6. (10 points) The distance between $A$ and $B$ is 100 kilometers. Car A takes $m$ hours to travel from $A$ to $B$, and Car B takes $n$ hours to travel from $A$ to $B$, where $m, n$ are integers. Now, Car A starts from $A$ and Car B starts from $B$ at the same time, heading towards each other, and they meet at point $C$ after 5 hours. If Car A has already traveled half of its distance, then the distance from $C$ to $A$ is $\qquad$ kilometers.
【Analysis】According to car A starting from $A$ and car B starting from $B$ at the same time, moving towards each other, and meeting at point $C$ after 5 hours, we can get $\frac{100}{\frac{100}{m}+\frac{100}{\mathrm{n}}}=5$, then $\frac{1}{\mathrm{~m}}+\frac{1}{\mathrm{n}}=\frac{1}{5}$. Let $m=5+p, n=5+q$. That is, $\frac{1}{5+\mathrm{p}}+\frac{1}{5+\mathrm{q}}=\frac{1}{5}$, removing the denominators of the fractional equation, we get $5(5+p+5+q)=(5+p)(5+q)$, simplifying to $50+5(p+q)=25+5(p+q)+p q, p q$ $=25$. Since $m, n$ are integers, we can set the values of $m n$ and see if they meet the requirements to solve the problem.

【Solution】Solve: $\frac{100}{\frac{100}{m}+\frac{100}{\mathrm{n}}}=5$, then $\frac{1}{\mathrm{~m}}+\frac{1}{\mathrm{n}}=\frac{1}{5}$. Let $m=5+p, n=5+q$. Then:
$$
\begin{array}{l}
\frac{1}{5+p}+\frac{1}{5+q}=\frac{1}{5}, \\
5(5+p+5+q)=(5+p)(5+q), \\
50+5(p+q)=25+5(p+q)+p q, \\
p q=25,
\end{array}
$$
(1) $p=5, q=5, m=10, A C=5 \times \frac{1}{10} A B=\frac{1}{2} A B$; does not meet the requirements.
(2) $p=25, q=1, m=30, A C=5 \times \frac{1}{30} A B=\frac{1}{6} A B$, does not meet the requirements.
(3) $p=1, q=25, m=6, A C=5 \times \frac{1}{6} A B=\frac{5}{6} A B=\frac{500}{6}=83 \frac{1}{3}$ kilometers $>\frac{1}{2} A B$, meets the requirements. Therefore, the distance from $C$ to $A$ is $83 \frac{1}{3}$ kilometers. The answer is: $83 \frac{1}{3}$.