2. 10 people go to the bookstore to buy books, it is known that: (1) each person bought three types of books; (2) any two people have at least one book in common.

How many people at most could have bought the book that was purchased by the fewest people?
2. Suppose among them, A bought three books, and since A has at least one book in common with each of the other 9 people, the book that the most people bought among A's three books is purchased by no less than 4 people. If the book that the most people bought is purchased by 4 people, then all three books bought by A are each purchased by 4 people, and each book bought by the other 9 people is also purchased by 4 people. Therefore, the total number of books bought by the 10 people is a multiple of 4, i.e., $4 \mid 30$, which is a contradiction. Thus, the book that the most people bought is purchased by at least 5 people. Consider the following purchasing method:
$$
\begin{array}{l}
\quad\left\{B_{1}, B_{2}, B_{3}\right\},\left\{B_{1}, B_{2}, B_{4}\right\},\left\{B_{2}, B_{3},\right. \\
\left.B_{5}\right\},\left\{B_{1}, B_{3}, B_{6}\right\},\left\{B_{1}, B_{4}, B_{5}\right\},\left\{B_{2}, B_{4},\right. \\
\left.B_{6}\right\},\left\{B_{3}, B_{4}, B_{5}\right\},\left\{B_{1}, B_{5}, B_{6}\right\},\left\{B_{2}, B_{5},\right. \\
\left.B_{6}\right\},\left\{B_{3}, B_{4}, B_{6}\right\}
\end{array}
$$