34. Consider the set $S=\{1,2,3, \cdots, 2010,2011\}$. A subset $T$ of $S$ is said to be a $k$-element RP-subset if $T$ has exactly $k$ elements and every pair of elements of $T$ are relatively prime. Find the smallest positive integer $k$ such that every $k$-element RP-subset of $S$ contains at least one prime number. (As an example, $\{1,8,9\}$ is a 3-element RP-subset of $S$ which does not contain any prime number.)
34. Answer: 16
Consider the subset $T=\left\{1,2^{2}, 3^{2}, 5^{2}, 7^{2}, \cdots, 43^{2}\right\}$ consisting of the number 1 and the squares of all prime numbers up to 43 . Then $T \subseteq S,|T|=15$, and all elements in $T$ are pairwise relatively prime; however, $T$ contains no prime number. Thus $k \geq 16$. Next we show that if $A$ is any subset of $S$ with $|A|=16$ such that all elements in $A$ are pairwise relatively prime, then $A$ contains a prime number. Suppose to the contrary that $A$ does not contain a prime number. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{16}\right\}$. We consider two cases:
Case 1. 1 is not in A. Then $a_{1}, a_{2}, \cdots, a_{16}$ are composite numbers. Let $p_{i}$ be the smallest prime factor of $a_{i}, i=1,2, \cdots, 16$. Since $\operatorname{gcd}\left(a_{i}, a_{j}\right)=1$ for $i \neq j$, $i, j=1,2, \cdots, 16$, we see that the prime numbers $p_{1}, p_{2}, \cdots, p_{16}$ are all distinct. By re-ordering $a_{1}, a_{2}, \cdots, a_{16}$ if necessary, we may assume that $p_{1}2011,
$$
which is a contradiction to the given fact that each element of $S$ is less than or equal to 2011 .

Case 2. 1 is in $A$. We may let $a_{16}=1$. Then $a_{1}, a_{2}, \cdots, a_{15}$ are composite numbers. As in Case 1, we have
$$
a_{1} \geq p_{1}^{2} \geq 2^{2}, a_{2} \geq p_{2}^{2} \geq 3^{2}, \cdots, a_{15} \geq p_{15}^{2} \geq 47^{2}>2011,
$$
which is a contradiction.
Thus we have shown that every 16-element subset $A$ of $S$ such that all elements in $A$ are pairwise relatively prime must contain a prime number. Hence the smallest $k$ is 16 .