(5) 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 bought at least one type of book in common.
How many people at most bought the most purchased type of book? Explain your reasoning.
Let the total number of books sold be $m$, with $x_{1}, x_{2}, \cdots, x_{m}$ people buying each type of book, and let the book with the most buyers have $x$ people purchasing it. Then we have
$$
\sum_{i=1}^{m} x_{i}=30, x_{i} \leqslant x, i=1,2, \cdots, m
$$

Considering the total number of pairs of people who bought the same book, we have
$$
\mathrm{C}_{10}^{2} \leqslant \sum_{i=1}^{m} \mathrm{C}_{x_{i}}^{2}
$$

Thus, $\quad 45 \leqslant \frac{1}{2} \sum_{i=1}^{m} x_{i}\left(x_{i}-1\right) \leqslant \frac{x-1}{2} \sum_{i=1}^{m} x_{i}=15(x-1)$
Therefore,
$$
x \geqslant 4
$$

If $x=4$, then $x_{i}=4, i=1,2, \cdots, m$, and $4 \mid \sum_{i=1}^{m} x_{i}$, i.e., $4 \mid 30$, which is a contradiction. Hence, $x \geqslant 5$, meaning the book with the most buyers has at least 5 people purchasing it.

On the other hand, there exists a scenario where the book with the most buyers is purchased by exactly 5 people: Let $A_{i}$ represent different types of books, $i=1,2, \cdots, 6$. The purchasing scenarios for 10 people are as follows:
$$
\begin{array}{l}
\left\{A_{1}, A_{2}, A_{3}\right\},\left\{A_{1}, A_{6}, A_{2}\right\},\left\{A_{2}, A_{4}, A_{3}\right\},\left\{A_{3}, A_{5}, A_{1}\right\} \\
\left\{A_{1}, A_{6}, A_{4}\right\},\left\{A_{1}, A_{5}, A_{4}\right\},\left\{A_{2}, A_{4}, A_{5}\right\} \\
\left\{A_{2}, A_{6}, A_{5}\right\},\left\{A_{3}, A_{5}, A_{6}\right\},\left\{A_{3}, A_{4}, A_{5}\right\}
\end{array}
$$

Therefore, the book with the most buyers has at least 5 people purchasing it.