How many different squares are there which are the product of six different integers from 1 to 10 inclusive?
A similar problem, asking how many different squares are there which are the product of six different integers from 1 to 9 inclusive is simple, as we are able to simply remove 5, 7 and 2 or 8 (odd powers of 2) to ensure that the number produced by the multiplication of the six integers is a square number, having only even powers of its prime factors. However, when the range is from one to ten, there are more possibilities as, now, the 2, 3 and 5 prime factors appear an even number of times.
Thanks.
Start with $1,2,3,4,5,6,7,8,9,10$. Discard $7$, as $7^2$ cannot appear in the product of any subset of those numbers. Set aside $1,4,9$, as they are squares, and can be included or removed from some other grouping without affecting whether the product is a square.
This leaves $2,3,5,6,8,10$. As it happens, the product of those six numbers is $120^2$, so that is one set of six numbers whose product is a square. If we remove one of those numbers from the product, the product of the remaining five numbers cannot be a square, because a square divided by a non-square cannot give a square quotient.
If we try to remove two members of that set, they must multiply to a square. If we remove a number with a factor of $5$, we must remove two numbers with a factor of $5$ to ensure that the product contains $5^2$. But the product of $5,10$ is not a square. Hence, removing two numbers containing factors of $5$ is impossible. Similar logic shows that you cannot remove two numbers containing a factor of $3$. This leaves only $2,8$, which do multiply to $4^2$, as a set of two removable numbers. The remaining $3,5,6,10$ multiply to $30^2$. We can fill out that set with any two of $1,4,9$ to make a set of six: $1,3,4,5,6,10$ multiply to $60^2$; $1,3,5,6,9,10$ multiply to $90^2$; and $3,4,5,6,9,10$ multiply to $180^2$. That gives three more sets of six numbers whose product is a square.
The last option is to see whether groups of three number can multiply to a square. Here we notice that if we identify any three such numbers, the three remaining numbers also must multiply to a square. So the six numbers will be segregated into sets of three. Since all numbers must be used, it is sufficient to identify sets of three numbers containing $2$, and identify other sets by complementation. Those sets are $2,3,6$ which multiply to $6^2$, leaving $5,8,10$ which multiply to $20^2$; and $2,5,10$ which multiply to $10^2$, leaving $3,6,8$ which multiply to $12^2$. Each of these sets of three can be augmented by including $1,4,9$ to give sets of six: $1,2,3,4,6,9$ which multiply to $36^2$; $1,4,5,8,9,10$ which multiply to $120^2$; $1,2,4,5,9,10$ which multiply to $60^2$; $1,3,4,6,8,9$ which multiply to $72^2$. These are four more sets of six numbers that multiply to a square.
In total, there are eight sets of six numbers taken from among $1,2,3,4,5,6,7,8,9,10$ that multiply to a square number.