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I've recently found a problem that I still can't solve: Dependency of the properties of numbers' subsets concerning subsets' sums of $n$ real numbers. A problem linked with it, that may be interesting not only for me has been proposed by @RossMillikan.

Having $2^{n-2}$ subsets of "unique" sums gives a hypothetical possibility that the remaining $3 \cdot 2^{n-2}$ subsets may have the same sum, which violates the given inequality (the second number is too big). What (and why) is instead the maximal number of the remaining subsets then?

I can't figure out why there may not be such a combination. I know that even for small $n=2$ sums of two subsets influence the third one, but I don't know what are the general dependencies. I would appreciate any help in this and/or linked problem.

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  • $\begingroup$ I dont think this is the correct approach to the original problem $\endgroup$ – Jorge Fernández Hidalgo Dec 7 '17 at 17:36
  • $\begingroup$ According to my calculations, the maximum value of $ab$ happens when $a$ is one of the integers nearest $\frac13 (2^n+2)$, and would correspond to $a-1$ of the equivalence classes being singletons. I don't know if this can be realized, though. $\endgroup$ – Fimpellizieri Dec 7 '17 at 19:02
  • $\begingroup$ If $n$ is even, $(2^n+2)/3$ is an integer, so we don't have to worry much. In this situation, the maximum 'possible' value of $ab$ happens when we have $a=(2^n+2)/3$ equivalence classes, $a-1$ of which contain a single subset. The other class should contain $2^n-(a-1)=(2^{n+1}+1)/3$ subsets that sum to the same value. For $n\geq 3$, this is less than the suggested value of $3\cdot2^{n-2}$ by $\sim11\%$. Still, I don't know if this can be realized. $\endgroup$ – Fimpellizieri Dec 7 '17 at 20:01
  • $\begingroup$ Even for small values like $n=4$ this does not look feasible; we're looking at $11$ subsets of a $4$-set of real numbers that all sum to the same value. $\endgroup$ – Fimpellizieri Dec 7 '17 at 20:08
  • $\begingroup$ @Fimpellizieri did you manage to get hypothetical ab=6^n? Maybe you could make it "backwards"? $\endgroup$ – user509680 Dec 7 '17 at 21:52

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