## Task 1 - 251231

Determine all pairs $(m ; n)$ of natural numbers $m$ and $n$ for which the following conditions (1) and (2) are satisfied:

(1) $m+n$ and $m \cdot n$ are two-digit numbers.

(2) If the digits of the number $m+n$ are reversed, the result is the (digit representation of the)

number $m \cdot n$.
According to the problem statement, there exist digits $c, d \in\{1 ; 2 ; \ldots ; 9\}$ such that $m+n=10 c+d$ and $m \cdot n=10 d+c$. No digit can be 0, as this would make the number with the corresponding tens place not a two-digit number. In particular,

$$
(m-1) \cdot(n-1)=(m \cdot n)-(m+n)+1=10 d+c-(10 c+d)+1=9(d-c)+1
$$

leaves a remainder of 1 when divided by 9. Therefore, the congruence class of $n$ can uniquely determine the congruence class of $m$ modulo 9:

If $n \equiv 0,2,3,5,6,8(\bmod 9)$, then $m \equiv 0,1,6,8,3,5(\bmod 9)$, while there is no solution if $n \equiv 1,4,7(\bmod 9)$, as then $n-1$ and thus $(n-1)(m-1)$ would be divisible by 3, which contradicts the congruence to 1 modulo 9.

Similarly, we obtain

$$
(m+1)(n+1)=(m \cdot n)+(m+n)+1=(10 d+c)+(10 c+d)+1=11(c+d)+1
$$

which leaves a remainder of 1 when divided by 11, allowing us to infer the congruence of $m$ from that of $n$ modulo 11:

If $n \equiv 0,1,2,3,4,5,6,7,8,9(\bmod 11)$, then $m \equiv 0,5,3,2,8,1,7,6,4,9(\bmod 11)$, where there is no solution for $n \equiv 10$ (mod 11) because then $(n+1)$ and thus $(n+1)(m+1)$ would be divisible by 11, contradicting the congruence condition.

Assume without loss of generality that $m \geq n$. Then $n \neq 0$, as otherwise $m \cdot n=0$ would not be a two-digit number. Furthermore, $n \neq 1,4,7$, as there is no solution for these values due to the congruence condition. Since $100>m \cdot n \geq n^{2}$, it follows that $n100$ is not a solution.
1. Case: $n=2$. Then $m \equiv 6(\bmod 9)$ and $m \equiv 3(\bmod 11)$, so $m=24$ must hold, which is a solution because $m n=48$ and $m+n=26$.
2. Case: $n=3$. Then $m \equiv 6(\bmod 9)$ and $m \equiv 2(\bmod 11)$, so $m=24$ must hold, which is a solution because $m n=72$ and $m+n=27$.
3. Case: $n=5$. Then $m \equiv 8(\bmod 9)$ and $m \equiv 1(\bmod 11)$, so $m=89$ must hold, which is not a solution because $m n>100$.
4. Case: $n=6$. Then $m \equiv 3(\bmod 9)$ and $m \equiv 7(\bmod 11)$, so $m=84$ must hold, which is not a solution because $m n>100$.
5. Case: $n=8$. Then $m \equiv 5(\bmod 9)$ and $m \equiv 4(\bmod 11)$, so $m=59$ must hold, which is not a solution because $m n>100$.
6. Case: $n=9$. Then $m \equiv 0(\bmod 9)$ and $m \equiv 9(\bmod 11)$, so $m=9$ must hold, which is a solution because $m n=81$ and $m+n=18$.

Thus, there are exactly three pairs that satisfy the conditions of the problem, namely $(m, n) \in\{(3,24),(24,3),(9,9)\}$.