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 ?

There are 10 numbers with only 1 digit. followed by 90 numbers with two digits. Thus it follows:
10 (1)
90 (2)
900 (3)
9000 (4)
90.000 (5) etc
10*1+90*2+900*3+9000*4= 38,890 (<40,000)
Thus we have to deal with 5 digit numbers to answer the question. WE need 40,000-38,890=110 more digits. Since the numbers contain 5 digits from now on, we have will have to look at the last number in the 110/5=22 th-5digit number.
Thus 10,022... will give us a 2.

Comments??

Posted by dennis wenne on 2003-09-15 12:31:01