Circular track problem with LCM

Question

I am not able to understand this question based on circular tracks:

$A$ and $B$ start together from the same point on a circular track, and walk in the same direction until they both again arrive together at the starting point. $A$ completes one round in $224$ seconds and $B$ in $364$ seconds. How many times $A$ might have passed $B$?

Firstly I took the LCM of $224$ and $364$ in order to find the meeting point: $2^5*7*13$. Then I divided this by $224$ to get the number of rounds taken by $A$ as $13$. Similarly I divided the LCM by $364$ to get the number of rounds taken by $B$ as $8$. So I think $A$ and $B$ must meet $8$ times, but the answer is $5$ i.e. difference between $13$ and $8$. Why? How can they meet just $5$ times? Can anyone explain?

Answer 1

A does 5 rounds more than B, so A will pass B 5 times

Answer 2

They meet whenever the difference between their individual total distances traveled, measured in revolutions, is an integer.

The difference of their distances starts at $0$, is strictly increasing, and ends at $5$, hence they meet when the difference is one of the numbers $1,2,3,4,5$.

Answer 3

Lets make a table, of laps completed, and the floor of laps ahead (times when they must have passed):

$\begin{array}{|c|c|c|} \hline A & B& \lfloor{\text {laps ahead}}\rfloor \\ \hline 1 & \frac{8}{13}& 0\\ \hline 2 & 1\frac{3}{13}&0\\ \hline 3 & 1\frac{11}{13}&1\\ \hline 4 & 2\frac{6}{13}&1\\ \hline 5 & 3\frac{1}{13}&1\\ \hline 6 & 3\frac{9}{13}&2\\ \hline 7 & 4\frac{4}{13}&2\\ \hline 8 & 4\frac{12}{13}&3\\ \hline 9 & 5\frac{7}{13}&3\\ \hline 10 & 6\frac{2}{13}&3\\ \hline 11 & 6\frac{10}{13}&4\\ \hline 12 & 7\frac{5}{13}&4\\ \hline 13 & 8 & 5\\ \hline \end{array}$

So, we can show, that the the number of laps ahead, hits 5 full laps when A did 13 laps, and B did 8 laps.