A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round 252 seconds, B in 308 seconds and C in 198 seconds all starting at the same point. After what time they will meet again at the starting point.

Hint: As it is asked that after what time will they meet again at the starting position and the times taken for each round of each A, B and C are given. So, you need to find the LCM of the three times to get the answer. To determine the LCM of the numbers, express the number in terms of the product of its prime factors and multiply all the prime factors the maximum number of times they occur in either number. This is the method of prime factorization.

Complete step-by-step answer:
As it is asked that after what time will they meet again at the starting position and the times taken for each round of each A, B and C are given. So, you need to find the LCM of the three times to get the answer. To further get the clarification, try to think that each person will be at the starting point in the multiples of their times taken to finish one round, and at the LCM time, the time will be the multiple of all of the three times taken to finish the round.
Let us find the prime factors each of the time using the prime factorisation method.
$\begin{align}
  \text{ }2\left| \!{\underline {\,
  252 \,}} \right. & \\
  2\left| \!{\underline {\,
  126 \,}} \right. & \\
  3\left| \!{\underline {\,
  063 \,}} \right. & \\
  3\left| \!{\underline {\,
  021 \,}} \right. & \\
  7\left| \!{\underline {\,
  007 \,}} \right. & \\
  \text{ 001} & \\
\end{align}$
\[\begin{align}
  & \text{ }2\left| \!{\underline {\,
  308 \,}} \right. \\
 & \text{ }2\left| \!{\underline {\,
  154 \,}} \right. \\
 & \text{ }7\left| \!{\underline {\,
  077 \,}} \right. \\
 & \text{ }11\left| \!{\underline {\,
  011 \,}} \right. \\
 & \text{ 001} \\
\end{align}\]

\[\begin{align}
  \text{ }2\left| \!{\underline {\,
  198 \,}} \right. & \\
  3\left| \!{\underline {\,
  099 \,}} \right. & \\
  3\left| \!{\underline {\,
  033 \,}} \right. & \\
  11\left| \!{\underline {\,
  011 \,}} \right. & \\
  \text{ 001} & \\
\end{align}\]

$252=2\times 2\times 3\times 3\times 7$
$308=2\times 2\times 7\times 11$
$198=2\times 3\times 3\times 11$
Now to find the LCM, we need to multiply all the prime factors the maximum number of times they occur in either number. So, LCM(252,308,198) is a product of two 2s, one 11, one 7 and two 3s.
$LCM\left( 252,308,198 \right)=2\times 2\times 3\times 3\times 11\times 7=2772$
Therefore, we can conclude that they will meet again at the starting point after 2772 seconds.

Note: Be careful while finding the prime factors of each number. Also, it is prescribed that you learn the division method of finding the LCM as well, as it might be helpful. The key to the above question is understanding that LCM of the time is to be found and the method is only valid if it is mentioned that they meet at the starting point.