Biggest integer obtained by deleting $100$ digits from $1234 \ldots 9899100$ Consider the following number $$x=1234 \ldots 9899100$$
Which obtained by listing the integers from $1$ through $100$. What is the largest integer we can get by deleting $100$ digits from $x$?
Let $y$ be this number. First there are $9+90×2+3=192$ digits in $x$ which means there are $92$ digits in $y$. Since $y$ is the largest, we want it to start by as many $9$'s as possible then $8$'s and so on.
There are $18$ nines in $x$ and $18$ eights... So $y$ would look like that $$y=9...98...87...7...5...544$$ It consists of $18$ digit of every number $\ge 4$ and since $y$ contains $92$ digits we're left to choose two more digits which happen to be $44$. Am I right? Of course here I'm not deleting $100$ digits instead I'm choosing $92$ digits from $x$. But the two methods are equivalent.
 A: Since you aren't allowed to rearrange the digits as pointed out in the comments, our best bet is to get as many $9$s in the beginning as we can.
So we first remove the digits 1 through 8, that's removing total 8 digits.
Then we remove digits 10, 11, and so on up to 18 and an additional 1, that's removing 19 digits.
We again remove digits in blocks of 19 like 202122...27282, 303132...37383, 404142...47484.
In total we have thus removed $8+19\times4=84$ digits, since we're to remove only 100 digits, we can't remove 19 more digits. In order to increase the first digit of the remaining number as much as possible, we see how many digits we have to remove in order to get what digit in the front:
get 9: remove 19 digits (as above)
get 8: remove 17 digits (505152...56575)
get 7: remove 15 digits (505152...55565)
so we can afford to get 7 in the front by removing 15 digits, we remove those digits and then we remove the succeeding 5 to get the final answer to be 9,9,9,9,9,7,8,59,60,61,62,63,64,...,98,99,100.
