If you can solve this then you get the brainliest!

Start with a blank grid

Place 3 x's to clump into one corner like this

Then fill out the opposite corner in a similar fashion to make sure that we don't get 3 in a row/column/diagonal anywhere

Each row has exactly 2 x's and each column has exactly 2 x's. The main diagonal has 2 x's, while the off diagonal has no x's at all.

I'm not sure if this is the only configuration where this is possible (if we ignored any rotated versions of what is shown above).

A: 10/12 ÷ 10

5/6 multipled by 2/2 will give you 10/12

5 x 2 = 10

6 x 2 = 12.

Of course you would keep the other numbers in the same places (10 in the numerator, 12 in the denominator.