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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 ?

See The Solution Submitted by Ravi Raja    
Rating: 2.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
The solution again | Comment 2 of 10 |
I agree.

A number with 40,000 digits is made from 9 x 1-digit, 90 x 2-digit, 900 x 3-digit, 9000 x 4-digit, N x 5-digit numbers. That equals 39889 + N x 5-digit numbers. The first 22 x 5-digit numbers give the last 110 digits of a 39999-digit number. The next digit to arrive is 1.

  Posted by retiarius on 2003-08-29 09:25:07
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