Number of vectors so that no two subset sums are equal Consider all $10$-tuple vectors each element of which is either $1$ or $0$. It is very easy to select a set $v_1,\dots,v_{10}= S$ of $10$ such vectors so that no two distinct subsets of vectors $S_1 \subset S$ and $S_2 \subset S$  have the same sum. Here $\sum_{v \in S_i} v$ assumes simple element-wise addition over $\mathbb{R}$.  For example, if we take the vectors that are the columns of the identity matrix as $S$ this will do.   
What is the maximum number of vectors one can choose that have this property?  Is there a counting argument that solves this?

A small clarification. The sum of two vectors in this problem is another vector.

Current records:


*

*Lower bound: $19$. First given by Brendan McKay over at MO.

*Upper bound: $30$. First given by Brendan McKay over at MO.



Cross-posted to https://mathoverflow.net/questions/157634/number-of-vectors-so-that-no-two-subset-sums-are-equal
 A: Given a particular subset $S'\subset S$, you can think of each position in the sum as a measurement: the $i$-th measurement tells you how many elements of $S'$ have a $1$ in the $i$-th position.  Equivalently, the $i$-th measurement gives you the size of $S'\cap A_i$, where $A_i=\{x\in S \;|\; \pi_i(x)=1\}$.  How large can $S$ be if you can identify an arbitrary subset using $n$ such measurements?  Well, if $|S|=k$, then each measurement can have $k+1$ different outcomes, and so the most information it can give you is $\log_2(k+1)$ bits.  Since you need to distinguish between $2^k$ subsets, you need to obtain $k$ bits of information in total, so
$$
n \log_2(k+1) \ge k,
$$
or
$$
\frac{k}{\log_2(k+1)} \le n.
$$
For $n=10$, for instance, this implies $|S| \le 59$.
A: A slight improvement to the upper bound showing that the answer is $\le 47$. Assume contrariwise that a set $S$ of $48$ such vectors would exist. The set $S$ has
$$
\sum_{k=1}^{24}{48\choose k}=156\,861\,290\,196\,877
$$
subsets of at most $24$ elements. The sum vectors of those subsets belong to the set $\{0,1,\ldots,24\}^{10}$ that has $25^{10}=95\,367\,431\,640\,625$ elements. Therefore a collision is inevitable by the pigeonhole principle. 
