We have a set A of numbers 1, 2, 3... to 200
The question is asking me to prove that if I choose 101 numbers from the set, there will be two such that one evenly divides the other.
I know this could be the pigeonhole principle question. I could prove by contradiction that no two numbers will evenly divide each other. Assume I take 101 numbers, I can't take all the odd numbers because there is only 100 of them, so there will be an even number. I think this goes no where.
Using a direct proof if I choose 101 numbers, I will get either 100 even + 1 odd or 100 odd + 1 even.
In order for two numbers to evenly divide each other I would choose the 100 even, and there is a big probability that two will be even, but if I have 100 odd + 1 even, there will be only 1 even... So I'm not sure how to solve this...