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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.

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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…~~

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Pick $101$ elements from $A$, label them $a_1,\ldots, a_{101}$. We can assume that $a_1 < \ldots < a_{100} < a_{101}$. Since we have $101$ distinct elements, $a_1 \leq 99$.

Consider the set of remainders upon division by $a_1$. Since $a_1 \leq 99$, there are at most 99 such remainders. Let $r_2$ be the remainder upon dividing $a_2$ by $a_1$, $r_3$ the remainder upon dividing $a_3$ by $a_1, \ldots, r_{101}$ the remainder upon dividing $a_{101}$ by $a_1$.

We have $100$ remainders $r_1,\ldots ,r_{101}$ (pigeons), and at most $99$ possible remainders (pigeonholes) upon dividing by $a_1$. Thus, by the pigeonhole principle, $r_i = r_j$ for some $2\leq i < j \leq 101$. Now what can you say about the number $a_j – a_i$?

**Hint:** The boxes will have labels $1$, $3$, $5$, and so on up to $199$. Odd labels! Note that there are $100$ boxes.

Box 1: Contains $1,2,4,8,16, 32,\dots$

Box 3: Contains $3,6,12,24, 48, \dots$

Box 5: Contains $5,10,20, 40,\dots$.

And so on.

Box 99: Contains $99,198$

Boxes 101, 103, and so on are pretty boring. Box 101 only contains the number $101$, Box 103 only contains $103$, and so on.

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