Kamila's triangle of odd numbers

First,we may know the position of the number $2013$ .

If $x$ is the $n^{th}$ number,then $\frac{x+1}{2}=n$

$\frac{2013+1}{2}=\frac{2014}{2}=1007$

$2013$ is the $1007^{th}$ number in the triangle.

Assume that $a$ is the quantity of numbers in the row which the number $2013$ appear,

$\frac{a(a+1)}{2}\geq1007$

$a(a+1)\geq2014$

Now,find the minimum value of $a$ .

$45^{2}=2025$

This square number is the closest to $2014$ .

$44\times45=1980$

$45\times46=2070>2014$

The minimum value of $a$ is $45$ .Hence, the answer is $45$ (the $45^{th}$ row).

The row in which any number in the triangle lies on is the square of that number rounded to the nearest whole number.

$\sqrt{2013}=44.866...\approx 45$

This can then be checked by a formula which describs the minimum and maximum numbers of each row ( $r$ ) of the triangle.

$min_{r}=1+2(r-1)+\sum_{i=1}^{r-2} 2i$

$max_{r}=1+2(r-1)+\sum_{i=1}^{r-1} 2i$

If $r=45$ then:

$2013 ≥ 1+2(45-1)+\sum_{i=1}^{45-2} 2i$

$2013 ≥ 1981$

And $2013 ≤ 1+2(45-1)+\sum_{i=1}^{45-1} 2i$

$2013 ≤ 2069$

Hence the row in which the number 2013 lies on is $\boxed{45}$ .

2013 is the 1007th odd number. Solve for n which satisfies $\frac{n-1}{2}$ (n)<1007< $\frac{n}{2}$ (n+1) and can get n= $\boxed{45}$

We can easily see that in the nth row, the last digit is n*n + n - 1. So at the 45th row, the first digit is 1937, while the last digit is 2025. Hence, 2013 lies in the 45th row.

As odd numbers, all elements of the triangle can be written as $2n-1$ where $n$ is a positive integer. Hence, the $n$ corresponding to the element $2013$ is $1007$ . Kamilla's triangle can then be written with elements $1,2,3,4,...$ with the last entry of the rows being $1,3,6,10,...,\frac{k(k+1)}{2})$ where $k$ is also an integer. Thus, we wish to find $k$ such that $\frac{(k-1)(k)}{2} \leq 1007 \leq \frac{k(k+1)}{2}$ .

$\Rightarrow \frac{(k-1)(k)}{2} \leq 1007 \leq \frac{k(k+1)}{2} \\ \Rightarrow (k-1)(k) \leq 2014 \leq (k)(k+1) \\ \Rightarrow (k-1)(k)+\frac{1}{4} \leq 2014+\frac{1}{4} \leq (k)(k+1)+\frac{1}{4} \\ \Rightarrow \frac{(2k-1)^2}{4} \leq \frac{8057}{4} \leq \frac{(2k+1)^2}{4} \\ \Rightarrow (2k-1)^2 \leq 8057 \leq (2k+1)^2$

* $k=45$ *

based on the given illustration,i've created this formula ; A=x(x-1)+1; where "A" is the number to be define which row is it located. and "x" is the row to which the certain number is located. thus,since we're searching which row the number "2013" is located. our "A" now is "2013"; hence, 2013=x(x-1)+1; i let x2 be x to the power of two since i don't know how to write a superscript. :D 0=x2-x-2012; 0=(x-45.86)(x+44.36); and we should use the principal value and not the extraneous one. so our x is equal to 45.86; thus the number "2013" should be located in the 45th row. :D

The last number on every line can be represented by the expression $n^2 + n - 1$ . So we continue to find such numbers until we find a number $n$ for which the expression evaluates to a number $>= 2013$ , which happens to be $\boxed{45}$

2013 is the 1007th odd number. Each row has one more number than the previous one.

Therefore, we find up to which row there are 1007 numbers or more, in total.

$\sum_{i=1}^n{i} = \frac{n^{2}+n}{2} >= 1007$

The smallest integer solution of this inequation is n=45, where n is the number of the row that contains 2013.

So the answer is $\boxed{45}$