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Find the lowest cardinality of $А\subseteq\Bbb{N}$ such that there are 2017 different partitions $B_i\subseteq A$ and $A\backslash B_i$ for which $\operatorname{lcm}(B_i) =\gcd(A\backslash B_i)$

How is it possible that we are looking for a GCD of a factor set?

And what are we really looking for? That the number of sets in the power set of A is 2017?

Here is my attempt [link]

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    It is the third time I've seen this problem in MSE. – 2017-01-04
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    @ajotatxe: [Here is the second](http://math.stackexchange.com/questions/2080237/lowest-cardinality-of-a-subset-of-n-being-able-to-be-divided-in-2017-ways). – 2017-01-04
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    @ajotatxe: [And here is the third](http://math.stackexchange.com/questions/2082764/find-the-lowest-cardinality-of-the-set-%D0%90-c-n-such-that-there-are-2017-different). – 2017-01-04
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    irrelevant, they are all deleted. – 2017-01-04
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    Hmmmm, I wonder why... – 2017-01-04
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    The OP's deleted them – 2017-01-04
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    Now why would they do that??? – 2017-01-04
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    Anyway, can any of you guys help instead of prosecuting my previous now-deleted post, please? – 2017-01-04
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    Sure, this problem is famous, it's IMO 2015 c5 – 2017-01-04

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For a valid partition we need in particular that every $x\in B_i$ divides every $y\in A\setminus B_i$. Hence for different indices $i,j$ we one of $B_i\setminus B_j$, $B_j\setminus B_i$ must be emtpy, i.e., we have $B_i\subseteq B_j$ or $B_j\subseteq B_i$. So the $B_i$ are linearly ordered by inclusion: $$\emptyset\subsetneq B_1\subsetneq B_2\subsetneq B_3\subsetneq \ldots \subsetneq B_{2017}\subsetneq A$$ Suppose for some $k$, $1\le k\le 2017$, we have $|B_{k+1}\setminus B_{k-1}|=2$ (with $B_0=\emptyset$ and $B_{2018}=A$ understood), so $B_{k}=B_{k-1}\cup \{m\}$, $B_{k+1}=B_{k}\cup\{n\}$ for some distinct numbers $m,n$. Then $\operatorname{lcm}(B_{k})=m$ and $\gcd(A\setminus B_{k})=n$, i.e., we should have $n=m$. We conclude $|B_{k+1}|\ge |B_{k-1}|+3$ for $1\le k\le 2017$. By induction, $|A|=|B_{2018}|\ge |B_0|+1009\cdot 3\ge 2027$.

But can this lower bound be achieved? Yes! For $n\in\Bbb N$ let $$S_n=\{\,2^a3^b: |a-b|\le 1, a+b2n$ or $m>2n+1$, respectively, $\gcd(S_m\setminus S_{2n})=\gcd(S_m\setminus S_{2n+1})=6^n$ as well. Also $|S_{2n}|=3n$. This allows us to pick $B_k=S_k$ for $1\le k\le 2017$ and $A=S_{2018}$ with $|A|=3027$.

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    Can you chat with me here? https://chatlink.com/#aquasolid – 2017-01-04