Question 15, Solution: Let $A=\{x|x \in M, 2| x\} \cup\{x|x \in M, 3| x\} \cup\{x|x \in M, 5| x\}$, then according to the principle of inclusion-exclusion, we have:
$$
|A|=\left[\frac{100}{2}\right]+\left[\frac{100}{3}\right]+\left[\frac{100}{5}\right]-\left[\frac{100}{2 \times 3}\right]-\left[\frac{100}{2 \times 5}\right]-\left[\frac{100}{3 \times 5}\right]+\left[\frac{100}{2 \times 3 \times 5}\right]=74
$$

Since every element in $A$ is a multiple of 2, 3, or 5, any four elements chosen from $A$ must include at least two elements that are not coprime.
$$
\begin{array}{l}
\text { On the other hand, when } n=75, \text { we can take } \\
A_{1}=\{1\} \cup\{\text { all primes not exceeding } 100\} ; \\
A_{2}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}\right\} ; \\
A_{3}=\left\{2^{3}, 3^{3}, 5 \times 19,7 \times 13\right\} ; \\
A_{4}=\left\{2^{4}, 3^{4}, 5 \times 17,7 \times 11\right\}
\end{array}
$$

Then all the numbers in these four sets are distinct, and any two numbers in each set are coprime. These four sets contain a total of 38 numbers. When 75 numbers are chosen from $M$, at least $75-(100-38)=13$ numbers must come from these four subsets. By the pigeonhole principle, at least one of these subsets must contain $\left\lceil\frac{13}{4}\right\rceil=4$ numbers, meaning these 4 numbers are pairwise coprime. In summary, the smallest value of $n$ is 75.
Question 15: Let the set $M=\{1,2, \ldots, 100\}$ be a 100-element set. If for any n-element subset $A$ of $M$, there are always 4 elements in $A$ that are pairwise coprime, find the minimum value of $\mathrm{n}$.

