I need to solve this for a job. There is no "1" nor "-1" on the answer card. Please help me

I need to solve this for a job. There is no "1" nor "-1" on the answer card. Please help me

5

explain?

Brainlet

Multiply the three numbers in the corners and then add 2 to the result.

10. Add the numbers in the top and left corners, then multiply that by two. 2*(1+0) = 2, 2*(1+3) = 8, 2*(2+3) = 10.

5. add the numbers in the left and top corners, and multiply them by the right one

what job

7
Corners multiplied + bottom right

theres like 5 different patterns going on here, wtf is on the answer card?

There are too many possible patterns. Not enough information.
Nobody's posted the one I came up with yet. A*B + 2C, where A and B are the bottom corners and C is the top. By that logic, X is also 8.

>there's like 5 different patterns
there are always infinite ways to continue a finite ""pattern"", whatever you think that is

Pointless pedantry aside, this problem expects you to figure out the operation(s) used to arrive at the answer in the center of each triangle, and then use that knowledge to find the answer for the third, unsolved one. There is ostensibly only one "correct" answer, but several valid answers can be easily found.

11
Sum of squares of all numbers minus bottom two.

This is just a retard filter to see if you can find patterns in things.

They want you to say 5. It's not just a pattern recognition test, it's to make sure you will be a good wagecuck and won't get any funny ideas like this guy here

it's shit. it's not pointless pedantry, this kind of "problem" is shit.

If a first grader writes "10" as the answer to "1 + 7 = __" because the book didn't specify it was supposed to be decimal instead of octal, that is wrong, not a "shit problem".

The middle one would be a 9, not an 8.

8 the third triangle is just the same as the second triangle but rotated.

I got 8, but I think multiple numbers work.
I did the bases multiplied plus 2 times the top.

I don't think there is only one solution.

5 is the cleanest answer that is what they want. 8 and 10 also work.

1^0=1 2^1=2
3^1=3 2^3=8
2^3=8 1^8=8

[math]\infty [/math]

Basically is the same as the 2nd triangle but rotated

Ahh, my bad, u good!
Then 8. Sum of squares minus all numbers.
Which boiles back to saying that two rightmost triangles are the same

Then it should be infinity

could also just be 2^(bottom_left)

>A*B + 2C
One could argue that it's more complex that (A+B)*C because it requires a coefficient.
You should always find the simplest rule.