5.55 10 people go to the bookstore to buy books, it is known that
(1) each person bought 3 types of books;
(2) any two people have at least one book in common. How many people at most bought the most purchased book? Explain your reasoning.
[Solution 1] Let $A$ be one of the 10 people. By the given information, $A$ bought 3 different books, and each of the other 9 people bought at least one book that $A$ also bought. Therefore, by the pigeonhole principle, among the 9 people, at least 3 people, plus $A$, a total of 4 people, bought the same book. Thus, the minimum value we are looking for is no less than 4.

If the book bought by the most people is bought by 4 people, then each book is bought by exactly 4 people. Let the 10 people buy a total of $n$ different books, then $4n = 30$. However, since $4 \nmid 30$, this is impossible, so the minimum value we are looking for is no less than 5.
We use numbers to represent the book codes and let the 10 people buy books as follows:
$$
\begin{array}{l}
(123),(123),(145),(167),(246), \\
(246),(257),(347),(356),(356) .
\end{array}
$$

It is easy to verify that the books they bought meet the requirements of the problem, and the book bought by the most people is bought by 5 people. Therefore, the minimum value we are looking for is 5.
[Solution 2] In the figure, the triangles formed by the center and two adjacent vertices of a regular pentagon total 5, and the triangles formed by three non-adjacent vertices of the regular pentagon also total 5. It is not difficult to see that any two of these 10 triangles share at least one vertex. Writing out the vertex number groups of these triangles and letting the book codes bought by the 10 people be these 10 triangle vertex number groups:
$$
\begin{array}{l}
(123),(134),(145),(156),(162), \\
(245),(356),(426),(523),(634) .
\end{array}
$$

Clearly, each book is bought by 5 people. Therefore, the minimum value we are looking for is no more than 5.
Assume the minimum value is 4, and the 10 people bought a total of $n$ different books, with the $i$-th book bought by $m_i$ people, so $m_i \leq 4$ and $m_1 + m_2 + \cdots + m_n = 30$. When two people buy the same book, it is called a "book pair". By the given information, each pair of people has at least 1 book pair, so there are at least $\binom{10}{2} = 45$ book pairs. On the other hand, the number of book pairs formed by the $i$-th book is $\binom{m_i}{2}$, so the total number of book pairs is $\binom{m_1}{2} + \binom{m_2}{2} + \cdots + \binom{m_n}{2}$. Therefore, we have
$$
\binom{m_1}{2} + \binom{m_2}{2} + \cdots + \binom{m_n}{2} \geq 45.
$$

Since $\binom{4}{2} = 6$, $\binom{3}{2} = 3$, and $\binom{2}{2} = 1$, we also have
$$
\binom{m_1}{2} + \binom{m_2}{2} + \cdots + \binom{m_n}{2} \leq 7 \binom{4}{2} + \binom{2}{2} = 43.
$$

Since (1) and (2) are contradictory, the minimum value we are looking for is 5.