Task A-1.5. (4 points)

How many elements at least need to be removed from the set $\{2,4,6,8,10,12,14,16\}$ so that the product of the remaining elements is a square of a natural number?
## Solution.

The product of all elements of the given set is

$$
\begin{aligned}
P & =2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 16 \\
& =2 \cdot 2^{2} \cdot(2 \cdot 3) \cdot 2^{3} \cdot(2 \cdot 5) \cdot\left(2^{2} \cdot 3\right) \cdot(2 \cdot 7) \cdot 2^{4} \\
& =2^{15} \cdot 3^{2} \cdot 5 \cdot 7
\end{aligned}
$$

The product is divisible by 5, but not by $5^{2}$, so it is clear that the number 10 needs to be removed.

The product is divisible by 7, but not by $7^{2}$, so it is clear that the number 14 also needs to be removed.

The product of the remaining numbers, $2^{13} \cdot 3^{2}$, is not a perfect square, so at least one more number needs to be removed. We see that it is sufficient to remove an odd power of 2, i.e., the number 2 or the number 8.

It is necessary to remove at least three numbers.

## Note.

It is not necessary for the student to list all possible ways to remove three elements, but it is necessary to show that it is not possible to remove only one or two numbers and for the product of the remaining to be a perfect square. The possible ways to remove three numbers are: $\{\mathscr{Z ,} 4,6,8,10,12,14,16\}$ and $\{2,4,6,8,10,12,14,16\}$.