10.5. A circle is inscribed with 103 numbers. It is known that among any five consecutive numbers, there are at least two positive numbers. What is the minimum number of positive numbers that can be among these 103 written numbers?
Answer: 42.

Solution. We will show that there will be 3 consecutive numbers, among which there are at least 2 positive ones. This can be done, for example, as follows. Consider 15 consecutive numbers. They can be divided into 3 sets of 5 consecutive numbers, so among them, there are at least 6 positive numbers. But these 15 numbers can also be divided into 5 sets of 3 consecutive numbers. Therefore, by the Pigeonhole Principle, in one of these triplets, there are at least 2 positive numbers.

Fix these 3 consecutive numbers. Among them, there are at least 2 positive ones. The remaining 100 numbers can be divided into 20 sets of 5 consecutive numbers. In each such set, there will be at least 2 positive numbers. Thus, the total number of positive numbers is at least $2 + 2 \cdot 20 = 42$. Such a situation is possible. Number the numbers in a circle. The positive numbers can be those with indices $1, 2, 6, 7, \ldots, 96, 97, 101, 102$.

Comment. A correct answer without justification - 0 points. Proving that the number of positive numbers is not less than 42 - 5 points. Providing an example with 42 positive numbers - 2 points.