9.3. From the natural numbers $1,2, \ldots, 101$, a group of numbers is chosen such that the greatest common divisor of any two numbers in the group is greater than two. What is the maximum number of numbers that can be in such a group?
Answer: 33. Solution: Estimation. Let's divide the numbers $1,2, \ldots, 101$ into 34 sets: $A_{0}=\{1,2\}, A_{1}=\{3,4,5\}$, $A_{2}=\{6,7,8\}, \ldots, A_{33}=\{99,100,101\}$ (i.e., $A_{k}$ for $k \geq 1$ consists of three numbers $3 k, 3 k+1,3 k+2$). In the desired group of numbers, there cannot be a number from $A_{0}$ (otherwise, the GCD of two numbers from the group, one of which belongs to $A_{0}$, would be $\leq 2$), and from each $A_{k}$ for $k \geq 1$, no more than one number can be in the group, since otherwise, if the group includes adjacent numbers, their GCD $=1$, and if the group includes numbers $3 k$ and $3 k+2$, their GCD $\leq 2$ (since their difference is 2). Therefore, the total number of numbers in the desired group is no more than 33 (no more than one from each $A_{k}$ for $k \geq 1$). Example. Consider the group of numbers $\{3,6,9, \ldots, 99\}$: this group contains 33 numbers, and all of them are divisible by 3.