2017 AMC 12A 第 21 题

先试着解答 2017 AMC 12A 第 21 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2017 AMC 12A 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

21.

集合 SS 按如下方式构造。开始时,S={0,10}S=\{0,10\}。之后尽可能重复下面的操作:如果 xx 是某个多项式 anxn+an−1xn−1a_nx^n+a_{n-1}x^{n-1} +⋯+a1x+a0+\cdots+a_1x+a_0 的整数根,其中 n≥1n\ge1 且所有系数 aia_i 都是 SS 中的元素,那么把 xx 加入 SS。当不能再向 SS 加入新元素时,SS 有多少个元素?
A set SS is constructed as follows. To begin, S={0,10}.S=\{0,10\}. Repeatedly, as long as possible, if xx is an integer root of some polynomial anxn+an−1xn−1a_nx^n+a_{n-1}x^{n-1} +⋯+a1x+a0+\cdots+a_1x+a_0 for some n≥1,n\ge1, all of whose coefficients aia_i are elements of S,S, then xx is put into S.S. When no more elements can be added to S,S, how many elements does SS have?
44
55
77
99
1111
答案:D
知识点:多项式整除性系统列举
小提示:
由有理根定理,系数在 SS 中的多项式的任意整数根都整除其常数项
By the Rational Root Theorem, any integer root of a polynomial with coefficients in SS divides its constant term
大提示:
从 {0,10}\{0,10\} 出发可得到 ±1,±2,±5,±10\pm1,\pm2,\pm5,\pm10;再说明没有别的数能进入
Starting from {0,10},\{0,10\}, reach ±1,±2,±5,±10;\pm1,\pm2,\pm5,\pm10; show nothing else can enter
解答:
使用多项式 10x+1010x+10,根 −1-1 进入 SS。然后 11 作为 −x10−x9−⋯−x+10-x^{10}-x^9-\cdots-x+10 的根进入,而 −10-10 由 x+10x+10 得到。
现在 x3+x−10x^3+x-10 有根 22,而 x+2x+2 给出 −2-2;接着 2x−102x-10 和 2x+102x+10 给出 ±5\pm5。此时 S={0,±1,±2,±5,±10}S=\{0,\pm1,\pm2,\pm5,\pm10\}。
不会再出现其他整数。用归纳法,SS 中每个非零元素都整除 1010。如果规则中使用的多项式常数项为 00,先提出最高可能次幂的 xx;任何非零根就成为一个新多项式的根,而其常数项是原多项式的第一个非零系数。由有理根定理,该根整除这个系数;归纳假设说明该系数整除 1010。因此 SS 有 99 个元素。
因此,正确答案是 D。
Using 10x+10,10x+10, the root −1-1 enters S.S. Then 11 enters as a root of −x10−x9−⋯−x+10,-x^{10}-x^9-\cdots-x+10, and −10-10 enters from x+10.x+10.
Now x3+x−10x^3+x-10 has root 2,2, and x+2x+2 gives −2;-2; then 2x−102x-10 and 2x+102x+10 give ±5.\pm5. At this point S={0,±1,±2,±5,±10}.S=\{0,\pm1,\pm2,\pm5,\pm10\}.
No further integer can appear. Inductively, every nonzero member of SS divides 10.10. If a polynomial used in the rule has constant term 0,0, factor out the largest possible power of x;x; any nonzero root is then a root of a polynomial whose constant term is the first nonzero original coefficient. The Rational Root Theorem shows that the root divides this coefficient, which by the inductive hypothesis divides 10.10. So SS has 99 elements.
Thus, the correct answer is D.
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