【Question 29】
8 different positive integers, among which there are at least: 6 multiples of 2, 5 multiples of 3, 3 multiples of 5, 1 multiple of 7. To make the maximum number among the 8 numbers as small as possible, then this maximum number is $\qquad$.
【Analysis and Solution】The problem of extremum.
There are at least $6+5-8=3$ numbers that are multiples of both 2 and 3, i.e., multiples of $[2,3]=2 \times 3=6$; thus, the maximum number is at least $6 \times 3=18$. Let's assume the maximum number is 18, then among the 8 numbers, there must be $(6,12,18)$;

Since there are at least 3 multiples of 5, but they cannot exceed 18, they can only be $(5,10,15)$;
The remaining 2 numbers must meet the following conditions: at least 2 multiples of 2, 1 multiple of 3, and 1 multiple of 7;
Thus, there must be a number that is a multiple of both 2 and 3, i.e., a multiple of $[2,3]=2 \times 3=6$, and it must be at least 24;
This contradicts the original assumption, so it is not valid.
Similarly, the maximum number being 19 is also not valid.
If the largest number is 20, for example, $(6,9,10,12,14,15,18,20)$ satisfies the conditions.
In conclusion, the smallest maximum number is 20.