2+x_3+x_4=9$ 的非负整数解个数为 $\binom{9+4-1}{4-1}=\binom{12}{3}=220$。每一个解都对应一个所求多项式。

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