, 2021

Solution:

In general, Bob can find the values of all $n$ integers asking only $\left\lfloor\log_{2} n\right\rfloor+1$ queries.

For each of Alice's numbers $x_{i}$, let $Q_{i}$ be the set of queries $S$ such that $i \in S$. Notice that all $Q_{i}$ must be nonempty and distinct. If there exists an empty $Q_{i}$, Bob has asked no queries that include $x_{i}$ and has no information about its value. If there exist $i, j, i \neq j$ such that $Q_{i}=Q_{j}$, $x_{i}$ and $x_{j}$ could be interchanged without the answer to any query changing, so there does not exist a unique sequence of numbers described by the answers to Bob's queries (Alice can make her numbers distinct).

From the above, $\left\lfloor\log_{2} n\right\rfloor+1$ is a lower bound on the number of queries, because the number of distinct nonempty subsets of $\{1, \ldots, n\}$ is $2^{n}-1$.

If Bob asks any set of queries such that all $Q_{i}$ are nonempty and disjoint, he can uniquely determine Alice's numbers. In particular, since the values $x_{1}, \ldots, x_{2021}$ are relatively prime, each prime factor of $x_{i}$ occurs in the answer to query $S_{j}$ iff $j \in Q(i)$ (and that prime factor will occur in each answer exactly to the power with which it appears in the factorization of $x_{i}$). Since all $Q(i)$ are unique, all $x_{i}$ can therefore be uniquely recovered by computing the product of the prime powers that occur exactly in the answers to queries $Q(i)$.

It is possible for Bob to ask $\left\lfloor\log_{2} n\right\rfloor+1$ queries so that each $i$ is contained in a unique nonempty subset of them. One possible construction is to include the index $i$ in the $j$th query iff the $2^{i-1}$-value bit is set in the binary representation of $j$. So the answer is $\left\lfloor\log_{2} 2021\right\rfloor+1=11$.