A special case says it all ... Let $ w\_1 < w\_2 < \ldots < w\_{12} $ be an increasing sequence of $12$ integers ("weights") such that the total weight $W=\sum\_{k=1}^{12}w\_k$ is even. 


Say that $I \subseteq \lbrace 1,2, \ldots ,12 \rbrace$ is an exact subset iff
the sum $\sum\_{k \in I}w\_k$ equals $\frac{W}{2}$. My question is : is there
a sequence for which $ \lbrace 1,2,5,7,10,12 \rbrace $ is exact and is
the only exact subset (up to complementation) ? 



Unique way to partition into two parts of equal weight
The answer is **yes**. Consider the sequence 


$100, 200, 201, 202, 500, 601, 700, 701, 801, 1000, 1194, 1200.$


It is easy to see that $X=\{1,2,5,7,10,12\}$ and $Y=\{3,4,6,8,9,11\}$ are exact. Moreover, we claim that they are the only exact subsets. To see this, note that for every subset $X'$ of $X$, the sum of the corresponding sequence values is divisible by $100$. However, for every proper subset $Y'$ of $Y$, the sum of the corresponding sequence values is not divisible by $100$. Thus, $X$ and $Y$ are the only exact subsets. 


