
05.1. Find all positive integers $k$ such that the product of the digits of $k$, in the decimal system, equals

$$
\frac{25}{8} k-211
$$


Solution. Let

$$
a=\sum_{k=0}^{n} a_{k} 10^{k}, \quad 0 \leq a_{k} \leq 9, \text { for } 0 \leq k \leq n-1,1 \leq a_{n} \leq 9
$$

Set

$$
f(a)=\prod_{k=0}^{n} a_{k}
$$

Since

$$
f(a)=\frac{25}{8} a-211 \geq 0
$$

$a \geq \frac{8}{25} \cdot 211=\frac{1688}{25}>66$. Also, $f(a)$ is an integer, and $\operatorname{gcf}(8,25)=1$, so $8 \mid a$. On the other hand,

$$
f(a) \leq 9^{n-1} a_{n} \leq 10^{n} a_{n} \leq a
$$

So

$$
\frac{25}{8} a-211 \leq a
$$

or $a \leq \frac{8}{17} \cdot 211=\frac{1688}{17}<100$. The only multiples of 8 between 66 and 100 are $72,80,88$, and 96. Now $25 \cdot 9-211=17=7 \cdot 2,25 \cdot 10-211=39 \neq 8 \cdot 0,25 \cdot 11-211=64=8 \cdot 8$, and $25 \cdot 12-211=89 \neq 9 \cdot 6$. So 72 and 88 are the numbers asked for.
