10. The distance between Port A and Port B does not exceed 5000 kilometers. Today, three ships depart from Port A to Port B at midnight simultaneously. Assuming the three ships sail at a constant speed for 24 hours a day, the first ship arrives at midnight several days later, the second ship arrives at 18:00 several days after that, and the third ship arrives at 8:00 a few days after the second ship. If the first ship travels 300 kilometers per day, the second ship travels 240 kilometers per day, and the third ship travels 180 kilometers per day, what is the actual distance between Port A and Port B in kilometers, and how long did each of the three ships travel?
10. Solution: Let the distance between ports A and B be $x$ kilometers. The distance the second ship travels in 18 hours is $240 \times \frac{18}{24}=180$ kilometers, and the distance the third ship travels in 8 hours is $180 \times \frac{8}{24}=60$ kilometers. According to the problem, we have
$$\left\{\begin{array}{l}
x \equiv 0(\bmod 300) \\
x \equiv 180(\bmod 240) \\
x \equiv 60(\bmod 180)
\end{array}\right.$$

Since
$$\begin{array}{l}
(300,240)=60, \text { and } 60 \mid(180-0) ; \\
(300,180)=60, \text { and } 60 \mid(60-0) ; \\
(240,180)=60, \text { and } 60 \mid(180-60) ;
\end{array}$$

the system of congruences has a solution. Because
$$\begin{aligned}
300 & =2^{2} \times 3 \times 5^{2} \\
240 & =2^{4} \times 3 \times 5 \\
180 & =2^{2} \times 3^{2} \times 5
\end{aligned}$$

the original system of congruences is equivalent to
$$\left\{\begin{array}{l}
x \equiv 0 \quad\left(\bmod 5^{2}\right) \\
x \equiv 180 \equiv 4 \quad\left(\bmod 2^{4}\right) \\
x \equiv 60 \equiv 6 \quad\left(\bmod 3^{2}\right)
\end{array}\right.$$

We solve this using the Chinese Remainder Theorem. Here,
$$\begin{array}{c}
b_{1}=0, \quad b_{2}=4, \quad b_{3}=6 \\
m_{1}=5^{2}, \quad m_{2}=2^{4}, \quad m_{3}=3^{2} \\
m=5^{2} \times 2^{4} \times 3^{2}=3600 \\
M_{1}=\frac{3600}{5^{2}}=144, \quad M_{2}=\frac{3600}{2^{4}}=225 \\
M_{3}=\frac{3600}{9}=400
\end{array}$$

From $144 M_{1}^{\prime} \equiv 1\left(\bmod 5^{2}\right)$, i.e., $19 M_{1}^{\prime} \equiv 1\left(\bmod 5^{2}\right)$, we get $M_{1}^{\prime}=4$. From $225 M_{2}^{\prime} \equiv 1\left(\bmod 2^{4}\right)$, i.e., $M_{2}^{\prime} \equiv 1\left(\bmod 2^{4}\right)$, we get $M_{2}^{\prime}=1$. From $400 M_{3}^{\prime} \equiv 1\left(\bmod 3^{2}\right)$, i.e., $4 M_{3}^{\prime} \equiv 1(\bmod 9)$, we get $M_{3}^{\prime}=-2$. Therefore,
$$\begin{aligned}
x & \equiv 4 \times 225 \times 1+6 \times 400 \times(-2) \\
& \equiv-3900 \equiv 3300(\bmod 3600)
\end{aligned}$$

Since the distance between ports A and B does not exceed 5000 kilometers, the actual distance is 3300 kilometers.

Also,
$$\begin{array}{l}
\frac{3300}{300}=11 \\
\frac{3300}{240}=13 \frac{18}{24} \\
\frac{3300}{180}=18 \frac{6}{18}
\end{array}$$

Answer: The distance between ports A and B is 3300 kilometers. The first ship takes 11 days, the second ship takes 13 days and 18 hours, and the third ship takes 18 days and 8 hours.