8. (4 points) With the number written on the board, the following operation is allowed: erase two adjacent digits whose sum does not exceed 9, and write this sum in their place. Initially, a 100-digit number $21162116 \ldots 2116$ was written. The operation was performed on the number on the board until it was no longer possible. What is the largest number that could end up on the board as a result?
Answer:

$46 \ldots 46$ - a 50-digit number.

Solution:

The sum of the digits of the original number is $(2+1+1+6) \cdot 25=250$. The sum of the digits of the final number is the same.

We will divide the digits of the final number into pairs: the first with the second, the third with the fourth, and so on. The last digit may remain without a pair. The sum of the digits in each pair is at least ten, so there cannot be more than 50 digits in the number. Since 50-digit numbers exceed all numbers with fewer digits, we should first look for the answer among them.

If the number has exactly 50 digits, and the sum of the digits in any pair is greater than 10, then the total sum of the digits is greater than 250. Therefore, in each 50-digit number that we can obtain, the sum of the digits in each pair is exactly 10.

Note: The statement that in each such number the sum of adjacent digits is 10 is generally incorrect: for example, the number $374646 \ldots 46$ can be obtained, in which this condition is violated.

In each pair, we need to maximize the first digit and minimize the second. Since each pair of digits is obtained by the operations specified in the condition from 2116, the second digit is not less than 6, and the largest variant for each pair is 46, from which we get the answer.