Let $p$ be a large prime and $n < p$. What is the smallest size of a set $A \subset \mathbb{Z} / p \mathbb{Z}$ such that $A \cdot \{1 , \ldots , n\} = \mathbb{Z} / p \mathbb{Z}$? Here $\cdot$ denotes the product set $X\cdot Y = \{xy : x \in X , y \in Y\}$. 


Trivial bounds are $$p/n \leq |A| \leq p,$$ and using the probabilistic method one can improve the upper bound to $|A| \ll (p/n) \log p$, just a logarithm away from the lower bound. 



Small set such that $\{1 , \ldots , n\} \cdot A = \mathbb{Z} / p \mathbb{Z}$
I believe that the answer to this question can be found in [this](https://arxiv.org/pdf/1310.0120v1.pdf) paper of Chen, Shparlinski and Winterhof. See Theorem 2 on page 6. It seems they give a contruction of a set with size $|A| \leq \frac{2p}{n}$ such that $A \cdot \{1,2,\dots,n\}=\mathbb Z\_p$.


