<s>[INST] Circular track problem with LCM 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?
 [/INST] 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. 
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