Problem 1. Determine the natural numbers that have the property of admitting exactly 8 positive divisors, three of which are prime numbers of the form $a, \overline{b c}$ and $\overline{c b}$, and $a+\overline{b c}+\overline{c b}$ is a perfect square, where $a, b$ and $c$ are digits with $b<c$.
Solution. Let $x$ be a number with the properties mentioned in the problem. $x$ has exactly 8 divisors, $a, \overline{b c}$, and $\overline{c b}$ are three prime divisors, so it follows that $x$ is divisible by $y=a \cdot \overline{b c} \cdot \overline{c b}$, and $y$ has exactly 8 divisors, hence $x=y$ $3 \text{p}$

The numbers $\overline{b c}$ and $\overline{c b}$ are distinct prime numbers, so $\overline{b c} \in\{13,17,37,79\}$ $1 \text{p}$

From the condition that $a+\overline{b c}+\overline{c b}$ is a perfect square and $a$ is prime, it follows that $a=5, \overline{b c}=13$ and $\overline{c b}=31$, thus $x=2015$