Find the $n^{\rm th}$ digit in the sequence $123456789101112\dots$ Basically, the question asks us to find the nth digit in the following sequence:
$$12345678910111213\dots9899100101\dots$$
where the 10th digit is $1$, the 11th digit is $0$, etc.
EDIT: Here are my workings:
I was thinking about defining intervals in some way, for example, I know that there are 9 digits in this interval $0\dots9$ and $10\dots99$, and later try to find a sum. However, I am stuck on this point.
 A: Consider a function $g(n)$, which is defined as:
$$g(n)=\sum_{1\leqslant k \leqslant n} 9 \times 10^{k-1} \times k  = \frac{ 9(n+1)10^n-10^{n+1}+1} {9} \qquad k,n \in \mathbb{Z^+}$$
You will probably see that the values can be calculated quite easily since there is a pattern:
$$g(1)=9$$
$$g(2)=189$$
$$g(3)=2889$$
$$g(4)=38889$$
Now, given we want to find the $n$th digit we have to first solve $p$:
$$ p=10^{\lceil a \rceil} -1 - \left\lfloor \frac{g( \lceil a \rceil) - g(a)}{\lceil a \rceil} \right\rfloor,  g(a) = n \qquad a \in \mathbb{R^+}$$
This, $p$, will give us the number that contains the $n$th digit. So, in order to find the $n$th digit, calculate:
$$r = g(\lceil a \rceil ) - g(a) \mod  \lceil a \rceil $$ The $r$ gives you the index of the $n$th digit in the number $p$.
$$p = (a_r\dots a_1a_0)$$
Reference:
Los., Artem. (2014). Finding the nth digit in a sequence of positive integers placed in a row in ascending order.. Available: https://myows.com/protects/copyright/67407_mathexploration-pdf. Last accessed 3rd Jan 2014.
