## Task B-2.2.

A finite sequence of numbers is excellent if each subsequent member of the sequence, except the first, is greater than the previous one, and if the product of all members of that sequence is a perfect square. For example, the sequence 2, 6, 27 is an excellent sequence. Determine the natural numbers $x$ and $y$ such that the sequence 28, $x, y, 65$ is excellent.
## Solution.

Since the number $28 \cdot 65 \cdot x \cdot y = 2^{2} \cdot 5 \cdot 7 \cdot 13 \cdot x \cdot y$ is a perfect square, it follows that at least one of the numbers $x$ and $y$ must be divisible by 5, 7, and 13.

Notice that a number divisible by 13 cannot be divisible by 5 or 7, otherwise we would get a number greater than or equal to 65.

Therefore, one of the numbers must be divisible by $7 \cdot 5 = 35$, i.e., of the form $35k$. It follows that $k=1$ because otherwise we would get a number greater than 65.

Thus, one of the desired numbers is 35. 1 point

The other number is of the form $13l$, and since $28 < 13l < 65$, we conclude that $l=3$ or $l=4$.

We discard the possibility $l=3$ because we would then get the number 39, for which the product $2^{2} \cdot 5 \cdot 7 \cdot 13 \cdot x \cdot y$ would not be a perfect square.

Therefore, $l=4$, and the other desired number is 52.

It holds that $28 < 35 < 52 < 65$ and $28 \cdot 35 \cdot 52 \cdot 65 = (2 \cdot 2 \cdot 5 \cdot 7 \cdot 13)^{2}$, which means we have obtained the required sequence, and the desired numbers are $x=35$ and $y=52$.

Note: If a student factors the numbers and mentions divisibility by 5, 7, and 13, and uses this to obtain the solutions 35 and 52 without justifying that these are the only solutions, 4 points should be awarded. For solutions obtained without justification and without the process, 2 points should be awarded if the verification of these solutions is visible, and 1 point if it is not.