1. Determine all pairs $(m, n)$ of natural numbers for which

$$
2 m+2 s(n)=n \cdot s(m)=70 \text{, }
$$

where $s(a)$ denotes the digit sum of the natural number $a$.

(Jaroslav Švrček)
SOLUTION. From the equation $2 m+2 s(n)=70$ simplified to $m+s(n)=35$, it follows that $m<35$, and thus $s(m) \leqq 11$ (equality occurs for $m=29$). According to the equation $n \cdot s(m)=70$, the number $s(m)$ is therefore a divisor of the number 70, which is at most 11, so $s(m) \in\{1,2,5,7,10\}$.

Now, for each value of $s(m)$ equal to $1,2,5,7,10$, we will sequentially calculate the values of $n=70 / s(m), s(n), m=35-s(n)$ and write them in the following table:

| $s(m)$ | 1 | 2 | 5 | 7 | 10 |
| :---: | ---: | ---: | ---: | ---: | ---: |
| $n$ | 70 | 35 | 14 | $\mathbf{1 0}$ | 7 |
| $s(n)$ | 7 | 8 | 5 | 1 | 7 |
| $m$ | 28 | 27 | 30 | 34 | 28 |

The numbers in each column do satisfy both given equations, but it is not guaranteed that $s(m)$ is actually the digit sum of the number $m$. We see that the first three cases, where $s(m) \in\{1,2,5\}$, do not lead to any solution, because $s(28) \neq 1, s(27) \neq 2$ and $s(30) \neq 5$. The remaining two cases, where $s(m)=7$ and $s(m)=10$, lead to two solutions of the problem, since $s(34)=7$ and $s(28)=10$. The corresponding $m$ and $n$ are printed in bold in the table.

Conclusion. The given problem has exactly two solutions, which are the pairs $(m, n)=(34,10)$ and $(m, n)=(28,7)$.

Note. If the inequality $s(m) \leqq 11$ is not derived beforehand, the problem can be solved by the method provided in the solution, by expanding the table to include columns for the values $s(m) \in\{14,35,70\}$, i.e., the remaining three divisors of the number 70:

| $s(m)$ | 14 | 35 | 70 |
| :---: | ---: | ---: | ---: |
| $n$ | 5 | 2 | 1 |
| $s(n)$ | 5 | 2 | 1 |
| $m$ | 30 | 33 | 34 |

These possibilities are excluded by the inequalities $s(30) \neq 14, s(33) \neq 35$ and $s(34) \neq 70$.

For a complete solution of the problem, award 6 points, of which: 2 points for listing all (previously unexcluded) possibilities for the values of $s(m)$ and $n$ from the factorization $70=n \cdot s(m)$, 3 points for excluding all unsuitable possibilities; 1 point for finding and verifying both solutions including the correct answer.

For merely guessing both suitable pairs, award 1 point.