5. Natural numbers divisible by 3 were painted in two colors: red and blue, such that the sum of a blue and a red number is red, and the product of a blue and a red number is blue. In how many ways can the numbers be colored so that the number 546 is blue?
Answer: 7.

Solution: We will prove that all blue numbers are divisible by the same number. First, we will prove that the product of two blue numbers is blue. Assume the statement is false. Let $a$ and $b$ be two such blue numbers that their product is a red number. Let $d$ be some red number (we can take $d = ab$). Then $a + d$ is red. From this, it follows that $a \cdot b + b \cdot d = b(a + d)$ is blue, since the blue number $b$ is multiplied by the red number $a + d$. The number $b \cdot d$ is also blue, and the number $a \cdot b$ is red by assumption. We have obtained that the sum of a blue and a red number is blue, which contradicts the condition. Therefore, the assumption is false, and the product of any two blue numbers is a blue number. Together with the condition, this means that the product of a blue number and any number is a blue number.

Let $a_0$ be the smallest blue number, and $a$ be any blue number. It can be represented as $a = k \cdot a_0 + r$, where $0 \leq r < a_0$. The remainder $r$ cannot be a blue number, since $a_0$ is the smallest blue number. The product $k \cdot a_0$, by the proven fact, is a blue number. The sum on the right, of a blue and a red number, is a red number by the condition, but the number on the left side of the equation is blue; a contradiction. Therefore, the remainder $r = 0$, and any blue number is divisible by the smallest blue number. The number $546 = 2 \cdot 3 \cdot 7 \cdot 13$. Only numbers divisible by 3 are considered, and among them, blue can be all numbers divisible by 6, or by 21, or by 39, or by 42, or by 78, or by 273, or by 546, a total of 7 ways.

Comment: 20 points for finding all ways. 15 points for establishing that blue numbers are divisible by divisors of the number 546, but the correct answer is not obtained. 5 points for finding at least one way in the absence of a general solution. 15 points for gaps in the justification. 2 points for starting the solution and making some progress.