8. (10 points) Five children $A, B, C, D,$ and $E$ participate in the "Happy Read Pinyin" competition. When they go on stage, the five stand in a row. They each have a contestant number on their chest, and the sum of the 5 numbers is 35. It is known that the sum of the numbers of the contestants standing to the right of $E, D, A,$ and $C$ are $13, 31, 21,$ and 7, respectively. What is the sum of the numbers of contestants $A, C,$ and $E$? $\qquad$ .
【Analysis】Since "the sum of the numbers of the players standing to the right of $E$, $D$, $A$, and $C$ are $13$, $31$, $21$, and $7$ respectively," the further left a child stands, the more people are to their right, and thus the larger the sum of the numbers. Since $31>21>13>7$, the order of the four children $E$, $D$, $A$, and $C$ from left to right is $D$, $A$, $E$, $C$. The sum of the numbers to the right of $C$ is 7, indicating that there is one more child $B$ to the right of $C$. Therefore, the five children from left to right are $D$, $A$, $E$, $C$, $B$. The sum of the numbers to the right of $D$ is 31, so $D$ is $35-31=4$. The sum of the numbers to the right of $A$ is 21, so $A$ is $35-21-4=10$. The sum of the numbers to the right of $E$ is 13, so $E$ is $35-13-4-10=8$. The sum of the numbers to the right of $C$ is 7, so $C$ is $35-7-4-10-8=6$. The sum of the numbers to the right of $C$ is 7, so $B$ is 7. Therefore, the sum of the numbers of the three players $A$, $C$, and $E$ is $10+8+6=24$. The answer can be derived accordingly.

【Solution】According to the analysis:
The position with a larger sum of numbers to the right is further to the left.
From the problem, we know:
$E$, $D$, $A$, $C$, from left to right, are $D$, $A$, $E$, $C$.
The number of the player to the right of $C$ is 7, which can only be $B$.
The number of the player at the far right, $D$, is:
$$
35-31=4
$$

So:
$$
A+C+E=35-(7+4)=24
$$

The answer is: 24.