Question 185: Consider every permutation of $1, 2, 3, 4, 5, 6, 7, 8$ as an eight-digit number. How many of these numbers are multiples of 11?
Question 185, Answer: For an eight-digit number $x=\overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} a_{7} a_{8}}$, if $11|x \Leftrightarrow 11|\left(a_{1}+\right.$ $\left.a_{3}+a_{5}+a_{7}\right)-\left(a_{2}+a_{4}+a_{6}+a_{8}\right)$. Let $A=\left\{a_{1}, a_{3}, a_{5}, a_{7}\right\}, B=\left\{a_{2}, a_{4}, a_{6}, a_{8}\right\}$, and denote $S(A)$ and $S(B)$ as the sums of the elements in the two sets, respectively. Then, $S(A)+S(B)=1+2+\ldots+8=36$, $|S(A)-S(B)| \leq(8+7+6+5)-(4+3+2+1)=16$, and $11 \mid S(A)-S(B)$, thus it is easy to know that $S(A)-S(B)=0$, i.e., $S(A)=S(B)=18$.

Considering the group where 8 is located, the other three numbers have only three cases: $\{1,2,7\}, \{1,3,6\}, \{1,4,5\}, \{2,3,4\}$. Once a set of numbers is determined, the other set of numbers is also determined. Considering $8 \in A$ or $8 \in B$ in two cases, we know that there are $2 \times 4 \times 4!\times 4!=4608$ numbers that satisfy the condition.