Guessing the value of $n$

On the first round ask the six questions. \begin{eqnarray*} \{ i \mid i \equiv 1 \pmod 7 \} \\ \{ i \mid i \equiv 1 \pmod 7 \text{ or } i \equiv 2 \pmod 7 \} \\ \vdots \\ \{ i \mid i \equiv 1 \text{ or } 2 \text{ or } 3 \text{ or } 4 \text{ or } 5 \text{ or } 6\pmod 7 \} \\ \end{eqnarray*}

On the second round ask the $10$ questions \begin{eqnarray*} \{ i \mid i \equiv 1 \pmod {11} \} \\ \{ i \mid i \equiv 1 \text{ or } 2 \pmod {11} \} \\ \vdots \\ \{ i \mid i \equiv 1 \text{ or } 2 \text{ or } 3 \text{ or } 4 \cdots 10 \pmod {11} \} \\ \end{eqnarray*}

On the third round ask the $12$ questions \begin{eqnarray*} \{ i \mid i \equiv 1 \pmod {13} \} \\ \{ i \mid i \equiv 1 \text{ or } 2 \pmod {13} \} \\ \vdots \\ \{ i \mid i \equiv 1 \text{ or } 2 \text{ or } 3 \text{ or } 4 \cdots 12 \pmod {13} \} \\ \end{eqnarray*}

The value can be deduced using the Chinese remainder theorem. and the optimal value of $k_1+k_2+k_3$ is ...

A way to prove a lower bound here is by observing that in the $i$-th round, there are $k_i+1$ potential answers given by $B$. Thus, a strategy using $k_1$, $k_2$, $k_3$ sets in the first three rounds and then winning for sure needs to satisfy that $(k_1+1)(k_2+1)(k_3+1) \geq 1001$, since it needs to be able to distinguish all $1001$ potential choices from $S$.