The traffic lights at three different road crossings change after 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 am, at what time will they change simultaneously again.

Hint: Find the Lcm of 48, 72 and 108. Let l be the lcm of 48,72 and 108. Think about what will happen exactly after l seconds. Draw a conclusion from the observation. Hence find the time taken for the lights to change simultaneously and hence the time at which the lights will change simultaneously.

Complete step-by-step answer:

Let the lights change simultaneously exactly x seconds after 7 am.
Since the first light changes every 48 seconds, we have 48|x.
Similarly, 72|x and 108|x.
Hence we have l|x, where l is the Lowest common multiple of 48,72 and 108.
Also, precisely after l seconds, the lights will change simultaneously.
Hence the smallest value of x is l.
Hence the lights will change simultaneously, l seconds after 7 am.
Now we have
$\begin{align}
  & 48={{2}^{4}}\times 3 \\
 & 72={{2}^{3}}\times {{3}^{2}} \\
 & 108={{2}^{2}}\times {{3}^{3}} \\
\end{align}$
Hence LCM (48,72,108) $={{2}^{4}}\times {{3}^{3}}=432$
Hence we have l = 432.
Hence the lights will change simultaneously after 432 seconds = 7 minutes and 12 seconds after 7 am.
Hence the lights will change simultaneously at 07:07:12 am(HH:MM: SS).

Note: In the above question we have used the property that if L is the LCM of a and b, then for any integer x such that a|x and b|x, we have l|x.
Proof:
Let l not divide x and let x leave non-zero remainder r on division by l.
So, we have
\[x=ql+r,0Now since a|x and a|l, we have a|r.
Similarly b|r.
Hence there exists r, such that r>0 and rHence l|x.
Hence proved.