## Subject 3.

Consider the numbers $1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$. What is the smallest number of numbers that must be excluded from these, so that with the remaining numbers we can form two sets with the same cardinality and the property that the product of the elements in the two sets is the same? Provide an example of such sets.
## Subject 3.

If the product of the elements in the first set is $\mathrm{x}$ and in the second set is also $\mathrm{x}$, then the product of all the numbers must be a perfect square.

We decompose the given numbers and calculate their product:

$1,2,3,2^{2}, 5,2 \cdot 3,7,2^{3}, 3^{2}, 2 \cdot 5,11,2^{2} \cdot 3,13,2 \cdot 7,3 \cdot 5$.

The product of these is $\mathrm{P}=2^{11} \cdot 3^{6} \cdot 5^{3} \cdot 7^{2} \cdot 11 \cdot 13$.

.. 2p.

We will exclude one factor from the product that has an odd power, i.e., $2,5,11,13$.

... 1p.

card $\mathrm{A}=$ card $\mathrm{B} \Rightarrow$ an odd number of numbers must be excluded: $10,11,13$.

$1 p$.

An example would be: $\mathrm{A}=\{2,3,5,8,9,14\}, \mathrm{B}=\{1,4,6,7,12,15\}$.

.. $1 p$.