For any finite set $T$ of real numbers, $s(T)$ is the sum of its elements. Let $S\subseteq \{1,2,…,15\}$. If $s(B)\ne s(C)$ for any disjoint subsets $B,C$ of $S$, then show that $|S|\le 5$.
Source : Principles and Techniques in Combinatorics
I know that $|S|\le 6$. Because for any $X\subseteq S$, $0\le s(X) \le 1+2+…+15=120$. If $|S|\ge 7$, 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 $S=\{x_1,x_2,…,x_6\}$ where $x_1<x_2<…<x_6$. If $x_6<14$, then $s(S)\le 13+12+…+8=63$. Since $S=\{8,9,…,12,13\}$ 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, $x_6\ge 14$. Then there are too many cases to consider for $x_5$.
Best Answer
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 $62<2^6$.