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

$$
m+s(n)=n+s(m)=70
$$

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

(Jaroslav Švrček)
SOLUTION. Both numbers $m$ and $n$ are clearly two-digit. Let then $m=10a+b$ and $n=10c+d$, where $a, b, c, d$ are digits of the decimal system $(a \neq 0 \neq c)$. According to the problem, we have

$$
\begin{aligned}
& m+s(n)=(10a+b)+(c+d)=70 \\
& n+s(m)=(10c+d)+(a+b)=70
\end{aligned}
$$

By subtracting the second equation from the first, we get $9(a-c)=0$, and thus $a=c$. By substituting $c$ with $a$, both equations (1) and (2) will turn into the same equation $11a+(b+d)=70$. Given that $0 \leq b+d \leq 18$, we obtain the estimates for $a$: $52 \leq 11a \leq 70$. From this, it immediately follows that either $a=5$ or $a=6$. In the first case (for $a=c=5$), from the condition $11a+(b+d)=70$ we get $b+d=15$, i.e., $(b, d) \in\{(6,9),(7,8),(8,7),(9,6)\}$. In the second case (for $a=c=6$), we get $b+d=4$, and thus $(b, d) \in\{(0,4),(1,3),(2,2),(3,1),(4,0)\}$. Verification is not necessary, as we are guaranteed the validity of both equations (1) and (2).

Conclusion. The problem has 9 solutions, which are the pairs $(m, n) \in\{(56,59),(57,58),(58,57),(59,56),(60,64),(61,63),(62,62),(63,61),(64,60)\}$.

## GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. Determine all two-digit numbers that are equal to three times their digit sum. [Let $n=\overline{ab}$, then $n=10a+b=3(a+b)$, i.e., $7a=2b$. Since the numbers 2 and 7 are coprime, $b=7$ and $a=2(a \neq 0)$, and thus $n=27$.]

N2. Determine all two-digit numbers that are equal to the sum of their tens digit and the square of their units digit. [For the desired number $n=\overline{ab}$, we have $n=10a+b=a+b^2$, i.e., $9a=b(b-1)$. Given that the numbers $b$ and $b-1$ are coprime, either $9 \mid b$ or $9 \mid (b-1)$. The problem is satisfied only by $n=89$.]

N3. Determine all natural numbers $n$ for which $n+s(n)=2019$, where $s(n)$ denotes the digit sum of the number $n$. [69-C-S-1]

N4. In the year 2000, Alena found out that her age was equal to the sum of the digits of the year in which she was born. How old is she in 2020? [The sum of the digits of her birth year is at most 28 $(=1+9+9+9)$. Her birth year is thus a four-digit number of the form $\overline{19xy}$, for which $2000-\overline{19xy}=1+9+x+y$. From this, after rearrangement, we have $90=11x+2y$, from which it is clear that $x=8$ and $y=1$. Alena's age in 2020 is therefore 39 years.]

D1. Determine all four-digit numbers that are four times smaller than the number written with the same digits but in reverse order. [Let $\overline{abcd}$ be the desired number. Then $4 \cdot \overline{abcd}=\overline{dcba}$. The digit $a$ must be even and at the same time $a \leq 2$. Thus, $a=2$, and therefore $d=8$. The only solution is then the number 2178.]

D2. Find all four-digit numbers $\overline{abcd}$ with a digit sum of 12 such that $\overline{ab}-\overline{cd}=1$. [69-C-I-1]