According to the task

$$
p p_{1}+p p_{2}+p p_{3}=I\left(p_{1}+p_{2}+p_{3}\right)=406=2 \cdot 7 \cdot 29
$$

where

$$
p>2 \text {, and } 1<p_{1}<p_{2}<p_{3}
$$

Thus, $p=7$, or $p=29$.

If $p=7$, then $p_{1}+p_{2}+p_{3}=2 \cdot 29=58$.

The sum of three prime numbers can only be even if one of them (naturally the smallest) is 2.

So $p_{1}=2$, and thus

$$
p_{2}+p_{3}=56
$$

This equality is only satisfied by the following three pairs of prime numbers:

$$
3+53, \quad 13+43 \text { and } 19+37 \text {. }
$$

If $p=29$, then $p_{1}+p_{2}+p_{3}=2 \cdot 7=14$. Since $p_{1}=2$ is also necessary in this case,

$$
p_{2}+p_{3}=12
$$

in which case only the pair of prime numbers $5+7$ fits.

Thus, there are 4 solutions in total:

$14,21,371$

$14,91,301$;

$14,133,259$

$58,145,203$.

The solutions to the problems posed in the advanced competition will be published in our next issue.
We know the following about three natural numbers:

a) all three are different;

b) their sum is 406;

c) their greatest common divisor is a prime number greater than 2 (denoted by $p$);

d) if the individual numbers are divided by $p$, we again get three prime numbers (let these be $p_{1}, p_{2}, p_{3}$). What are these numbers?

We note that 1 is not a prime number.