12. Let $S=\{1,2,3, \cdots, 100\}$, find the smallest positive integer $n$, such that every $n$-element subset of $S$ contains 4 pairwise coprime numbers.
12. Let $A_{2}=\{a|a \in S, 2| a\}, A_{3}=\{a|a \in S, 3| a\}, A_{5}=\{a|a \in S, 5| a\}$, then $\left|A_{2}\right|=50,\left|A_{3}\right|=\left[\frac{100}{3}\right]=33,\left|A_{5}\right|=20,\left|A_{2} \cap A_{3}\right|=\left[\frac{100}{2 \times 3}\right]=16,\left|A_{3} \cap A_{5}\right|=$ $\left[\frac{100}{3 \times 5}\right]=6,\left|A_{5} \cap A_{2}\right|=\left[\frac{100}{5 \times 2}\right]=10,\left|A_{2} \cap A_{3} \cap A_{5}\right|=\left[\frac{100}{2 \times 3 \times 5}\right]=3$, so $\left|A_{2} \cap A_{3} \cap A_{5}\right|=\left|A_{2}\right|+\left|A_{3}\right|+\left|A_{5}\right|-\left|A_{2} \cap A_{3}\right|-\left|A_{3} \cap A_{5}\right|-\left|A_{5} \cap A_{2}\right|+$ $\left|A_{2} \cap A_{3} \cap A_{5}\right|=74$. Therefore, for the 74-element set $A_{2} \cup A_{3} \cup A_{5}$, if we choose any 4 numbers, there must be two numbers that are both multiples of 2 or 3 or 5, and they are not coprime, thus $n \geqslant 75 . n=75$ is feasible, constructed as follows: $B_{1}=\{1$ and the first 25 prime numbers $\}, B_{2}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}\right\}, B_{3}=\left\{2^{3}, 3^{3}, 5 \times 19,7 \times 13\right\}, B_{4}=\left\{2^{4}, 3^{4}, 5 \times 17\right.$, $7 \times 11\}$, these 4 sets have an empty intersection with each other, and any two numbers in each set are coprime, $\mid B_{1} \cup B_{2} \cup$ $B_{3} \cup B_{4}|=| B_{1}|+| B_{2}|+| B_{3}|+| B_{4} \mid=38$. Let $X \subseteq S,|X| \geqslant 75$, then $X$ contains at least $75-$ $(100-38)=13$ (elements) from $B_{1} \cup B_{2} \cup B_{3} \cup B_{4}$. By the pigeonhole principle, at least $\left[\frac{13}{4}\right]+1=4$ (numbers) are taken from some $B_{i}(1 \leqslant i \leqslant 4)$, and the numbers in $B_{i}$ are pairwise coprime. In summary, the minimum value of $n$ is 75.