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Given these squares of numbers (maybe you can even solve it with just one!):

 9   6   2
 8   x   7
 7   7   1

 11   9   7
  7   x   6
 12   4   3

What is the value of x?

While the accepted answer is correct, the reasoning is actually a bit more involved:

All calculations are in the field $\mathbb{F}_{13}$ or $\mathbb{Z}/13\mathbb{Z}$.
The first two numbers in each row sum to the third.
The first two numbers in each column multiply to the third

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I think value of x is :

12 for both puzzle

Reasoning(for first puzzle) :

For first puzzle :

9   6   2
 8   x   7
 7   7   1
In first row in we need to add first two columns number and then subtract 13 to get the number in last column .i.e:
 for first row : 9+6=15-13=2
 for third row : 7+7=14-13=1
 for second row : 8+x-13=7 ( let ? be x )
                   -5+x=7 => x=7+5=> x=12

Reasoning(for second puzzle) :

For second puzzle :

 11   9   7
  7   x   6
 12   4   3
In first row in we need to add first two columns number and then subtract 13 to get the number in last column .i.e:
 for first row : 11+9=20-13=7
 for third row : 12+4=16-13=3
 for second row : 7+x-13=6 ( let ? be x )
                   -6+x=6 => x=6+6 => x=12

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I'm not 100% sure here but my pattern seems to work. The value of x is

1 in both squares.

Explanation :

in the following square :
a b c
d e f
g h i
we have a=h+c and g=b+i.
notice we are adding numbers on the 2nd and 3rd columns to get the number in the first column.
Since d, e and f were the only unused number, we can assume d=e+f
Therefore, 8=x+7 and 7=x+6

I might have entirely missed the pattern and found a wrong answer, but this is the best I could come up with.

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