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 vector addition where element addition takes place 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 has this
> property?

 

I previously asked this question on [MSE][1] . An explicit construction of $17$ vectors was given by Oleg567 using computer search and an upper bound of $45$  was given by jpvee simply using the observation that $\sum_{k=1}^{17} {46 \choose k} > (17+1)^{10}$ implies that $46$ vectors is impossible.

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Lower bound improved to $18$ by Oleg567. Upper bound still stuck at $45$ although it seems implausible the true value is far from the current lower bound.

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Upper bound of $36$ given by Seva.

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**Conjecture**. The optimal solution size  is  $\lfloor \frac{1}{2} (n+1) \log_2(n+1) \rfloor$ 

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New lower bound of $19$ by Brendan McKay.
  [1]: http://math.stackexchange.com/questions/667257/number-of-vectors-so-that-no-two-subset-sums-are-equal