The seventy-seventh question: Divide a circle with a circumference of 24 into 24 equal segments. Choose 8 points from the 24 points such that the arc length between any two points is not equal to 3 or 8. How many different ways are there to select such a group of 8 points? Explain your reasoning.
The 77th problem,
Solution: Label these points in a clockwise direction as $1, 2, 3, \ldots, 24$, and arrange them in the following $3 \times 8$ table:
$$
\begin{array}{cccccccc}
1 & 4 & 7 & 10 & 13 & 16 & 19 & 22 \\
9 & 12 & 15 & 18 & 21 & 24 & 3 & 6 \\
17 & 20 & 23 & 2 & 5 & 8 & 11 & 14
\end{array}
$$

In the table, the arc length between two adjacent numbers in the same column is 8, and the arc length between two adjacent numbers in the same row is 3 (the first column and the eighth column are also adjacent, as are the first row and the third row). Therefore, in the table, no two adjacent numbers can be taken simultaneously, meaning that only one number can be taken from each column, and exactly one number must be taken.

Let the number of ways to take one number from each column of a $3 \times n$ number table, such that any two adjacent columns (including the $n$-th column and the first column) do not take numbers from the same row, be denoted as $x_{n}$.

Taking one number from the first column has 3 ways. After the first column is determined, the number taken from the second column cannot be in the same row as the first column, leaving only two different ways. Each subsequent column also has two ways, resulting in a total of $3 \times 2^{n-1}$ ways. However, all the ways where the numbers taken from the first column and the last column are in the same row do not meet the requirements. In this case, these two columns can be considered as one column, i.e., the number of ways for $n-1$ columns, so $x_{n} + x_{n-1} = 3 \times 2^{n-1}$. It is also easy to see that $x_{2} = 6$.

Rewriting $x_{n} + x_{n-1} = 3 \times 2^{n-1}$ as $(-1)^{n} x_{n} - (-1)^{n-1} x_{n-1} = -3 \times (-2)^{n-1}$. Let $y_{n} = (-1)^{n} x_{n}$, then $y_{n} - y_{n-1} = -3 \times (-2)^{n-1}$, and $y_{2} = 6$. It is easy to solve that $y_{n} = (-2)^{n} + 2$, thus $x_{n} = 2^{n} + 2 \times (-1)^{n}$.
Therefore, $x_{8} = 2^{8} + 2 \times (-1)^{8} = 258$, meaning the total number of ways that meet the requirements of the problem is 258.