AMC10 2024 B

AMC10 2024 B · Q12

AMC10 2024 B · Q12. It mainly tests Pigeonhole principle, Logic puzzles.

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?

来自不同国家的 $100$ 名学生参加数学竞赛聚会。每个学生会说相同数量的语言，而且对于每对学生 $A$ 和 $B$，学生 $A$ 会说某种学生 $B$ 不会说的语言，学生 $B$ 会说某种学生 $A$ 不会说的语言。所有学生所说的语言总数的可能最小值为多少？

(A) 9 9

(B) 10 10

(D) 51 51

(E) 100 100

Answer

Correct choice: (A)

正确答案：（A）

Solution

We think of this problem like boxes. First start with 9. We see that we can arrange the groups of people into the 9 boxes. We take 9 people of different languages and arrange them in each of the boxes. This means we have $100 - 9 \times 9 = 19$ people remaining. We then take 18 people of different language (as they can be put in a pair and still qualify) and put them in the boxes. This gives us 1 person left over. Now, this is where MAA wants you to choose 10, but one can quickly see that the last person can be put into any group of three (except a group that includes their language) and still qualify whilst maintaining all the constraints. Therefore our answer is $\boxed{\textbf{(A) } 9}$. This solution is a basic introduction/summary of the Pigeonhole Principle

我们可以将这个问题想象成盒子。先从 $9$ 开始。我们可以将人分成 $9$ 个盒子。我们取 $9$ 个人，每人不同语言，放入每个盒子。这意味着还剩 $100 - 9 \times 9 = 19$ 人。然后取 $18$ 个人不同语言（他们可以成对放入仍满足条件），放入盒子，还剩 $1$ 人。现在，最后一个人可以放入任何三人组（除包含其语言的组），仍满足所有约束。因此答案是 $\boxed{\textbf{(A) } 9}$。这是一个鸽巢原理的基本介绍/总结。

Topics