2025 AMC 12A 第 23 题

先试着解答 2025 AMC 12A 第 23 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 2025 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。

23.

若一个正整数不重复使用任何数字,没有数字 00,并且没有任何数字同时与两个更大的数字相邻,则称它为公平数。例如,196196,2323,和 1246312463 是公平数,但 15461546,320320,和 3432134321 不是。共有多少个公平正整数?
Call a positive integer fair if no digit is used more than once, it has no 00s, and no digit is adjacent to two greater digits. For example, 196,196, 23,23, and 1246312463 are fair, but 1546,1546, 320,320, and 3432134321 are not fair. How many fair positive integers are there?
511511
25842584
98419841
1771117711
1968219682
答案:C
知识点:有限制的排列二项式定理
小提示:
公平数没有任何内部数字小于它的两个相邻数字
A fair number has no interior digit smaller than both of its neighbors
大提示:
对选定的数字集合,从大到小构造有效排列,每个新数字放在任一端:nn 个数字有 2n−12^{n-1} 种
Build a valid arrangement of a chosen digit set by placing digits largest-to-smallest, each at either end: 2n−12^{n-1} ways for nn digits
解答:
数字互不相同,并且来自 {1,…,9}\{1, \ldots, 9\},条件“没有数字同时与两个更大的数字相邻”等价于没有内部数字小于它的两个相邻数字。
对固定的 nn 个数字,从大到小插入来构造排列;每个新的较小数字必须放在当前排列的两个端点之一,因此有 2n−12^{n-1} 个有效排列。
对所有非空数字子集求和,∑n=19(9n)2n−1=12∑n=19(9n)2n=39−12=196822=9841。 \begin{gathered} \sum_{n=1}^{9}\binom{9}{n}2^{n-1} \\ = \frac{1}{2}\sum_{n=1}^{9}\binom{9}{n}2^{n} \\ = \frac{3^9 - 1}{2} = \frac{19682}{2} = 9841 \end{gathered}\text{。}
因此,正确答案是 C。
The digits are distinct and drawn from {1,…,9},\{1, \ldots, 9\}, and “no digit adjacent to two greater digits” means no interior digit is smaller than both neighbors.
For a fixed set of nn digits, build the arrangement by inserting digits from largest to smallest; each new (smaller) digit must go to one of the two ends, giving 2n−12^{n-1} valid arrangements.
Summing over all nonempty digit subsets, ∑n=19(9n)2n−1=12∑n=19(9n)2n=39−12=196822=9841. \begin{gathered} \sum_{n=1}^{9}\binom{9}{n}2^{n-1} \\ = \frac{1}{2}\sum_{n=1}^{9}\binom{9}{n}2^{n} \\ = \frac{3^9 - 1}{2} = \frac{19682}{2} = 9841. \end{gathered}
Thus, the correct answer is C.
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