Running Out of Digits, Level 3

Question

_{The challenge idea is credited to HelloWorld1337.}

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You initially have x of each digit from 0 to 9. This means you have x * 10 digits in total. This count for each digit is shown in the table below.

Digit	0	1	2	3	4	5	6	7	8	9
# Remaining	x	x	x	x	x	x	x	x	x	x

Now start counting by ones, from 1. Each time you say a number you must remove the digits required to make the number from your stockpile of digits. For example, after you have counted from 1 to 13, the above table now looks like:

As an example, if you counted from 1 to 13 and didn't skip any numbers, the above table will look like:

Digit	0	1	2	3	4	5	6	7	8	9
# Remaining	x-1	x-6	x-2	x-2	x-1	x-1	x-1	x-1	x-1	x-1

What is the largest number x you can count to without running out of the digits needed to form the number?

Answer 1

I believe the answer is:

199,981. Mmm. Not this. You can add more and still work: 199,990

Details:

Short answer: I wrote a script. I know, I know.
Roughly, though, 1 in 10 of each digit is a 1. Consider 199,990. Take the last 5 digits of 199,990 and it gives you ~99,995 ones. To this add the 99,990 ones for the hundred-thousands place and you get ~199,985.
At this value, each additional number is adding 1 additional 1 (in the hundred-thousands place), so you can increase/decrease until you get a second 1 somewhere else. So this suggests that for $x\in [199,981;199,990]$ you use exactly $x$ 1s to write out the numbers from 1 to $x$.
When you go to 199,980 you will have 1 one left over. When you get to 199,991, you will be 1 one short.

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Edit: Paul Panzer is correct. He should get the tick. Here's a justification:

If $x = k\times10^n-1$ then the number of ones needed is $kn10^{n-1}$ plus an additional $10^n$ if $k\geq2$.
This can be used to show that $1\times10^9<x<2\times10^9$.
With a bit of care, you can now add $2\times10^8$ and discover that $x$ is less than that. But you can add $1\times10^8$, so $x>1.1\times10^9$.
Continue with that process and you get 1,111,111,110 as the maximum $x$

Answer 2

I'm pretty sure it's:

1,111,111,110

No proof, though, but a few remarks and a suggestive plot:

the number of ones will never be smaller than that of any other digit
for x of the form 99...99 (k digits) the distribution of 1,2,3,... is even each having been used exactly k x 10^{k-1} times; zeros will have occurred 11...12 (k digits) times less often.
for x of the form 11...10 (k digits) the number of ones used is k x 11...11 (k-1 digits)