6. Black and white balls are arranged in a circle, with black balls being twice as many as white ones. It is known that among pairs of adjacent balls, there are three times as many monochromatic pairs as polychromatic ones. What is the smallest number of balls that could have been arranged? (B. Trushin)
Answer: 24. Solution: Since the number of black balls is twice the number of white balls, the total number of balls is divisible by three. Let's denote it by $n$. All the balls are divided into alternating groups of consecutive balls of the same color (a group can consist of just one ball). Since the colors of the groups alternate, the total number of groups is even. Let the number of groups of each color be $k$. Then the number of adjacent balls of different colors will be $2k$, and the number of adjacent balls of the same color will be $n-2k$. From the condition of the problem, we get that $n-2k = 3 \cdot 2k$. Therefore, $n = 8k$. Thus, the total number of balls is divisible by both three and eight. Hence, $n$ is divisible by 24, and so $n \geq 24$. An example can be any circle of 8 white and 16 black balls, in which there are three white and black groups. For example, white - black - white - black - six white - fourteen black.

Criteria: Only the answer - 0 points. The answer with an example of 24 balls without proof that fewer than 24 balls is impossible - 2 points. Proof that fewer than 24 balls is impossible without an example of 24 balls - 4 points. Using the fact that the number of different-colored pairs is even without proper justification reduces the score by 1 point. Proving that the number of balls is divisible by 12 without further substantial progress - 1 point.