Happy weekend folks,
The intersections of the three circles divide their interiors into seven "regions" By complete non-coincidence, we have seven tiles
A "valid" placement is one where
For any valid placement, what are A, B, and C?
The solution is
A = 5, B = 4, C = 3
Here's the (probably too wordy) explanation
The total of the numbers we can place is $42$. This number can also be computed as the sum of the top two circles, minus the overlap, plus the number outside the circle. In formulae $$42 = 2X-A-B+C$$
After this we can work out which numbers can enter in the bottom circle: we know that they must multiply to a square, so it cannot contain $5$ or $7$ since there's no more number with $5$ or $7$ as factors. The remaining options are $(3,4 = 2^2,6 = 2*3,8 = 2^3,9 = 3^2)$. It is easy to see that $3$,$6$ and $8$ must be in the circle, since excluding them will lead to either an odd number of $2$ factors or an odd number of $3$ factors.
This means that there are two possibilities: either the bottom circle is $(3,4,6,8)$, that implies $X=24$, or it is $(3,6,8,9)$, that implies $X=36$.
We can exclude the second possibility since the first equation would give $$A+B=C+30$$ that is clearly not possible with the given numbers, since $A+B\leq 17$. So we have found that $X=24$ and the bottom circle is $(3,4,6,8)$. This set also contains $B$ and $C$. The top three numbers are $(5,7,9)$ and they contain $A$.
We can see that these last numbers are all odd. To compute $X$ in either top circles we have to sum two of these numbers and two numbers from the bottom circle. The first sum is even, since it is the sum of two odd numbers, and the total must be $24$, so even. This means that the only odd number of the bottom circle must be outside any intersections with the top, so $C = 3$.
The first equation now gives $A+B = 9$ and the only possibility, given the choices we figured out earlier, is $A=5$ and $B=4$.
