# Find the $n^{\rm th}$ digit in the sequence $123456789101112\dots$ [duplicate]

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.

• Although it is OK to ask and answer your own question, I can't up-vote this, as you should at least show an effort to answer in the question. +1 for the answer though. Commented Jan 3, 2014 at 19:19

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 $p$.
$$p = (a_r\dots a_1a_0)$$