Solution. Let's observe that two numbers differing by 5 have the same color. Indeed, if they were of different colors, their difference should be red, according to rule a). But their difference is 5, which is blue. Therefore, it is enough to know the color of the first 4 numbers. Here, we distinguish two cases: 1) the tile 1 is blue, and 2) the tile 1 is red.

Case 1) If 1 is blue, the rest of the tiles must be blue, by rule b). This is because if the tile $k \neq 1$ were red, then by b), $k = k \cdot 1$ would have to be blue, which contradicts that $k$ is red.

Case 2) If 1 is red, by rule a) $4 = 5 - 1$ is red. To determine the color of 2 and 3, suppose 3 is blue. Since $2 = 3 - 1$ and 3 and 1 are of different colors, then 2 is red. Now, $3 = 5 - 2$ and 5 is blue and 2 is red, so 3 is red. This cannot be, therefore 3 cannot be blue and is red, which means 2 is also red. Thus, 1, 2, 3, and 4 are red, as are the rest of the tiles that are not multiples of 5.

Therefore, there are only two possible colorings, either all tiles are blue or all are red, except for the multiples of 5 which are blue.
1. We have 50 chips numbered from 1 to 50, and they need to be colored red or blue. We know that chip 5 is blue. For the coloring of the rest of the chips, the following rules are followed:

a) If the chip with the number $x$ and the chip with the number $y$ are of different colors, then the chip with the number $|x-y|$ is painted red.

b) If the chip with the number $x$ and the chip with the number $y$ are of different colors and $x \cdot y$ is a number between 1 and 50 (inclusive), then the chip with the number $x \cdot y$ is painted blue.

Determine how many different colorings can be performed on the set of chips.