$5 \cdot 5610$ people go to the bookstore to buy books, given that:
（1）each person buys 3 different books;
（2）any two people have at least 1 book in common;
（3）no more than 5 people buy the same book;
（4）10 people buy a total of $n$ different books.

Find all possible values of $n$.
[Solution] From (1) and (3), we know that $n \geqslant 6$. We use natural numbers $1,2,3, \cdots$ to represent the numbers of the book types. When $n=6$, 10 people can buy books as follows:
$$
\begin{array}{l}
(1,2,3),(1,3,4),(1,4,5),(1,5,6),(1,2,6), \\
(2,4,5),(3,5,6),(2,4,6),(2,3,5),(3,4,6) .
\end{array}
$$

When $n=7$, 10 people can buy books as follows:
$$
\begin{array}{l}
(1,2,3),(1,4,5),(2,4,6),(2,4,6),(2,5,7), \\
(2,5,7),(3,4,7),(3,4,7),(3,5,6),(3,5,6) .
\end{array}
$$

It is easy to verify that both sets of book purchasing scenarios meet the requirements of the problem.
If at least 8 types of books are purchased, then by the pigeonhole principle, the book $\alpha$ purchased by the fewest people has at most 3 people purchasing it.
(i) If $\alpha$ is purchased by only 1 person $A$, then $A$ also needs to buy two other books $\beta$ and $\gamma$. By (2), the other 9 people each need to buy one of $\beta$ or $\gamma$. Thus, at least 5 people will buy the same one of these two books, and adding $A$ makes 6 people buying the same book, which contradicts (3).
(ii) If book $\alpha$ is purchased by exactly two people, let the three books these two people buy be $(\alpha, \beta, \gamma)$ and $(\alpha, \delta, \varepsilon)$. Then $\beta, \gamma, \delta, \varepsilon$ are four different books. Otherwise, it would lead to 6 people buying the same book. By (2), the other 8 people each need to buy one from $\{\beta, \gamma\}$ and one from $\{\delta, \varepsilon\}$. And by (3), the other 8 people each buy $\beta, \gamma, \delta, \varepsilon$ exactly 4 times. The 8 people can be divided into two groups: those buying $(\beta, \delta, \chi)$ and $(\gamma, \varepsilon, \chi)$ form the first group, and those buying $(\beta, \varepsilon, \chi)$ and $(\gamma, \delta, \chi)$ form the second group. By (2), the third book bought by each group must be the same, but this means only 7 types of books are bought, which is a contradiction.
(iii) Suppose each type of book is purchased by at least 3 people, and the $i$-th type of book is purchased by $m_{i}$ people. When two people buy the same type of book, we say there is 1 book pair between them. Thus, by (2), there is at least 1 book pair between any two people. Therefore,
$$
C_{m_{1}}^{2}+C_{m_{2}}^{2}+\cdots+C_{8}^{2} \geqslant C_{10}^{2}=45 .
$$

On the other hand, since $C_{3}^{2}+C_{5}^{2}>2 C_{4}^{2}$, we also have
$$
C_{m_{1}}^{2}+C_{m_{2}}^{2}+\cdots+C_{m_{8}}^{2} \leqslant 5 C_{3}^{2}+3 C_{5}^{2}=45 .
$$

By
and
(2)
$$
C_{m_{1}}^{2}+C_{m_{2}}^{2}+\cdots+C_{m_{8}}^{2}=45 .
$$
(3) This means that each pair of the 10 people has exactly 1 book pair, and among the 8 types of books, 5 types are each purchased by 3 people, and 3 types are each purchased by 5 people. In this way, the last 3 types of books are purchased a total of 15 times, so among the 10 people, there must be two people who bought the same two of these 3 types of books, i.e., there are 2 book pairs between these two people, which is a contradiction.
In summary, $n$ can only take two values: 6 and 7.