8. (4 points) In an $8 \times 8$ table, integers from 0 to 10 are arranged (naturally, numbers can repeat, not all specified numbers necessarily appear). It is known that in each rectangle $3 \times 2$ or $2 \times 3$, the sum of the numbers is 10. Find the smallest possible value of the sum of the numbers in the entire table.

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# Answer: 105

## Solution:

Let's look at the left part of the figure below. Consider the rectangles $3 \times 2$ and $2 \times 3$, which contain the left upper $2 \times 2$ square. The remaining parts of these two rectangles are the leftmost upper rectangles $1 \times 2$ and $2 \times 1$, shaded in gray. Since the sum of the numbers in both rectangles $3 \times 2$ and $2 \times 3$ is 10, the sum of the numbers in these two gray rectangles is equal.

Similarly, it can be proven that the sums of the numbers in all gray rectangles in the picture, as well as in all white $2 \times 2$ squares, are equal. Let the sum of the numbers in each white square be $x$, then the sum of the numbers in each gray rectangle is $10-x$. Therefore, the total sum of the numbers in the table is at least $9x + 12(10-x) = 120 - 3x$. If $x > 5$, the result is a number greater than 105.

Thus, the sum of all numbers in the table is at least 105.

An example is provided in the right part of the figure, cells filled with zeros are left blank for convenience.

![](https://cdn.mathpix.com/cropped/2024_05_06_f3802874779b2cbb336bg-3.jpg?height=377&width=511&top_left_y=371&top_left_x=113)

|  |  | 5 |  |  | 5 |  |  |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
|  | 5 |  |  | 5 |  |  | 5 |
| 5 |  |  | 5 |  |  | 5 |  |
|  |  | 5 |  |  | 5 |  |  |
|  | 5 |  |  | 5 |  |  | 5 |
| 5 |  |  | 5 |  |  | 5 |  |
|  |  | 5 |  |  | 5 |  |  |
|  | 5 |  |  | 5 |  |  | 5 |