7.3. On a board, all natural numbers $1,2,3, \ldots, 15$ are written. Determine the minimum number of numbers that can be erased from the board so that with the remaining numbers we can form two non-empty sets of numbers that simultaneously satisfy the following properties:

a) they have no common elements;

b) they have the same number of elements;

c) the product of the elements of the two sets is the same.
## Solution:

The product of all the numbers initially written on the board is $n=1 \cdot 2 \cdot 3 \cdot 2^{2} \cdot 5 \cdot 2 \cdot 3 \cdot 7 \cdot 2^{3} \cdot 3^{2} \cdot 2 \cdot 5 \cdot 11 \cdot 2^{2} \cdot 3 \cdot 13 \cdot 2 \cdot 7 \cdot 3 \cdot 5=2^{11} \cdot 3^{6} \cdot 5^{3} \cdot 7^{2} \cdot 11 \cdot 13$. To obtain the same product for the elements of the two sets, it follows that the product of all the remaining numbers must be a perfect square. Thus, we need to delete a factor from $n$ that has an odd exponent, i.e., $2, 5, 11, 13$. Since the cardinality of the two sets formed is the same, it follows that an even number of numbers must remain on the board. As we initially have an odd number of numbers on the board, it follows that an odd number of numbers must be deleted. Thus, we delete the numbers $10, 11, 13$ from the board. The numbers remaining on the board are $1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 14, 15$. From the remaining numbers, we deduce that 5 and 15 do not belong to the same set, just as the numbers 7 and 14 do not belong to the same set. As a result, we can obtain the sets $M=\{2,3,5,8,9,14\}$ and $N=\{1,4,6,7,12,15\}$ or $M=\{1,4,5,9,12,14\}$ and $N=\{2,3,6,7,8,15\}$.