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i have read in the Proofs by Jay Cummings, i came across this proposition:
Proposition 1.11. Given any 101 integers from {1,2,...,200}, at least one of these numbers will divide another. If i do it correctly, i found that its just need about 48 numbers to make sure that i found a number that would divided another, my method is to found that there were 46 prime numbers and the number 1(in case that we picked 1, cause 1 cant divided by any numbers but itself) which is 47.
Which lead to the conclusion that we only need 48 numbers to qualified the proposition, but the confusing came when i read that
'Alas, and perhaps surprisingly, our procedure is indeed optimal. Even if you chose just 1 fewer number — 100 — you would not be guaranteed that one divides another. You really do need 101.'
but i found it just only need 48 numbers ??

Tankut Beygu
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Vietnamese IPhO Competitant -
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  • You are answering a different question. The given questions asks you to prove the claim for any specified set of $101$ integers from that list. You are looking at a particular list. (And if you get to pick numbers in the list, you can do a lot better than $48$. After all, if the list contains $1$, then any other value must be divisible by $1$). – lulu Aug 13 '24 at 11:29
  • If your hundred chosen numbers are ${101,102,103,\ldots, 199,200}$ then none can be divided by another exactly to give an integer quotient since you would have $0 < \frac ab <2$ and $\frac ab \not=1$. – Henry Aug 13 '24 at 11:40

1 Answers1

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As mentioned in the comment above, if we pick 100 numbers from {1,2,3,4,…,200},we can pick { 101,102,103,...200 } Which cannot be divided by one another. The Pigeonhole Principle guarantees a least solution that the statement is true. I.E. You cannot pick any number from 1-100 that is not divisible by the set { 101,102,103,...200 }

Overnight FYT
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