Hi

http://i301.photobucket.com/albums/n...5/Noether8.jpg

I have absolutely no idea where to start.

Please help!

Thanks, BG

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- Jun 23rd 2009, 07:09 AMBG5965Finding the specific digit of a large number
Hi

http://i301.photobucket.com/albums/n...5/Noether8.jpg

I have absolutely no idea where to start.

Please help!

Thanks, BG - Jun 23rd 2009, 09:56 AMTheAbstractionist
There are 9 1-digit numbers, 90 2-digit numbers, 900 3-digit numbers and 9000 4-digit numbers. Hence the last digit of the last 4-digit number is in position $\displaystyle 9+2\times90+3\times900+4\times9000=38889.$

The last digit of the last 5-digit number will be in position $\displaystyle 38889+5\times90000=488889$ which is greater than 206787. Hence the number we are looking for is a 5-digit number. The position of the last digit of the 5-digit number $\displaystyle N$ is $\displaystyle 5(N-9999)+38889=5N-11106.$ Now find the largest $\displaystyle N$ such that $\displaystyle 5N-11106\le206787.$

$\displaystyle 5N-11106\,\le\,206787$

$\displaystyle \implies\ 5N\,\le\,217893$

$\displaystyle \implies\ N\,\le\,43578.6$

So that number is $\displaystyle N=43578,$ whose last digit is in position 206784. The next 5-digit number is 43579 and the 206787th digit in the sequence is the 3rd digit of this number, which is 5.