Question 37502: Beginning with 1, all the postive intgers are written successively, beginning 123456789101112. What digit appears in the 2006th position?
Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website! To count digits of the counting numbers, I started with 0
0,1,2,3,4,5,6,7,8,9 add up to 10 digits
10 -> 19 add 10 X 2 = 20 more digits
20 -> 29 add 20 more
30 -> 39 add 20 more
40 -> 49 add 20 more
- - - - -
90 -> 99 total of 20 X 9 = 180 digits (10 -> 99)
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so far, I have
10 (0 -> 9)
180 (10 -> 99)
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190 total digits
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but the sequence is 1234567...., not 01234567....
so the 190th digit is really the first digit of the
next number, which is 100
Now I count 3-digit numbers until I get to 2006 digits
There are 3 X 100 = 300 digits from 100 to 199
190 + 300 puts me at the 2 in 200
190 + 600 puts me at the 3 in 300
190 + 900 puts me at the 4 in 400
190 + 1200 puts me at the 5 in 500
190 + 1500 puts me at the 6 in 600
190 + 1800 puts me at the 7 in 700
If I add 300 more, I'll be up to 190 + 2100 = 2290
which is over the 2006th digit I'm looking for
I'm up to 190 + 1800 = 1990
If I add 3 more, I'm at the 7 in 701
I'm up to 1993
If I add 12 more past the 7 in 701,
I'm at the 7 in 705
701702703704705
Now I'm up to 2005 digits
The next one, the 2006th, is the 0 in 705
So 0 is the 2006th digit
unless I made a mistake
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