TITLE: Number of vectors so that no two subset sums are equal
QUESTION [25 upvotes]: 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

REPLY [4 votes]: 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.