LIVE by Po-Shen Loh标识

LIVE

中文

2021 AMC 12B Spring 第 19 题

先试着解答 2021 AMC 12B Spring 第 19 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 2021 AMC 12B Spring 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。

19.

掷两个公平骰子，每个骰子至少有 66 个面。每个骰子的各面分别标着从 11 到该骰子面数的不同整数。掷出和为 77 的概率是掷出和为 1010 的概率的 34\dfrac34，且掷出和为 1212 的概率是 112\dfrac{1}{12}。两个骰子的面数总和最小可能是多少？

Two fair dice, each with at least 66 faces are rolled. On each face of each die is printed a distinct integer from 11 to the number of faces on that die, inclusive. The probability of rolling a sum of 77 is 34\dfrac34 of the probability of rolling a sum of 10,10, and the probability of rolling a sum of 1212 is 112.\dfrac{1}{12}. What is the least possible number of faces on the two dice combined?

1616

1717

1818

1919

2020

答案：B
知识点：骰子（概率）最优化
难度评级：2120
小提示：

因为每个骰子至少有 66 个面，和为 77 总有 66 种方式，所以和为 1010 有 88 种方式

With at least 66 faces each, a sum of 77 always has 66 ways, so a sum of 1010 has 88 ways

大提示：

设两个骰子分别有 a≤ba\le b 个面；数和为 1010 与 1212 的结果数，并使用 方法数(12)=ab12\text{方法数}(12)=\tfrac{ab}{12}

Let the dice have a≤ba\le b faces; count outcomes for sums 1010 and 1212 and use ways(12)=ab12\text{ways}(12)=\tfrac{ab}{12}

解答：

设两个骰子的面数为 a≤ba\le b。因为两个骰子都至少有 66 个面，和为 77 恰有 66 种方式，所以和为 1010 有 6÷34=86\div\tfrac34=8 种方式。

掷出 1010 的方式数为 min⁡(a,9)\min(a,9) −max⁡(1,10−b)+1=8-\max(1,10-b)+1=8。和为 1212 的概率为 112\tfrac{1}{12}，所以它有 ab12\tfrac{ab}{12} 种方式。

和为 1010 有 88 种结果，要求 a≥8a\ge8 且 b≥9b\ge9，所以必有 a+b≥17a+b\ge17。当 (a,b)=(8,9)(a,b)=(8,9) 时，和为 1010 恰有 88 种结果，而和为 1212 恰有第一枚骰子掷出 3,4,…,83,4,\ldots,8 时的 66 种结果。因为 68⋅9=112\frac{6}{8\cdot9}=\frac{1}{12}，两个条件都成立，从而下界 1717 可以达到。

所以正确答案是 B。

Let the dice have a≤ba\le b faces. Since both have at least 66 faces, a sum of 77 occurs in exactly 66 ways, so a sum of 1010 occurs in 6÷34=86\div\tfrac34=8 ways.

The number of ways to roll 1010 is min⁡(a,9)\min(a,9) −max⁡(1,10−b)+1=8.-\max(1,10-b)+1=8. A sum of 1212 has probability 112,\tfrac{1}{12}, so it occurs in ab12\tfrac{ab}{12} ways.

Having 88 outcomes for a sum of 1010 requires a≥8a\ge8 and b≥9,b\ge9, so necessarily a+b≥17.a+b\ge17. For (a,b)=(8,9),(a,b)=(8,9), a sum of 1010 has 88 outcomes, while a sum of 1212 has the 66 outcomes with the first die showing 3,4,…,8.3,4,\ldots,8. Since 68⋅9=112,\frac{6}{8\cdot9}=\frac{1}{12}, both conditions hold and the lower bound 1717 is attained.

Thus, the correct answer is B.

第 18 题#18
完整试卷
第 20 题#20

其他年份的第 19 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 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 12A · 2025 AMC 12B