8. (10 points) There are 11 children standing in a circle playing a game, numbered clockwise as $1, 2, 3, \cdots 11$. The game rule is that starting from number 1, each child says a two-digit number, with the requirement that the sum of the digits of this number cannot be 6 or 9, and the subsequent child cannot repeat a number that has already been said. The game continues until someone cannot come up with a new number, at which point the game ends and the person who cannot say a new number is the loser. Who is the last one to be the loser of the game?
8. (10 points) There are 11 children playing a game in a circle, numbered clockwise as $1,2,3, \cdots 11$. The game rule is to start from No. 1, and each person says a two-digit number, with the requirement that the sum of the digits cannot be 6 or 9, and the next person cannot repeat a number that has already been said. The game continues until someone cannot say a new number, and that person is the loser of the game. So, the last $\qquad$ number is the loser of the game.

【Solution】Solution: According to the problem:
Numbers whose digit sum is 6 are: $15,24,33,42,51,60$, a total of 6 numbers.
Numbers whose digit sum is 9 are: $18,27,36,45,54,63,72,81,90$, a total of 9 numbers.
The numbers that meet the criteria are $90-6-9=75$ (numbers);
$75 \div 11=6 \cdots 9$.
After 6 rounds, there are only 9 numbers left that meet the criteria.
Therefore, the answer is: 10.