## Task B-1.2.

Natural numbers are sequentially written to form a spiral of numbers as shown in the figure:

| 21 | 22 |  | 23 |  | 24 |  | 25 | 26 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  |  |  |  |  |  |  |  | $\vdots$ |
| 20 | 7 | $\rightarrow$ | 8 | $\rightarrow$ | 9 | $\rightarrow$ | 10 |  |
|  | $\uparrow$ |  |  |  |  |  |  |  |
| 19 | 6 |  | 1 | $\rightarrow$ | 2 |  | 11 |  |
|  | $\uparrow$ |  |  |  | $\downarrow$ |  |  |  |
| 18 | 5 | $\leftarrow$ | 4 | $\leftarrow$ | 3 | 12 |  |  |
| 17 | 16 |  | 15 |  | 14 | 13 |  |  |

The position of a number is determined relative to the number 1. For example, the number 24 is located 1 place to the right and 2 places up from the number 1. Determine the position of the number 2015 relative to the number 1.
## Solution.

Notice that the numbers on the diagonal to the right and up are squares of odd natural numbers $1, 9, 25, 49, \ldots$

![](https://cdn.mathpix.com/cropped/2024_05_30_1020492bf6e1d2c949fdg-02.jpg?height=494&width=866&top_left_y=410&top_left_x=595)

The number $9=(2 \cdot \mathbf{1}+1)^{2}$ and its position is 1 to the right and 1 up.

The number $25=(2 \cdot \mathbf{2}+1)^{2}$ and its position is 2 to the right and 2 up.

The number $49=(2 \cdot \mathbf{3}+1)^{2}$ and its position is 3 to the right and 3 up...

The square of the odd natural number closest to the number 2015 is $45^{2}=2025$.

Since $45=2 \cdot \mathbf{22}+1$, the position of the number 2025 is 22 places to the right and 22 places up.

Since $2025=2015+10$, the position of the number 2015 relative to the number 1 is $22-10=12$ places to the right and 22 places up.