1. Place the numbers $1,2,3,4,5,6,7,8$ and 9 in the nine cells of the figure shown in the diagram, so that the sum of the numbers in each column, starting from the second, is 1 more than in the previous one. It is sufficient to find at least one such arrangement. In your answer, indicate the number in the first column.

![](https://cdn.mathpix.com/cropped/2024_05_06_edcb2754fbb412b0df86g-2.jpg?height=70&width=131&top_left_y=903&top_left_x=997)
Answer: 7.

Solution: For now, we will not pay attention to the order of numbers in one column.

The sum of the given numbers is 45. Let $x$ be the number in the bottom-left cell. Then $5x + 10 = 45$, from which $x = 7$. Therefore, the sum of the numbers in the second column is $8 = 5 + 3 = 6 + 2$. If the second column contains 3 and 5, then the third column must contain 1 and 8, the fourth column must contain 6 and 4, and the last column must contain 2 and 9. If the second column contains 6 and 2, then the third column can contain 1 and 8 or 4 and 5. It can be shown that if the third column contains 1 and 8, it is impossible to select numbers for the fourth column. Therefore, the third column must contain 4 and 5, then the fourth column must contain 1 and 9, and the last column must contain 3 and 8. This results in 2 arrangements without considering the order of the numbers.

Notice that in each column (except the first), the numbers can be swapped, which gives 16 options for each arrangement. In the end, we get 32 options considering the order of the numbers in the columns.