Fill in the five blank cells of the given $3 \times 3$ table with the numbers from 3 to 8, without repeating them, so that the sums of the four numbers written in the four $2 \times 2$ subtables are always the same.

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-031.jpg?height=165&width=209&top_left_y=1688&top_left_x=1666)
Let $a, b, c, d, e$ and $f$ be the numbers that need to be filled in the table, that is, consider the table

| 1 | $a$ | 2 |
| :--- | :--- | :--- |
| $b$ | 9 | $c$ |
| $d$ | $e$ | $f$ |

In the four $2 \times 2$ subtables we can form, namely,

| 1 | $a$ |
| :--- | :--- |
| $b$ | 9 |


| $a$ | 2 |
| :--- | :--- |
| 9 | $c$ |


| $b$ | 9 |
| :--- | :--- |
| $d$ | $e$ | and | 9 | $c$ |
| :--- | :--- |
| $e$ | $f$ |

we should have

$$
\left\{\begin{array} { l } 
{ 1 + a + b + 9 = a + 2 + 9 + c } \\
{ 1 + a + b + 9 = b + 9 + d + e } \\
{ a + 2 + 9 + c = 9 + c + e + f }
\end{array} \quad \text { or, equivalently, } \left\{\begin{array}{l}
b-1=c \\
a+1=d+e \\
a+2=e+f
\end{array}\right.\right.
$$

Subtracting the second equality obtained from the third, we get $1=f-1$, or $f=1+d$. Our table, then, is given as follows.

| 1 | $a$ | 2 |
| :---: | :---: | :---: |
| $b$ | 9 | $b-1$ |
| $d$ | $e$ | $d+1$ |

For the numbers $a, b, c, d, e$ and $f$, we only have the options $3,4,5,6,7$ and 8, without repetition. If $a=3,4$ or 5, then $d+e=a+1=4,5$ or 6, which is impossible for distinct integers $d$ and $e$ greater than 2. If $a=7$, then $d+e=a+1=8$ and we could have $(d, e)=(3,5)$ or $(5,3)$. If $d=3$, then $e=5, f=d+1=4$ and, necessarily, $b=6$ or 8: however, this is impossible, as it would imply $c=b-1=5=e$ or $c=b-1=7=a$.

Thus, $a$ cannot be $3,4,5$ or 6, leaving only the options $a=6$ or $a=8$. Using the relations $a+1=d+e$ and $c=b-1$, we obtain the two unique options for filling the given table, as follows.

| 1 | 6 | 2 |
| :--- | :--- | :--- |
| 8 | 9 | 7 |
| 4 | 3 | 5 |

and

| 1 | 8 | 2 |
| :--- | :--- | :--- |
| 5 | 9 | 4 |
| 6 | 3 | 7 |