## C3 ALB

Positive integers are put into the following table

| 1 | 3 | 6 | 10 | 15 | 21 | 28 | 36 |  |  |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 |  |  |
| 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 |  |  |
| 7 | 12 | 18 | 25 | 33 | 42 |  |  |  |  |
| 11 | 17 | 24 | 32 | 41 |  |  |  |  |  |
| 16 | 23 |  |  |  |  |  |  |  |  |
| $\ldots$ |  |  |  |  |  |  |  |  |  |
| $\ldots$ |  |  |  |  |  |  |  |  |  |

Find the number of the line and column where the number 2015 stays.


## Solution 1:

We shall observe straights lines as on the next picture. We can call these lines diagonals.

| 1 | $\sqrt{3}$ | 6 | 10 | 15 | 21 | 28 | 36 |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 |  |
| 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 |  |
|  | 12 | 18 | 25 | 33 | 42 |  |  |  |
| 11 | 17 | 24 | 32 | 41 |  |  |  |  |

![](https://cdn.mathpix.com/cropped/2024_06_05_07a1b610f48bd7c35addg-16.jpg?height=152&width=799&top_left_y=301&top_left_x=327)

On the first diagonal is number 1 .

On the second diagonal are two numbers: 2 and 3 .

On the 3rd diagonal are three numbers: 4,5 and 6

.

On the $n$-th diagonal are $n$ numbers. These numbers are greater then $\frac{(n-1) n}{2}$ and not greater than $\frac{n(n+1)}{2}$ (see the next sentence!).

On the first $n$ diagonals are $1+2+3+\ldots+n=\frac{n(n+1)}{2}$ numbers.

If $m$ is in the $k$-th row $l$-th column and on the $n$-th diagonal, then it is $m=\frac{(n-1) n}{2}+l$ and $n+1=k+l$. So, $m=\frac{(k+l-2)(k+l-1)}{2}+l$.

We have to find such numbers $n, k$ and $l$ for which:

$$
\begin{gathered}
\frac{(n-1) n}{2}<2015 \leq \frac{n(n+1)}{2} \\
n+1=k+l \\
2015=\frac{(k+l-2)(k+l-1)}{2}+l
\end{gathered}
$$

(1), (2), (3) $\Rightarrow n^{2}-n<4030 \leq n^{2}+n \Rightarrow n=63, k+l=64,2015=\frac{(64-2)(64-1)}{2}+l \Rightarrow$ $t=2015-31 \cdot 63=62, k=64-62=2$

Therefore 2015 is located in the second row and 62 -th column.

