Number: 103 : Series 28382th Term 12345678910101213.....

find the 28383rd term of series: 1234567891011121314....

First observe that there are 9 1-digit numbers which take 9*1 = 9 digits, 90 2-digit numbers which take 90*2 = 180digits, 900 3-digit numbers which take 900*3 = 2700 digits, 9000 4-digit numbers which take 9000*4 = 36000 digits.
So it can be concluded easily that our required digit will be some digit in a 4-digit number. Now we need to find out that which digit of what 4-digit number, right?
That can be easily done if I make all numbers 4-digit ones by inserting some 0's before them. Like insert 3 0's before every 1-digit number (i.e. 3*9 = 27 0's), 2 0's before every 2-digit number (i.e. 2*90 = 180 0's) and 1 zero before every 3-digit number (i.e. 1*900 = 900 0's) to meke every number a 4-digit one.

Now we have inserted 27 + 180 + 900 = 1107 zeros in all so we need to find out 28383 + 1107 = 29490th digit in the following sequence: 0001 0002 0003 0004 0005 0006 0007 0008 0009 0010 0011 0012 .....

Which can be easily found as 2nd digit of 7373 i.e. 3 (because 29490 = 4*7372 + 2).