2. The numbers 1 through 15 are each colored blue or red. Determine all possible colorings that satisfy the following rules:

- The number 15 is red.
- If the numbers $x$ and $y$ have different colors and $x+y \leqslant 15$, then $x+y$ is blue.
- If the numbers $x$ and $y$ have different colors and $x \cdot y \leqslant 15$, then $x \cdot y$ is red.
## 2. Version for Grade 6

First, let's consider the case where 1 is red. Then all numbers from 1 to 15 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 $k=1 \cdot k$ 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.

$\mathrm{Now}$ let's consider the case where 1 is blue. Note that if two numbers add up to 15, then those numbers must have the same color according to the second rule. The following pairs must therefore always have the same color: 1 and 14, 2 and 13, 3 and 12, 4 and 11, 5 and 10, 6 and 9, 7 and 8.

The number 2 is blue. Suppose 2 is red. Then from the second rule, $3=1+2$ must be blue. Repeated application of the second rule gives that $5=3+2$ is blue, that $7=5+2$ is blue, and eventually that 15 is blue. Since 15 is not blue, 2 cannot be red.

The number 7 is blue. Suppose 7 is red. Then from the second rule, $8=1+7$ must be blue. However, 7 and 8 must have the same color, so that cannot be.

The number 4 is blue. Suppose 4 is red. Then from the second rule, $11=4+7$ must be blue. But 4 and 11 have the same color, so 4 cannot be red.

The pairs 3 and 12 and 6 and 9 must have the same color. Otherwise, $9=3+6$ would be blue according to the second rule, and also $12=3+9$ would be blue according to the second rule.

So far, we know that 15 is red, that 1, 2, 4, 7, 8, 11, 13, 14 are blue, that $3, 6, 9, 12$ have the same color, and that 5, 10 have the same color. The numbers 3 and 5 cannot both be red, because if 3 is red, then $5=2+3$ would be blue according to the second rule. The three remaining possibilities for the colors of 3 and 5 give the following colorings:

(1) Only 15 is red.

(2) Only 5, 10, 15 (the numbers divisible by 5) are red.

(3) Only 3, 6, 9, 12, 15 (the numbers divisible by 3) are red.

It is easy to verify that these three colorings are indeed valid. We will check this for the third coloring, the other two are verified in the same way. That the sum of a red and a blue number is always blue follows from the fact that the sum of a number divisible by 3 and a number not divisible by 3 is also not divisible by 3. That the product of a blue and a red number is always red follows from the fact that the product of a number divisible by 3 and a number not divisible by 3 is again divisible by 3.

We conclude that there are a total of 4 valid colorings: the coloring where all numbers are red and the colorings (1), (2), and (3).