The cells of a $3 \times 3$ grid are to be filled with integers so that the average value of the entries along each row, each column, and each diagonal is the same. The integers 10, 64 and 70 are entered, as shown. When the remaining six squares are filled in to complete the grid, what integer replaces $x$ ?
(A) 78
(B) 82
(C) 86
(D) 90
(E) 94

| 64 | 10 |  |
| :--- | :--- | :--- |
| 70 |  |  |
|  | $x$ |  |
Suppose that the integer in the bottom left corner is $n$.

In this case, the sum of the integers in the first column is $64+70+n$ or $n+134$.

Thus, the sum of the integers in each row, in each column, and on each diagonal also equals $n+134$. (This means that this square is in fact a magic square, since the sum of the numbers in each row, in each column, and on each diagonal is the same.)

Using the top row, the top right integer equals $(n+134)-64-10$ or $n+60$.

Using the northeast diagonal, the centre integer equals $(n+134)-n-(n+60)$ or $74-n$.

Using the second row, the middle integer in the right column equals $(n+134)-70-(74-n)$ or $2 n-10$.

Using the southeast diagonal, the bottom right integer equals $(n+134)-64-(74-n)$ or $2 n-4$.

Using the third row, the middle integer is $x=(n+134)-n-(2 n-4)=138-2 n$.

| 64 | 10 | $n+60$ |
| :---: | :---: | :---: |
| 70 | $74-n$ | $2 n-10$ |
| $n$ | $138-2 n$ | $2 n-4$ |

Using the third column,

$$
\begin{aligned}
(n+60)+(2 n-10)+(2 n-4) & =n+134 \\
5 n+46 & =n+134 \\
4 n & =88 \\
n & =22
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-033.jpg?height=179&width=1456&top_left_y=908&top_left_x=278)

ANSWER: (E)