Five, 10 people go to the bookstore to buy books, it is known that
(1) Each person bought three 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.
(Na Chengzhang provided the question)
Solution 1: Let's assume that each person buys at most one copy of the same book. Thus, by condition (1), 10 people buy a total of 30 books.

Let the minimum number of people who buy the most popular book be $n$. Assume $A$ buys books 甲, 乙, and 丙.

Clearly, $n > 3$. Because if $n \leqslant 3$, then at most two more people can buy 甲, two more can buy 乙, and two more can buy 丙. Therefore, at most 7 people can buy 甲, 乙, and 丙, and the remaining 3 people cannot share a book with $A$, which is a contradiction.

If $n=4$. In this case, we can prove that each book is bought by exactly 4 people. Because otherwise, at least one book is bought by at most three people. Assume the 甲 book bought by $A$ is bought by at most three people. That is, apart from $A$, 甲 is bought by at most two more people, and 乙 and 丙 are each bought by at most three more people. Thus, 甲, 乙, and 丙 are bought by at most $1+2+3+3=9$ people, and the remaining one cannot share a book with $A$, which is a contradiction.

The above argument shows that when $n=4$, each book is bought by exactly 4 people. However, 4 does not divide 30, which is a contradiction. Therefore, $n \geqslant 5$. In fact, $n=5$.

Below is an example for $n=5$: Assume there are 1, 2, ..., 7 seven types of books. The 10 people's book purchase situation is as follows:
\begin{tabular}{|c|c|c|c|c|}
\hline 2 & $3),(1$ & 4 & $5),(2$ & 4 \\
\hline 4 & $6),(2$ & 5 & $7),(2$ & 5 \\
\hline 4 & $7),(3$ & 4 & $7),(3$ & 5 \\
\hline
\end{tabular}

Solution 2: Using numbers to represent the book types, then when 10 people buy books as follows:
$$
\begin{array}{l}
(123),(134),(145),(156),(126), \\
\text { (235), (245), (246), (346), (356) }
\end{array}
$$

it satisfies the requirements of the problem and each type of book is bought by exactly 5 people, so the minimum value is no more than 5.

Assume the minimum value is 4, and 10 people buy a total of $n$ types of books, with the $i$-th type of book being bought by $m_{i}$ people. Thus, $m_{i} \leqslant 4$ 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 condition, each pair of people has at least one book pair, so there are at least $C_{10}^{2} = 45$ book pairs. On the other hand, the number of book pairs formed by the $i$-th type of book is $C_{m_{i}}^{2}$, so there are a total of $C_{m_{1}}^{2} + C_{m_{2}}^{2} + \cdots + C_{m_{n}}^{2}$ book pairs. Therefore, there must be
$$
C_{m_{1}}^{2} + C_{m_{2}}^{2} + \cdots + C_{m_{n}}^{2} \geqslant 45.
$$

However, since $C_{4}^{2} = 6$, $C_{3}^{2} = 3$, and $C_{2}^{2} = 1$, we also know
$$
C_{m_{1}}^{2} + C_{m_{2}}^{2} + \cdots + C_{m_{n}}^{2} \leqslant 7 C_{4}^{2} + C_{2}^{2} = 43.
$$

Since (1) and (2) are contradictory, the minimum value must be 5.