4. Select five more natural numbers to the number 34 so that two conditions are met:

(1) among the six numbers, there are no three numbers that have a common divisor (greater than one);

(2) among any three numbers from these six, there will be two numbers that have a common divisor (greater than one).
Answer. For example, $6,34,35,51,55,77$.

Solution. Since $34=2 \cdot 17$, let's take the prime number 3 and add the numbers $2 \cdot 3=6$ and $17 \cdot 3=51$. The numbers $6,34,51$ satisfy the conditions. We will construct another set of three numbers using the prime numbers $5,7,11$, obtaining the numbers $35,55,77$. These numbers also satisfy the conditions. Combining the sets of numbers, condition (1) is satisfied. When choosing three numbers, at least two numbers will be from the same set, so they will have a common divisor, meaning condition (2) is also satisfied.

Comment. Correct solution - 20 points. Correct solution with an arithmetic error - 15 points. Only the correct answer without explanations - 1 point. Solution started but with fundamental errors - 1 point.