10. (10 points) On a circular track, 2015 flags are evenly spaced. A and B start from a certain flag at the same time and in the same direction. When A and B return to the starting point at the same time again, A has run 23 laps, and B has run 13 laps. Excluding the starting flag position, how many times does A catch up to B exactly at a flag position?
【Analysis】Assume the distance between every two flags is 1, i.e., the track circumference is 2015. Since the time is constant, the speed ratio equals the lap ratio (i.e., the distance ratio). Because $v_{\text {甲: }} v_{\text {乙 }}=23: 13$, let $v_{\text {甲 }}=23 x, v_{\text {乙 }}=13 x$. For 甲 to catch up with 乙, 甲 needs to run $n$ more laps, $(23 x-13 x) t=2015 n, 10 x \times t=2015 n$, i.e., the time 甲 spends to catch up with 乙 is $t=\frac{403 \mathrm{n}}{2 \mathrm{x}}$. Then, the distance 甲 runs when catching up with 乙 is $23 x \times \frac{403 \mathrm{n}}{2 \mathrm{x}}=\frac{23 \times 403}{2} \mathrm{n}$. To catch up exactly at the flag position, the distance run must be an integer, i.e., $n$ must be even. Then, based on $n=2,4,6,8,10$ (a maximum of 10 more laps), solve accordingly.

【Solution】Solution: Assume the distance between every two flags is 1, i.e., the track circumference is 2015.
Since $v_{\text {甲: }} v_{\text {乙 }}=23: 13$,
let $v_{\text {甲 }}=23 x, v_{\text {乙 }}=13 x$. For 甲 to catch up with 乙, 甲 needs to run $n$ more laps, and the time 甲 spends to catch up with 乙 is $t$,
then $(23 x-13 x) t=2015 n$
$$
\begin{aligned}
10 x \times t= & 2015 n \\
t & =\frac{403 \mathrm{n}}{2 \mathrm{x}},
\end{aligned}
$$

then the distance 甲 runs when catching up with 乙 is:
$$
23 x \times \frac{403 \mathrm{n}}{2 \mathrm{x}}=\frac{23 \times 403}{2} \mathrm{n} \text {, }
$$

to catch up exactly at the flag position, the distance run must be an integer, i.e., $n$ must be even,
so $n=2,4,6,8,10$ (a maximum of $23-13=10$ more laps);
i.e., 甲 catches up with 乙 5 times when 甲 runs 2, 4, 6, 8, 10 more laps, exactly at the flag position. In summary, 甲 catches up with 乙 5 times exactly at the flag position.

Answer: 甲 catches up with 乙 5 times exactly at the flag position.