It is possible to arrange eight of nine numbers 2,3,4,5,7,10,11,12,13 in the vacant squares of the 3 by 4 array shown below so that the arithmetic average of the numbers in each row and column is the same integer.

Which one of the nine numbers must be left out when completing the array?

Let the average of the number of each row and column be integer denoted by $$x$$

Then the sum of numbers in Row 1 = $$4x$$.

Similarly for row 2 and row 3 the sum will be $$4x$$ each. Hence the sum of all the numbers in the table is $$12x$$.

Let $$N$$ be the number only number that is left out of the table.

Then, (1+2+3+4+5+7+9+10+11+12+13+14+15 -N) = $$12x$$ 

$$106-N\ =\ 12x$$

Since $$x$$ is integer then $$Remainder\left(\ \frac{\ 106-N}{12}\right)=0$$

$$Remainder\left(\ \frac{\ 10-N}{12}\right)=0$$

Hence N has to be 10 which is left out. And average of each row and column will be $$\ \frac{\ 106-10}{12}=\ \frac{\ 96}{12}=8$$

Upon solving the table will look as follows