## 1. Determine all natural numbers n for which

$$
n+p(n)=70
$$

where $p(n)$ denotes the product of all digits of the number $n$.

(Jaroslav Švrček)
SOLUTION. First, we realize that each sought number $n$ must be two-digit. Indeed, for a one-digit number $n$, we have $n+p(n)=2 n \leqq 18$ and at the same time from the given equation $n+p(n)=70$ it follows that $n \leqq 70$.

Now, let's substitute a two-digit $n=\overline{a b}$ with the first digit $a \leqq 7$ into the equation and immediately transform it into a product form

$$
n+p(n)=(10 a+b)+a b=70 \text{, after rearrangement }(a+1)(b+10)=80 \text{. }
$$

Since $2 \leqq a+1 \leqq 8$ and simultaneously $10 \leqq b+10 \leqq 19$, the number $80=2^{4} \cdot 5$ can be decomposed into a product of two factors $a+1, b+10$ in only two ways, namely as $5 \cdot 16$ or $8 \cdot 10$ (depending on which factor contains the prime factor 5). In the first case, $a=4$ and $b=6$, in the second $a=7$ and $b=0$.

Conclusion. The only two numbers that satisfy the condition are $n=46$ and $n=70$.

Remark. The equation $(10 a+b)+a b=70$ can also be solved by expressing one of the unknowns $a, b$ as a fractional expression using the other and then performing the appropriate division of binomials (with remainder), for example

$$
b=\frac{70-10 a}{a+1}=-10+\frac{80}{a+1} .
$$

Now it is enough to find out for which $a \in\{1,2, \ldots, 7\}$ it holds that $(a+1) \mid 80$ and at the same time the corresponding $b$ lies in $\{0,1, \ldots, 9\}$.

Let's add that the equation $(10 a+b)+a b=70 \mathrm{~s}$ with unknown digits $a, b$ can also be solved by a shorter method. We rearrange it to the form $b(a+1)=10(7-a)$, according to which at least one of the numbers $b, a+1$ is divisible by five. Therefore, one of the cases $b=0, b=5$, $a=4$ or $a=9$ must occur. The rest of the solution is straightforward.