5. In an $m \times 10$ grid, each cell is filled with either 0 or 1, such that each column contains exactly three 1s. The sum of the numbers in the $i(i=1,2, \cdots, m)$-th row is denoted as $x_{i}$, and for any two columns, there always exists a row where the cells at the intersection with these two columns are both filled with 1. Let $x=\max _{1 \leqslant i \leqslant m}\left\{x_{i}\right\}$. Then $x_{\text {min }}=$
5.5.

Constructing the application model: "10 people go to a bookstore to buy $m$ kinds of books, each person buys 3 books, and any two people have at least one book in common. Find the sales volume of the book that is purchased by the most people."

Let the $i(i=1,2, \cdots, m)$-th kind of book be purchased by $x_{i}$ people, $x=\max _{1 \leqslant i \leqslant m}\left\{x_{i}\right\}$. Then
$$
\sum_{i=1}^{m} x_{i}=30 \text {, and } x_{i} \leqslant x(i=1,2, \cdots, m) \text {. }
$$

Considering the total number of "same book pairs" between any two people, we get the inequality
$$
\begin{array}{l}
\sum_{i=1}^{n} \mathrm{C}_{x_{i}}^{2} \geqslant \mathrm{C}_{10}^{2} \\
\Rightarrow \sum_{i=1}^{m} \frac{x_{i}\left(x_{i}-1\right)}{2} \geqslant 45 \\
\Rightarrow 45 \leqslant \frac{1}{2}(x-1) \sum_{i=1}^{m} x_{i}=15(x-1) \\
\Rightarrow x \geqslant 4 .
\end{array}
$$

If $x=4$, then all $x_{i}=4(i=1,2, \cdots, m)$, which contradicts $\sum_{i=1}^{m} x_{i}=30$.
Therefore, $x \geqslant 5$.
Below, we construct a case where $x=5$, and there exists a book that is purchased by exactly five people, denoted as $B_{i}$. The book purchasing situation for the 10 people is as follows:
$$
\begin{array}{c}
B_{1} B_{2} B_{3}, B_{3} B_{4} B_{5}, B_{1} B_{5} B_{6}, B_{1} B_{3} B_{5}, B_{1} B_{2} B_{4}, \\
B_{1} B_{4} B_{6}, B_{2} B_{4} B_{5}, B_{2} B_{5} B_{6}, B_{2} B_{3} B_{6}, B_{3} B_{4} B_{6} .
\end{array}
$$

Therefore, the book that is purchased by the most people is bought by at least five people.