## Task 3 - 070933

Imagine the natural numbers from 1 to 100, written in ascending order.

The digits written in this way are combined in the same order to form the sequence of the number

$$
1234567891011121314 \ldots 979899100
$$

From this sequence, exactly 100 digits are to be removed so that the remaining digits, in the same order, form the largest possible number.

What is this number?
First, we observe that the result always has the same number of digits, regardless of which digits are struck out.

To ensure that the resulting number starts with at least five nines, one must definitely strike out the numbers

$$
1,2, \ldots, 8,10,11, \ldots, 18,20,21, \ldots, 28,30, \ldots, 38,40,41, \ldots, 48
$$

and the digit 4 from 49. This amounts to $8+4 \cdot 19=84$ digits.

It is not possible for the result to start with six nines, because for that, one would also have to strike out the numbers $50,51, \ldots, 58$ and the digit 5 from 59, which would already be $84+19>100$ digits. It is also not possible for the result to start with the digits 999998, because for that, one would additionally have to strike out $50,51, \ldots, 57$ and the 5 from 58, which would total $84+17>100$ digits.

By additionally striking out the numbers $50,51, \ldots, 56$ and the digit 5 from 57, one obtains a number that starts with 999997 and has so far struck out $84+15=99$ digits.

If one then also strikes out the digit 5 from 58, one has struck out 100 digits and obtains

$$
\epsilon:=999997859606162 \ldots 99100
$$

as the result.

If one does not strike out the digit 5 from 58, one gets a number that starts with 9999975, which is smaller than $\epsilon$.

Therefore, $\epsilon$ is the largest possible number.