## Task A-2.5.

Fill in the blanks in the table with numbers so that in each row, column, and diagonal, the number in the middle is the arithmetic mean of the other two numbers. Explain!

|  | 8 |  |
| :--- | :--- | :--- |
| 11 |  |  |
|  |  | 29 |
## Solution.

Let $a$ be the number in the upper left corner, $b$ the number in the upper right corner, and $c$ the number in the lower left corner.

| $a$ | 8 | $b$ |
| :---: | :---: | :---: |
| 11 |  |  |
| $c$ |  | 29 |

From the top row, we have $8=\frac{a+b}{2}$, or $a+b=16$.

From the left column, we have $11=\frac{a+c}{2}$, or $a+c=22$.

Since the arithmetic mean of the first and third numbers in both diagonals is equal to the number in the middle, the sums of the first and third numbers on both diagonals are also equal, i.e., $a+29=b+c$.

We have a system of 3 equations with 3 unknowns. By substituting, for example, $b=16-a$ and $c=22-a$ from the first two equations into the third equation, we get

$$
a+29=16-a+22-a
$$

which simplifies to $3a=38-29=9$, or $a=3$.

From this, we get $b=16-3=13$ and $c=22-3=19$.

It remains to fill in the remaining values in the table. The number in the middle is the arithmetic mean of the numbers 3 and 29 (or 13 and 19), so it is 16.

1 point

The remaining two numbers are 21 and 24. The final solution:

| 3 | 8 | 13 |
| :---: | :---: | :---: |
| 11 | $\underline{16}$ | $\underline{21}$ |
| 19 | $\underline{24}$ | 29 |