1. A necklace consists of 175 beads of red, blue, and green colors. It is known that each red bead has neighbors of different colors, and on any segment of the necklace between two green beads, there is at least one blue bead. What is the minimum number of blue beads that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last one is adjacent to the first.)
Answer: 30.

Solution 1. We will show that any block of six consecutive beads contains a blue bead. We can assume that there is no more than one green bead in it, otherwise there is nothing to prove. If the block contains 5 red beads, then at least 3 of them are consecutive, and the middle one does not satisfy the problem's condition. Therefore, the block contains no more than four red beads, and thus, there is a blue one.

Fix a blue bead in the necklace, and divide the rest into 29 blocks of 6 beads each. By the proven fact, each block contains at least one blue bead. Therefore, there are at least 30 blue beads in the necklace.

It remains to provide an example of a necklace with exactly 30 blue beads:

$$
\text {RRGRBS; RRGRBS; ..; RRGRBS (29 times); B. }
$$

Solution 2. Fix any green or blue bead, and divide the rest into 58 consecutive triplets. Each triplet contains no more than two red beads, otherwise the middle red bead would have same-colored neighbors. Therefore, there are no more than $2 \cdot 58 = 116$ red beads in the necklace, and thus, there are at least $175 - 116 = 59$ blue and green beads together. If we mentally remove all the red beads, we notice that the blue beads now make up at least half, as there is a blue bead between any two green beads. Therefore, there are at least $\frac{59}{2} = 29 \frac{1}{2}$ blue beads, meaning there are at least 30.

It remains to provide an example of a necklace with exactly 30 blue beads:

$$
\text {RRGRBS; RRGRBS; ...; RRGRBS (29 times); B. }
$$