Answer:

(a) (2 points) 12.

(b) (2 points) 24.

Solution. By the condition, there will be two red balls, between which lies exactly one ball. Number the balls clockwise with numbers \(1, 2, \ldots, 36\) such that the 1st and 3rd balls are red. From the condition, it follows that the 35th ball is blue (otherwise, the first condition is false for the 1st ball), the 5th ball is blue (otherwise, the first condition is false for the 3rd ball), the 7th ball is red (otherwise, the second condition is false for the 3rd ball), and the 9th ball is red (otherwise, the first condition is false for the 7th ball). Thus, assuming that the 1st and 3rd balls are red, we have that the 5th ball is blue, and the 7th and 9th balls are red. Continuing this reasoning, we get that the 11th ball is blue, the 13th and 15th balls are red, and so on.

![](https://cdn.mathpix.com/cropped/2024_05_06_d4c5d21501884c712855g-2.jpg?height=42&width=1268&top_left_y=1598&top_left_x=93)
Two-thirds of the balls are red, and one-third are blue.

![](https://cdn.mathpix.com/cropped/2024_05_06_d4c5d21501884c712855g-3.jpg?height=521&width=521&top_left_y=81&top_left_x=466)

(a) Suppose there are no two adjacent red balls. It is easy to see that then there are no red balls among the balls with even numbers. Therefore, the total number of red balls is \(\frac{36}{2} \cdot \frac{2}{3} = 12\). An example with 12 red balls is shown below.

![](https://cdn.mathpix.com/cropped/2024_05_06_d4c5d21501884c712855g-3.jpg?height=519&width=519&top_left_y=786&top_left_x=467)

(b) Suppose there are two adjacent red balls. It is easy to see that then among the balls with even numbers, red balls also constitute two-thirds. Therefore, the total number of red balls is \(36 \cdot \frac{2}{3} = 24\). An example with 24 red balls is shown below.

![](https://cdn.mathpix.com/cropped/2024_05_06_d4c5d21501884c712855g-4.jpg?height=533&width=523&top_left_y=81&top_left_x=465)
Problem 10.2. There are 36 balls lying in a circle, each of which is either red or blue (balls of each of these colors are present). It is known that:

- for any red ball, there is exactly one red ball such that there is exactly one ball between them;
- for any red ball, there is exactly one red ball such that there are exactly three balls between them.

(a) (2 points) Suppose there are no two adjacent red balls. How many red balls can lie in a circle? Indicate all possible options.

(b) (2 points) Suppose there are two adjacent red balls. How many red balls can lie in a circle? Indicate all possible options.