For any finite set $T$ of real numbers, $s(T)$ is the sum of its elements. Let $S \subseteq\{1,2, \ldots, 15\}$. If $s(B) \neq s(C)$ for any disjoint subsets $B, C$ of $S$, then show that $|S| \leq 5$
Source : Principles and Techniques in Combinatorics
I know that $|S| \leq 6$. Because for any $X \subseteq S, 0 \leq s(X) \leq 1+2+\ldots+15=120$. If $|S| \geq 7$, then $S$ has $2^{7}>121$ subsets, so there are two subsets $B, C$ of $S$ with equal element sum. We can assume that $B$ and $C$ are disjoint (otherwise, remove the intersection from $B, C$ ).
I also know that there is an example with $|S|=5$. My example is \{8,11,13,14,15\} . But how to exclude the case $|S|=6$ ?
Oh, if you know which contest this problem came from, that’d be great. Apparently, the book cited a wrong competition because it says USAMO 1986 , but this problem is not in USAMO 1986 . Thank you very much.
BTW here is what I know about the case $|S|=6$. Let $S=\left\{x_{1}, x_{2}, \ldots, x_{6}\right\}$ where $x_{1}<x_{2}<\ldots<x_{6} .$ If $x_{6}<14$, then $s(S) \leq 13+12+\ldots+8=63$. Since
$S=\{8,9, \ldots, 12,13\}$ doesn’t work we must have $s(S)<63$. So $s(X)$ for $X \subseteq S$ can have at most 63 possible values but $63<2^{6}$, so this is a contradiction. Therefore, $x_{6} \geq 14$. Then there are too many cases to consider for $x_{5}$.
Suppose $|S|=6$. Note that the maximum sum of any 4 -element (or smaller) subset is $12+13+14+15=54,$ so at most 55 different sum values are taken by subsets of $S$ of size at most $4 .$ But at most 7 different values are taken by subsets of $S$ of size 5 or 6 (since it only has this many subsets of size 5 or 6 ). So in total the number of sum values for subsets of $S$ is at most $62<2^{6}$