Product of numbers in a set

Question

What is the least number of elements we have to delete from the set {10, 20, 30, 40, 50, 60, 70, 80, 90} so that the product of the elements remaining in the set is a perfect square?

Answer 1

Note that $$\prod S_i=10^{10}*2^8*3^4*5^2*7$$ To make this a square you must remove the 7 of course, so removing the factor of 70 yields:

$$10^{9}*2^8*3^4*5^2$$

So now you have an extra power of ten, so removing a ten along with any perfect square (1,4, or 9) should suffice.

So the answer is 2, removing either (70,10) (70,40) or (70,90)

Answer 2

You obviously have to remove the $70$, as there is no other power of $7$ to make a perfect square.

But now, you also have to remove one among the remaining numbers as there is an odd power of $5$ overall.

Removing $30$ or $60$ will disrupt the already even power of $3$. Power of $2$ in the remaining product (after removing $70$) is odd. Also, an odd power of $5$ remains.

Removing $90$, $10$ or $40$ suffices (none of them disrupts any remaining powers as they are a multiple of $5 \times 2 \times \text{some perfect square}$ )

Hence $2$ is the answer