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, 2018

Algebra Difficulty 5.7 AIME, harder Prove it United States

Problem:

In a game, $N$ people are in a room. Each of them simultaneously writes down an integer between $0$ and $100$ inclusive. A person wins the game if their number is exactly two-thirds of the average of all the numbers written down. There can be multiple winners or no winners in this game. Let $m$ be the maximum possible number such that it is possible to win the game by writing down $m$ . Find the smallest possible value of $N$ for which it is possible to win the game by writing down $m$ in a room of $N$ people.

Solution

Solution:

Since the average of the numbers is at most $100$ , the winning number is an integer which is at most two-thirds of $100$ , or at most $66$ . This is achieved in a room with $34$ people, in which $33$ people pick $100$ and one person picks $66$ , so the average number is $99$ .

Furthermore, this cannot happen with less than $34$ people. If the winning number is $66$ and there are $N$ people, the sum of the numbers must be $99N$ . Then we must have that $99N \leq 66 + 100(N-1)$ , which reduces to $N \geq 34$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.