2. Determine the numbers $\overline{ab}$ for which there exists a prime number $k$ such that

$$
a \cdot b=k(a+b)
$$

Cătălin Ciupală
## Solution.

The relation from the hypothesis can be written as $(a-k)(b-k)=k^{2}$. (2p)

Then, since $a \geq 1$ and $k$ is a prime number, it follows that $(a-k) \in \{1, k, k^{2}\}$. (2p)

If $a-k=1$, then $a=k+1$ and $b=k^{2}+k$. Since $k$ is prime and $b \leq 9$, we deduce $k=2$. We obtain the number 36. (1p)

If $a-k=k$, then $a=b=2k$. Since $k$ is prime and $a, b \leq 9$, we deduce $k \in \{2,3\}$. We obtain the numbers 44 and 66. (1p)

If $a-k=k^{2}$, then $a=k^{2}+k$ and $b=k+1$. Since $k$ is prime and $a \leq 9$, we deduce $k=2$. We obtain the number 63. (1p)