LIVE by Po-Shen Loh标识

LIVE

中文

2025 AMC 12A 第 13 题

先试着解答 2025 AMC 12A 第 13 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。

13.

令 C={1,2,3,…,13}C = \{1, 2, 3, \ldots, 13\}。令 NN 为最大的整数，使得存在一个有 NN 个元素的 CC 的子集，并且该子集不含五个连续整数。现在从 CC 中随机不放回地选出 NN 个整数。所选元素不包含五个连续整数的概率是多少？

Let C={1,2,3,…,13}.C = \{1, 2, 3, \ldots, 13\}. Let NN be the greatest integer such that there exists a subset of CC with NN elements that does not contain five consecutive integers. Suppose NN integers are chosen at random from CC without replacement. What is the probability that the chosen elements do not include five consecutive integers?

3130\dfrac{3}{130}

3143\dfrac{3}{143}

5143\dfrac{5}{143}

126\dfrac{1}{26}

578\dfrac{5}{78}

答案：D
知识点：组合基本概率极端原理
难度评级：1660
小提示：

去掉两个合适的元素就能打断每一段五连整数，所以 N=11N = 11。

Removing two well-placed elements breaks every run of five, so N=11N = 11

大提示：

选 1111 个等价于去掉 22 个；数一数哪两个被去掉的元素能碰到每一段五个连续整数

Choosing 1111 means removing 22; count removals whose two elements hit every block of five consecutive integers

解答：

要避免五个连续整数，只需去掉两个元素（例如 55 和 1010），而去掉一个元素不可能同时击中两个不相交的区块 {1,…,5}\{1,\ldots,5\} 和 {9,…,13}\{9,\ldots,13\}。因此 N=11N = 11。

从 1313 个元素中选 1111 个，等价于去掉 22 个，共有 (132)=78\binom{13}{2} = 78 种。所选集合不含五个连续整数，当且仅当两个被去掉的元素合起来与每个窗口 {t,t+1,t+2,t+3,t+4}\{t, t+1, t+2, t+3, t+4\} 相交，其中 t=1,…,9t = 1, \ldots, 9。

这迫使一个被去掉的元素在 {1,…,5}\{1,\ldots,5\} 中，另一个在 {9,…,13}\{9,\ldots,13\} 中，并且两者相距不超过 55。有效的去法为 {4,9}\{4,9\}、{5,9}\{5,9\} 和 {5,10}\{5,10\}，共 33 种。

概率为 378=126\dfrac{3}{78} = \dfrac{1}{26}。

因此，正确答案是 D。

To avoid five consecutive integers, it suffices to remove two elements (for example 55 and 1010), and no single removal can hit both disjoint blocks {1,…,5}\{1,\ldots,5\} and {9,…,13}.\{9,\ldots,13\}. Thus N=11.N = 11.

Choosing 1111 of 1313 elements is the same as removing 2,2, which can be done in (132)=78\binom{13}{2} = 78 ways. The chosen set avoids five consecutive integers exactly when the two removed elements together intersect every window {t,t+1,t+2,t+3,t+4}\{t, t+1, t+2, t+3, t+4\} for t=1,…,9.t = 1, \ldots, 9.

This forces one removed element in {1,…,5},\{1,\ldots,5\}, the other in {9,…,13},\{9,\ldots,13\}, and the two within 55 of each other. The valid removals are {4,9},\{4,9\}, {5,9},\{5,9\}, and {5,10},\{5,10\}, giving 33 of them.

The probability is 378=126.\dfrac{3}{78} = \dfrac{1}{26}.

Thus, the correct answer is D.

第 12 题#12
完整试卷
第 14 题#14

其他年份的第 13 题

1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1953 AMC 12 · 1954 AMC 12 · 1955 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12 · 1960 AMC 12 · 1961 AMC 12 · 1962 AMC 12 · 1963 AMC 12 · 1964 AMC 12 · 1965 AMC 12 · 1966 AMC 12 · 1967 AMC 12 · 1968 AMC 12 · 1969 AMC 12 · 1970 AMC 12 · 1971 AMC 12 · 1972 AMC 12 · 1973 AMC 12 · 1974 AMC 12 · 1975 AMC 12 · 1976 AMC 12 · 1977 AMC 12 · 1978 AMC 12 · 1979 AMC 12 · 1980 AMC 12 · 1981 AMC 12 · 1982 AMC 12 · 1983 AMC 12 · 1984 AMC 12 · 1985 AMC 12 · 1986 AMC 12 · 1987 AMC 12 · 1988 AMC 12 · 1989 AMC 12 · 1990 AMC 12 · 1991 AMC 12 · 1992 AMC 12 · 1993 AMC 12 · 1994 AMC 12 · 1995 AMC 12 · 1996 AMC 12 · 1997 AMC 12 · 1998 AMC 12 · 1999 AMC 12 · 2000 AMC 12 · 2001 AMC 12 · 2002 AMC 12A · 2002 AMC 12B · 2003 AMC 12A · 2003 AMC 12B · 2004 AMC 12A · 2004 AMC 12B · 2005 AMC 12A · 2005 AMC 12B · 2006 AMC 12A · 2006 AMC 12B · 2007 AMC 12A · 2007 AMC 12B · 2008 AMC 12A · 2008 AMC 12B · 2009 AMC 12A · 2009 AMC 12B · 2010 AMC 12A · 2010 AMC 12B · 2011 AMC 12A · 2011 AMC 12B · 2012 AMC 12A · 2012 AMC 12B · 2013 AMC 12A · 2013 AMC 12B · 2014 AMC 12A · 2014 AMC 12B · 2015 AMC 12A · 2015 AMC 12B · 2016 AMC 12A · 2016 AMC 12B · 2017 AMC 12A · 2017 AMC 12B · 2018 AMC 12A · 2018 AMC 12B · 2019 AMC 12A · 2019 AMC 12B · 2020 AMC 12A · 2020 AMC 12B · 2021 AMC 12A Spring · 2021 AMC 12B Spring · 2021 AMC 12A Fall · 2021 AMC 12B Fall · 2022 AMC 12A · 2022 AMC 12B · 2023 AMC 12A · 2023 AMC 12B · 2024 AMC 12A · 2024 AMC 12B · 2025 AMC 12B