AMC12 2017 A

AMC12 2017 A · Q21

AMC12 2017 A · Q21. It mainly tests Polynomials, Divisibility & factors.

A set $S$ is constructed as follows. To begin, $S = \{0, 10\}$. Repeatedly, as long as possible, if $x$ is an integer root of some polynomial $a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ for some $n \geq 1$, all of whose coefficients $a_i$ are elements of $S$, then $x$ is put into $S$. When no more elements can be added to $S$, how many elements does $S$ have?

一个集合 $S$ 的构造如下。开始时，$S = \{0, 10\}$。反复地，只要可能，如果 $x$ 是某个多项式 $a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ 的整数根，其中 $n \geq 1$，且所有系数 $a_i$ 都是 $S$ 的元素，则将 $x$ 加入 $S$。当无法再添加元素到 $S$ 时，$S$ 有多少个元素？

(A) 4 4

(B) 5 5

(D) 9 9

(E) 11 11

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Because $-1$ is a root of $10x+10$, $-1$ is added to $S$. Then $1$ is also added to $S$, because it is a root of $(-1)x^{10}+(-1)x^{9}+\cdots+(-1)x+10$. At this point $-10$, a root of $1\cdot x+10$, can be added to $S$. Because $2$ is a root of $1\cdot x^{3}+0\cdot x^{2}+1\cdot x+(-10)$, and $-2$ is a root of $1\cdot x+2$, both $2$ and $-2$ can be added to $S$. The polynomials $2x+(-10)$ and $2x+10$ allow $5$ and $-5$ into $S$. At this point $S=\{0,\pm 1,\pm 2,\pm 5,\pm 10\}$. No more elements can be added to $S$, because by the Rational Root Theorem, any integer root of a polynomial with integer coefficients whose constant term is a factor of $10$ must be a factor of $10$. Therefore $S$ contains $9$ elements. Note: It is not true that in general if $S$ starts with $\{0,c\}$ then all factors of $c$ can be added to $S$. For example, applying the procedure to $\{0,35\}$ gives only $\{0,\pm 1,\pm 35\}$, although of course it takes some argument to rule out $\pm 5$ and $\pm 7$.

答案（D）：因为$-1$是$10x+10$的一个根，所以把$-1$加入$S$。然后也把$1$加入$S$，因为它是$(-1)x^{10}+(-1)x^{9}+\cdots+(-1)x+10$的一个根。此时，$-10$（它是$1\cdot x+10$的一个根）可以加入$S$。因为$2$是$1\cdot x^{3}+0\cdot x^{2}+1\cdot x+(-10)$的一个根，而$-2$是$1\cdot x+2$的一个根，所以$2$和$-2$都可以加入$S$。多项式$2x+(-10)$和$2x+10$使得$5$和$-5$可以加入$S$。此时$S=\{0,\pm 1,\pm 2,\pm 5,\pm 10\}$。不能再向$S$加入更多元素，因为根据有理根定理，若一个整系数多项式的常数项是$10$的因子，则它的任何整数根都必须是$10$的因子。因此$S$包含$9$个元素。注：一般来说，并不成立“如果$S$从$\{0,c\}$开始，那么$c$的所有因子都能加入$S$”。例如，把该过程应用到$\{0,35\}$只得到$\{0,\pm 1,\pm 35\}$，当然需要一些论证才能排除$\pm 5$和$\pm 7$。

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