2. The numbers 1 through 25 are each colored blue or red. Determine all possible colorings that satisfy the following rules:

- The number 5 is red.
- If the numbers $x$ and $y$ have different colors and $x+y \leqslant 25$, then $x+y$ is blue.
- If the numbers $x$ and $y$ have different colors and $x \cdot y \leqslant 25$, then $x \cdot y$ is red.
## 2. Version for class $5 \&$ class 4 and lower

First, we consider the case where 1 is red. Then all numbers from 1 to 25 must be red. Suppose a certain number $k$ is blue. Then 1 and $k$ have different colors, so according to the third rule, the number $1 \cdot k = k$ must be red. But this number was blue, which is a contradiction. Note that a coloring where all numbers are red is indeed a valid coloring.

Now we consider the case where 1 is blue. Since 5 is not blue, it follows from the second rule that 1 and 4 have the same color. Therefore, 4 is blue. Also, it follows that 2 and 3 have the same color. If 2 were red, then $3 = 1 + 2$ would have to be blue according to the second rule, so 2 and 3 are both blue.

We now know that 1, 2, 3, and 4 are blue. Applying the second rule again, we find that $6 = 5 + 1, 7 = 5 + 2, 8 = 5 + 3$, and $9 = 5 + 4$ are also blue. Applying the second rule again, we find that 11, 12, 13, 14 are blue, that 16, 17, 18, 19 are blue, and that 21, 22, 23, 24 are blue.

According to the third rule, $10 = 2 \cdot 5, 15 = 3 \cdot 5$, and $20 = 4 \cdot 5$ are red. Only the color of 25 is still undetermined. We find two possible colorings:

(1) Only 5, 10, 15, 20, 25 (the numbers divisible by 5) are red.

(2) Only 5, 10, 15, 20 are red.

We check that coloring (1) is indeed correct. The second rule is satisfied because the sum of a number divisible by 5 and a number not divisible by 5 is itself not divisible by 5. The third rule is satisfied because the product of a number divisible by 5 and a number not divisible by 5 is itself divisible by 5.

Coloring (2) differs from coloring (1) only for the number 25. To check that coloring (2) is correct, we only need to check the rules in the case where one of the involved numbers is 25. We check that 25 is not the product of a blue and a red number (which is true). We also need to check that there is no red number $k$ for which $25 + k$ is red or $25k$ is blue. This is fine because numbers greater than 25 are not colored.

We conclude that there are a total of 3 valid colorings: the coloring where all numbers are red and the colorings (1) and (2).