2. Let A, B be sets of positive integers such that the sum of any two different numbers from $A$ belongs to $B$ and the quotient of any two different numbers from $B$ (the larger divided by the smaller) lies in $A$. Determine the maximum possible number of elements in the set $A \cup B$.


2. First, we will prove that the set $\mathrm{A}$ can contain at most two numbers. Assume that $\mathrm{A}$ contains three numbers $a < b < c$. Then the set $\mathrm{B}$ contains the numbers $a+b < a+c < b+c$, and thus the set $\mathrm{A}$ must contain the number

$$
\frac{b+c}{a+c}=1+\frac{b-a}{a+c}
$$

which is not an integer, since $0 < b-a < a+c$.

If the set $\mathrm{B}$ contained four numbers $k < l < m < n$, then the set $\mathrm{A}$ would contain three different numbers $n / k, n / l, n / m$. Therefore, the set $\mathrm{B}$ has at most three elements, and $\mathrm{A} \cup \mathrm{B}$ cannot have more than five elements.

This number is achieved precisely when $\mathrm{A}=\{a, b\}, \mathrm{B}=\{k, l, m\}$, where $a < b$ and $l / k = m / l = a, m / k = b$. Then $b = a^2 (a \geq 2)$ and one of the elements of the set $\mathrm{B}$ is $a + a^2$; the other two are then either $a^2 + a^3$ and $a^3 + a^4$, or $1 + a$ and $a^2 + a^3$. For example, the sets $\mathrm{A}=\{2,4\}, \mathrm{B}=\{3,6,12\}$ together have five elements.

For a complete solution, award 6 points. Award 3 points for proving the inequality $|A| \leq 2$, 1 point for proving the inequality $|B| \leq |A| + 1$, and 2 points for finding (one is sufficient) examples of sets $\mathrm{A}$ and $\mathrm{B}$ such that $|A \cup B| = 5$.