5. Petya wants to test the knowledge of his brother Kolya, the winner of the "Higher Test" mathematics Olympiad. For this, Petya thought of three natural numbers $a, b, c$, and calculated $x = \text{GCD}(a, b)$, $y = \text{GCD}(b, c)$, $z = \text{GCD}(c, a)$. Then he wrote three rows of five numbers each on the board:

$$
\begin{gathered}
6, 8, 12, 18, 24 \\
14, 20, 28, 44, 56 \\
5, 15, 18, 27, 42
\end{gathered}
$$

Petya informed Kolya that one of the numbers in the first row is $x$, one of the numbers in the second row is $y$, and one of the numbers in the third row is $z$, and asked him to guess the numbers $x, y, z$. After thinking for a few minutes, Kolya managed to solve the problem, correctly naming all three numbers. Name them and you. Prove that there is a unique such triplet $(x, y, z)$.
Answer. $x=8, y=14, z=18$.

Solution. We will use the following statement: If two of the numbers $x, y, z$ are divisible by some natural number $m$, then the third is also divisible by $m$.

Proof. Let's say, for example, $x$ and $y$ are divisible by $m$.

$$
\left\{\begin{array} { l } 
{ x : m \Rightarrow a : m , b : m } \\
{ y : m \Rightarrow b : m , c : m . }
\end{array} \Rightarrow \left\{\begin{array}{l}
a: m \\
c: m
\end{array} \Rightarrow z: m\right.\right.
$$

Corollary: If one of the numbers $x, y, z$ is not divisible by $m$, then at least one of the remaining two is also not divisible by $m$.

Now consider the numbers given in the problem:

$$
\begin{gathered}
6,8,12,18,24 \\
14,20,28,44,56 \\
5,15,18,27,42
\end{gathered}
$$

Notice that in the first two rows, all numbers are even, i.e., $x: 2, y: 2 \Rightarrow z: 2 \Rightarrow z=18$ or $z=42$. Further, both numbers 18 and 42 are divisible by 3, i.e., $z: 3$. In the second row, there are no numbers divisible by 3, i.e., $y \not 3 \Rightarrow x ; 3 \Rightarrow x=8$. Further, $x: 4, z \Varangle 4 \Rightarrow y \% 4 \Rightarrow y=14$. Finally, $y: 7, x ; 7 \Rightarrow z ; 7 \Rightarrow z=18$.

The values $x=8, y=14, z=18$ are possible, for example, when $a=72, b=56, c=126$.

Criteria for evaluating solutions.

## $(-)$

(-.) Divisibility by one number is considered.

$(-/+)$ Divisibility by two numbers is considered

OR

a lemma analogous or equivalent to the lemma in the provided solution is formulated and proven.
$(+/ 2)$ Divisibility by three numbers is considered.

(+.) A lemma analogous or equivalent to the lemma in the provided solution is used in the reasoning but is not proven.

$(+)$ Correct solution