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Find the $2000^{th}$ digit of the series $1234567891011121314\cdots $

where $123456789\underbrace{1}_{10\text{th digit}}~~\underbrace{0}_{11\text{th digit}}~~\underbrace{1}_{12\text{th digit}}~~\underbrace{1}_{13\text{th digit}}~~\underbrace{1}_{14\text{th digit}}~~\underbrace{2}_{15\text{th digit}} \cdots $ and so on.

I really have no clue how to begin, as it is totally different from the other sequence and series problems ,thanks .

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    $\begingroup$ Clue how to begin: the sequence is naturally composed of a $1$-digit section, followed by a $2$-digit section, followed by a $3$-digit section, etc. Find out which section the 2000th digit belongs to. How can you be sure? $\endgroup$ – Erick Wong Mar 3 '15 at 20:02
  • $\begingroup$ ok, and how will i find that ... $\endgroup$ – R K Mar 3 '15 at 20:03
  • $\begingroup$ This is one of those problems about a sequence which starts with a particular pattern, and it is supposed to be assumed that the pattern will persist. Erick's interpretation is presumably the intended one, but personally I don't like problems like this. There are infinitely many possible continuations, and possibly a much more sophisticated algorithm could kick in at a later stage to generate the remainder of the sequence. $\endgroup$ – Geoff Robinson Mar 3 '15 at 20:12
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    $\begingroup$ @GeoffRobinson What exactly is it that you don't like about this? There may be infinitely many possible continuations, but there's one that is by far the most natural, and it's so much clearer to write it this way than it'd be to say 'the sequence formed by writing down the natural numbers in order and reading along successive digits'. That was the clearest complete definition I could come up with, and I still think it's a bit ambiguous. Mathematicians are not computers; we can recognize the 'obvious' pattern without needing it to be explicitly spelled out. $\endgroup$ – John Gowers Mar 3 '15 at 21:08
  • $\begingroup$ @Donkey_2009: It's just a matter of taste about this sort of problem, not this one in particular. I guess we are indeed genetically programmed to look for patterns, and it generally stands us in good stead, but also, as Mathematicians, we can sometimes recognise that patterns can be more subtle than our first instinct tells us. $\endgroup$ – Geoff Robinson Mar 3 '15 at 21:18
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The sequence you appear to have given is the concatenation of all natural numbers. As others have noted, there is no guarantee that this is the case, but if it is, the following solution should work.

There are 9 single digit numbers, 90 with 2 digits, 900 with 3 digits, and 9000 with 4 digits, and, generally, $9(10)^m$ total numbers with $m$ digits.

$9+90\cdot2+900\cdot3=2889$, so you are in the 3 digit 'realm'.

$9+90\cdot2+x\cdot3=2000$ gives $x=603.\overline{6}$, so you are looking for the $2^\text{nd}$ digit of the $604^\text{th}$ $3$ digit natural number. The first 3 digit number is $100$, therefor the $604^\text{th}$ 3 digit number is $100+603=703$.

The last digit of that number would be represented by $x=604$, and $x=603.\overline{6}$ is the middle digit, and $x=603.\overline{3}$ would be the first digit.

So the digit you are looking for is $0$.

It is in the sequence here: $$1\ 2\ 3\ 4\ 5\ldots \ 700\ \underset{604^\text{th}\text{ 3 digit natural number}}{701 \ 702\underbrace{703} 704\ 705\ 706\ldots} \ .$$


In general, the $N^\text{th}$ digit in that sequence can be found in the following way. Find the smallest $m$ such that $$N\leq\sum_{k=0}^{m}9k(10)^k.$$ If the above equation is true with equality, then you are looking for the last digit of the last natural number with $m$ digits. If the above equation an inequality, then solve for $x$: $$N=mx+\sum_{k=0}^{m-1}9k(10)^k.$$

The solution will be $$x=\left\lfloor x\right\rfloor+\frac{j}{m}.$$ And the digit you are looking for is the $j^\text{th}$ digit of the $\left\lceil x\right\rceil^\text{th}$ $m$ digit natural number.

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  • $\begingroup$ u concluded the 2000th digit is in 3 digit realm because ,$9+90*2<2000<9+90*2+900*3$ is that so... $\endgroup$ – R K Mar 3 '15 at 20:11
  • $\begingroup$ That's my reasoning, yes. $\endgroup$ – jdods Mar 3 '15 at 20:13
  • $\begingroup$ ok thanks , i get that realm trick $\endgroup$ – R K Mar 3 '15 at 20:15
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    $\begingroup$ It should be the 604th 3-digit number, so the second digit of $100+603=703$, which is actually also $0$. $\endgroup$ – String Mar 3 '15 at 20:35
  • $\begingroup$ I figured I was close enough to just call it since it was somewhere in the 700's. I'll fix the answer. Thanks! $\endgroup$ – jdods Mar 3 '15 at 23:11
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This is not a mathematical answer, but just for the record, the Python code:

length=0
n=0
while length<2000:
    n+=1
    length+=len(str(n))
    print "\t", n, "\t", length

printet a list ending with

700     1992
701     1995
702     1998
703     2001

so the digit $3$ of the number $703$ is the $2001^{th}$ digit. Thus the $2000^{th}$ digit is confirmed to be the second digit of $703$, which is indeed $0$.

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The above number is just a sequence of natural numbers.

Hence, from $1$ to $9$, there are $9$ numbers and $9$ digits; from $10$ to $99$, there are $90$ two-digit numbers and 180 digits (So, from 1 to $99$ there are $180 + 9 = 189$ digits) and from $100$ to $999$, there are $900$ three-digit numbers and $2700$ digits.

So, total number of digits from $1$ to $999$ = $9+180+2700 = 2889$

Starting from $1$, i.e. the $1^{st}$ digit (so $2$ is $2^{nd}$, $3$ is $3^{rd}$ and so on); we have to find the $2000^{th}$ digit of the given number

i.e. $2000-189 = 1811^{th}$ digit out of the three-digit numbers (Here I am removing all $1$-digit and $2$-digit numbers)

$1811 = (3)(603)+2$; which implies that our required digit is $2^{nd}$ digit of the $603^{rd}$ $3$-digit number, i.e. $2^{nd}$ digit of the $703^{rd}$ natural number from the starting, i.e. the $2^{nd}$ digit of $703$ i.e. $0$. Q.E.D.

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