Track / Stage 5 / 326 of 400 #926 of 2000

Problem 926

AIME late

Combinatorics Difficulty 5.8 Prove it

Example 3 - (Beijing Initial Number Competition Question) A cinema has a total of 985 seats, with performances in the morning and afternoon. Schools A and B each have 1985 students watching the movie (either the morning or the afternoon session). Prove: There must be such a seat in the cinema where a student from a different school sits in the morning and afternoon.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Proof: Assuming no seat is occupied by students from different schools in the morning and afternoon, let's say students from School A sit in $n$ seats in the morning, then students from School B sit in $1985-n$ seats. The remaining $n$ students from School B must sit in the $1985-n$ seats occupied by School B in the morning, thus we have $n=1985-n$ , which means $2n=1985$ . The left side of this equation is even, while the right side is odd, leading to a contradiction. The original problem is proved.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.