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Natural Problem (Posted on 2003-08-29) Difficulty: 3 of 5
The Natural Numbers are written successively as shown below:
12345678910111213141516..........., such that the 4th digit is '4' the 9th digit is '9' but the 11th digit is '0', the 15th digit '2', the 17th '3', and so on.
What is the 40,000th digit that appears in this list ?

  Submitted by Ravi Raja    
Rating: 2.0000 (3 votes)
Solution: (Hide)
The digit in question is 1.

First, we need to realize that the first 9 one-digit numbrs provide 9 digits. 180 digits come from the 90 two-digits numbers, 2700 digits come from the 900 three-digit numbers, and so on. The set of n-digit numbers will provide 9n×10^n digits to the list.

Based on this, we get:

Digits 1-9 come from the 1-digit numbers.
Digits 10-189 come from the 2-digit numbers.
Digits 190-2889 come from the 3-digit numbers.
Digits 2890-38889 come from the 4-digit numbers.
Digits 38890-488889 come from the 5-digit numbers.

So, the 40000th digit must be the 1111th digit of the sequence of 5 digit numbers.

We also know that 1111=222×5+1, so the 1111th digit of this set is the first digit of the 223rd 5-digit number. 10222 is the 223rd 5 digit number, which means that 1 is the 40000th digit of the overall sequence.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionPuzzle SolutionK Sengupta2007-11-29 04:45:20
spreadsheetAxorion2004-03-13 18:18:48
re: Solutionabc2003-09-15 18:55:27
Solutiondennis wenne2003-09-15 12:31:01
SolutionLawrence2003-08-30 00:39:45
re(2): Agreed (with code)Brian Wainscott2003-08-29 14:55:54
re: AgreedSilverKnight2003-08-29 11:36:04
AgreedBrian Wainscott2003-08-29 11:11:01
The solution againretiarius2003-08-29 09:25:07
SolutionSolutionfwaff2003-08-29 08:37:23
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