Numbers in a Row

$999999=9+(10^2-10)\times2+(10^3-10^2)\times3+(10^4-10^3)\times4+(10^5-10^4)\times5+(1.85185\times10^5-10^5)\times6$ with 185185 is the last number, so $\boxed{1}$ is $1000000^{th}$ of this sequence.

Break this up into numbers by digit

There are 9 numbers with 1 digit. (1-9) -> 9 total digits

There are 90 numbers with 2 digits (10-99) -> 189 total digits

There are 900 numbers with 3 digits (100-999) -> 2889 total digits

There are 9000 numbers with 4 digits (1000-9999) -> 38889 total digits

There are 90000 numbers with 5 digits (10000-99999) -> 488889 total digits

Now we have to work with 6 digit numbers. There are 511111 digits remaining for us to use. Divide this by 6 to see how many numbers we have left to look at (85185) and then the first digit of the next number.

99999 (last number we accounted for) + 85185 = 185184

We need to look at the first digit of the next number (185185), giving us the answer 1.