# Problem 8. (2 points)

Can the natural numbers from 1 to 48 (each used exactly once) be arranged in a $6 \times 8$ rectangular table so that in every $1 \times 3$ rectangle (vertical or horizontal) the sum of the numbers is even?
Answer: Yes

## Solution:

To come up with an example, first note the following: numbers in cells that are two cells apart horizontally or vertically must have the same parity, because together with the numbers written in the two intermediate cells, they must sum to an even number. Thus, in each row and each column, the remainders of the numbers when divided by two repeat with a period of three.

Let's highlight a $3 \times 3$ rectangle in the upper left corner of our table and mark in each of its cells how many numbers must have the same remainder according to the previous reasoning.

| 6 | 6 | 4 | $\ldots$ |
| :---: | :---: | :---: | :---: |
| 6 | 6 | 4 | $\ldots$ |
| 6 | 6 | 4 | $\ldots$ |
| $\ldots$ | $\ldots$ | $\ldots$ | $\ldots$ |

Since there are 24 even and 24 odd numbers from 1 to 48, we need to divide the numbers obtained in the table into two groups, each with a total of 24. In each row and each column of our $3 \times 3$ table, there should be either 2 odd remainders or none. Based on these considerations, we will highlight in yellow the cells corresponding to odd numbers.

Accordingly, an example is constructed (we will not write the actual numbers, but only indicate their parity):

| O | O | E | O | O | E | O | O |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| O | O | E | O | O | E | O | O |
| E | E | E | E | E | E | E | E |
| O | O | E | O | O | E | O | O |
| O | O | E | O | O | E | O | O |
| E | E | E | E | E | E | E | E |