>>14032405
is it 4?

>>14032411
Checked
Idk why do you think so?

>>14032405
bump

Aight heres a bit of a cop out but here we go

X is the upper left number of each, Y is the upper right, Z is the bottom.

Z=(-0.0625*(X-Y)*Y+0.875*(X-Y)*Y+0.5)*(X-Y)

So Z=0 when X is 8 and Y is 8

Technically, it works

if x=y z=0 and x different from y then z=x+y. anything above 10 reduced to first digit

>>14032540
>if x=y and x different from y
How is that possible?

>>14032540
i think that may be it. i mean, it fits. idk the correct answer

>>14032487
how did you come up with the formula for Z?

>>14032487
>Z=(-0.0625*(X-Y)*Y+0.875*(X-Y)*Y+0.5)*(X-Y)
Shit my bad guys, it's actually
Z=(-0.0625*(X-Y)^2*Y^2+0.875*(X-Y)*Y+0.5)*(X-Y)

>>14032609
It's two functions I thought up combined together.
U=(X-Y)*Y
and
Z=0.5*(-0.125*U^2+1.75*U+1)*U
Ok, so for the first function, I tried getting a function of the top two numbers that would equal 0 when the two numbers were equal, but also be a unique number for any other combination of number there. I tried U=(X-Y), but then the U of circle 1, 2 and 4 would all equal 2, and you need them to be different numbers because the bottom answers are all different. Then I tried (X-Y)*Y, which gave a U of 4,14,0,2,0 respectively for circles 1,2,3,4,5. This is perfect for the next step.
I made ordered pairs of U and Z for each circle that didn't have the top numbers equal the same thing. These are the ordered pairs: (4,6),(14,1),(2,4). Here is where I cheated a little. A quadratic can perfectly fit any three points in it's graph, so I ran a regression to get
Z=-0.125*U^2+1.75*U+1
This perfectly fits circles 1,2, and 4. However, we still need circle 3 and 5 to equal 0, which they don't right now. They equal 1 because their U is 0, so only the +1 in the quadratic is counted for Z. So I took the whole function of Z=-0.125*U^2+1.75*U+1 and multiplied it by (X-Y), which is either 0 when they match or a non-zero number when they don't match. This is where I got a bit lucky, because X-Y for circles 1,2, and 4 all equal 2. So (X-Y)/2 will either equal 0 if the numbers above are the same, or 1 if it's circles 1,2, or 4. Multiplying this by the original quadratic we got from the regression will essentially do nothing to Z unless X=Y, in which case it will make it 0.
The end result:
Z=(-0.125*U^2+1.75*U+1)*(X-Y)/2
Z=0.5*(-0.125*U^2+1.75*U+1)*(X-Y)
Now I substitute U for (X-Y)*Y because that's what I defined U as before.
Z=(-0.0625*(X-Y)^2*Y^2+0.875*(X-Y)*Y+0.5)*(X-Y)

>>14032658
>>14032609
>>14032487
Here is how you can check it if you want
Paste this into Wolfram Alpha
Z=(-0.0625*(X-Y)^2*Y^2+0.875*(X-Y)*Y+0.5)*(X-Y), X=4, Y=2
and change the X and Y values to the ones in the circles, and you get the proper Z value.