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My reasoning so far is:

If the number 1 is chosen, we are done since 1 divides any number.

If 2 is chosen, then the remaining n numbers to choose will either be all odd (including 1) or contain an even number.

I do not know how to progress from here. Does it require the pigeonhole principle? Thank you in advance.

TreeGuy
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1 Answers1

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Note that any $n \in \mathbb{Z}^{+}$ can be expressed in the form $2^{k} \times O$, where $k \in \mathbb{Z}_{\geq 0}$ and $O$ is the greatest positive odd divisor of $n$. This can be quite easily proved.

Now, consider any integer $z$ from $1,2,3, \ldots, 2n$. The greatest positive odd divisor of $z$ must be one of $1,3,5,7,\dots,2n - 1$. Hence the greatest positive divisor of $z$ can take any one of $n$ such values.

Now, let the pigeons be the $n + 1$ integers selected and the $n$ pigeonholes correspond to the possible values of the greatest positive odd divisor. Place each integer in the sequence $1,2,3,\ldots,2n$ into its unique pigeonhole.

By the pigeonhole principle, when $n+1$ integers are selected from $1,2,3,\ldots,2n$ it is guaranteed that two integers are selected from the same pigeonhole. Let $a,b$ be the integers selected from the same pigeonhole and let $O$ be the greatest positive odd divisor of $a,b$. Assume WLOG $a \lt b$ such that $a = 2^{r} \times O$ and $b = 2^{s} \times O$ where $r \lt s$ and $r,s \in \mathbb{Z}_{\geq 0}$. Now,

$$ \frac{b}{a} = \frac{2^{s} \times O}{2^{r} \times O}$$ $$ \implies b = 2^{s - r}\times a$$

By the closure of integers under subtraction, $s - r \in \mathbb{Z}$ and hence $2^{s - r} \in \mathbb{Z}$. Then it immediately follows by the definition of divisibility that $a|b$.

anon
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  • That is very interesting. Thank you. Can you show me a proof of the 2^k * O property? – TreeGuy Apr 14 '23 at 04:06
  • Look at https://math.stackexchange.com/questions/25914/proof-by-strong-induction-n-2a-b-b-odd-every-natural-a-product-of-an .Looking at this proof, you can easily get inspiration to prove it for the greatest positive odd divisor. – anon Apr 14 '23 at 04:16
  • So speaking informally, if I took an even number, I can just continue to divide it by two until I reached an odd number? – TreeGuy Apr 14 '23 at 04:27
  • And additionally, that first odd number you reach after continually dividing by two would be the greatest odd divisor? – TreeGuy Apr 14 '23 at 04:36
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    In some sense yes, suppose $n$ is even and $O$ is the greatest positive odd divisor. Then for some $k_{0} \in \mathbb{Z}$, $n = k_{0} \times O$. Surely $k_{0}$ cannot be odd since then $k_{0} \times O$ would be odd and would be greater than $O$ and still be a positive odd divisor contradicting that $O$ is the greatest positive odd divisor. Hence $k_{0}$ must be even, so by definition $k_{0} = 2 \times k_{1}$ where $k_{1} \in \mathbb{Z}$. So now $n = 2 \times k_{1} \times O$. Now apply the same argument for $k_{1}$. – anon Apr 14 '23 at 04:49
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    Cont... This continues for some finite number of times (since ultimately $n$ is finite) and when it terminates (then $k_{m}$ should have been $1$) it should be of the desired form of $2^{k} \times O$ – anon Apr 14 '23 at 04:50