2024 AMC 10B Problems/Problem 12

Problem

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?

$\textbf{(A) } 9 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 51 \qquad\textbf{(E) } 100$

Solution 1

Let's say we have some number of languages. Then each student will speak some amount of those languages, and no two people can have the same combination of languages or else the conditions will no longer be satisfied. Notice that $\dbinom94=126\geq100$ . So each of the $100$ students can speak some $4$ of the $9$ languages. Thus, $\boxed{9}$ is our answer.

~lprado ~LaTeX edits by Elephant200

Solution 2 (Rigorous)

Let $n$ be the total number of languages spoken by all the students, and let $k$ be the number of languages that each student speaks. Since all students speak the same number of languages, the condition given in the question can be modified as -

"If $N$ is the $n$ -element subset containing the union of the languages spoken by all students, each of the 100 students speaks a different (unique) $k$ -element subset of $N$ combination of languages."

In more mathematical terms, this means ${n \choose k} \ge 100$ . (Using PHP)

Because we need the least value of $n$ , $k$ must be the closest integer to $\frac{n}{2}$

$\Rightarrow {n \choose \frac{n}{2}} \ge 100$ for $n = 2m$ , ${n \choose \frac{n-1}{2}} \ge 100$ for $n=2m-1$ , $m \in \mathbb{N}$

In the context of this question, it will be fastest to just substitute $\normalsize n$ to each of the option values to find the least value possible as $\boxed{\textbf{(A) } 9}$

For a more rigorous proof, where options are not an option, one can manually count up from $n=1$ as $100$ is a small number.

~laythe_enjoyer211

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/YqKmvSR1Ckk?feature=shared

~ Pi Academy

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search

2024 AMC 10B Problems/Problem 12

Contents

Problem

Solution 1

Solution 2 (Rigorous)

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

See also