## Task 2 - 080512

Place the multiples of the number 3 from 3 to 27 in the individual fields of the square so that the sums of each row (horizontally), each column (vertically), and each diagonal (from top left to bottom right and from top right to bottom left) are equal!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0040.jpg?height=257&width=246&top_left_y=934&top_left_x=1473)
The sum of the multiples of 3 up to 27 is: $3+6+9+12+15+18+21+24+27=135$.

Since the square has 3 rows and columns, the sum along a row, column, or diagonal must be equal to $\frac{135}{3}=45$.

By placing the middle number 15 of the multiples of 3 in the center of the square, one possible solution can be found through trial and error:

| 18 | 3 | 24 |
| :---: | :---: | :---: |
| 21 | 15 | 9 |
| 6 | 27 | 12 |