1. (10 points) In workshop $B$, there are beads of two shapes - "cubes" and "pyramids". The "cubes" are green and blue, and the "pyramids" are red and blue. Traditions require that in any jewelry, adjacent beads must be of different shapes and different colors. The jeweler made a necklace in the form of a ring, using all four types of beads and in accordance with traditions. How many beads can be in such a necklace?

a) 7 ; b) 8 ; c) 10 ; d) 11.
Answer: b) and c).

Solution: We will prove that the solution to the problem is numbers of the form $6+2k$, where $k$ is a non-negative integer. Let's agree to denote green and blue cubes as $\mathrm{K}$ and $\mathrm{K}$, and red and blue pyramids as П and П.

Firstly, note that the number of beads is even. Indeed, cubes and pyramids cannot stand next to each other. Therefore, they are equally represented in the necklace, and thus the total number of beads is twice the number of cubes.

We will show that the minimum number of beads is 6. According to the condition, the necklace must contain a blue cube K. To the left and right of it, only red pyramids can stand, so the necklace contains the combination П K П. Also, according to the condition, a green cube $K$ and a blue pyramid П must be used. Therefore, the necklace must contain at least 5 beads, and considering the evenness, at least 6. The combination П K П K П K gives an example of a traditional necklace with 6 beads.

It remains to note that between П and $\mathrm{K}$, pairs of these beads can be inserted any number of times. Therefore, numbers of the form $6+2k$ for non-negative integers $k$ are solutions to the problem.