Question

来自不同国家的 $100$ 名学生在一次数学竞赛中相遇。每名学生会说相同数量的语言，并且对任意两名学生 $A$ 和 $B$ ，学生 $A$ 会说某种学生 $B$ 不会说的语言，同时学生 $B$ 也会说某种学生 $A$ 不会说的语言。所有学生会说的语言种类总数最少可能是多少？

A group of $100$ students from different countries meet at a mathematics competition. Each student speaks the same number of languages, and, for every pair of students $A$ and $B,$ student $A$ speaks some language that student $B$ does not speak, and student $B$ speaks some language that student $A$ does not speak. What is the least possible total number of languages spoken by all the students?

$9$

$10$

$12$

$51$

$100$

答案：A

知识点：子集组合极端原理

难度评级：1500

解答：

把每名学生会说的语言看作一个集合。条件表示没有人的集合包含另一个人的集合。每个人会说同样数量 $k$ 种语言，而两个不同的 $k$ 元集合不可能互相包含，所以只需要有 $100$ 个不同的 $k$ 元子集。设语言总数为 $n$ ，需要 $\binom{n}{k} \ge 100$ 。当 $n = 8$ 时最多为 $\binom{8}{4} = 70$ ，不足 $100$ 。但 $\binom{9}{4} = 126 \ge 100$ 。因此 $9$ 种语言既足够也必要。正确答案是 A。

Give each student the set of languages they speak. The condition says no one's set sits inside another's. Everyone speaks the same number $k$ of languages, and two distinct $k$ -element sets can never contain each other, so all we need is $100$ different $k$ -subsets of the $n$ languages, i.e. $\binom{n}{k} \ge 100.$ With $n = 8$ the best we can manage is $\binom{8}{4} = 70,$ short of $100.$ But $\binom{9}{4} = 126 \ge 100.$ So $9$ languages are both enough and necessary. Therefore, the answer is A.

2024 AMC 10B 第 12 题

答案：A

难度评级：1500

解答：

其他年份的第 12 题