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The traffic lights at three different road crossings change after every 48 sec, 72 sec, and 108 sec respectively. If they all change simultaneously at 8:20:00hours, then at what time will they change again simultaneously?

okay seriously how would someone even guess we have to do this by LCM GCD method ?

nban
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    Well, the first one changes at every multiple of $48$, the second at every multiple of $72$, and the third at every multiple of $108$...thus you want the least (positive) number which is a multiple of all three...which is the definition of the LCM. – lulu Sep 02 '17 at 22:46
  • Why least number ? – nban Sep 02 '17 at 22:48
  • Well, I took the question to be asking for the first time (after 8:20) that they all changed simultaneously. Doesn't matter much...if you want all the times then you still need to find the LCM (and the answer will then be all multiples of the LCM). – lulu Sep 02 '17 at 22:50
  • That is what I am not understanding, why the concept of LCM ? – nban Sep 02 '17 at 22:52
  • What part of my comment did you not understand? Can you see where the lights change simultaneously precisely at the common multiples of $48,72,108$? – lulu Sep 02 '17 at 22:55
  • because that's what's needed to solve it ... –  Sep 02 '17 at 22:59
  • Maybe the point you aren't getting is that all common multiples are multiples of the LCM. That is true, but it isn't obvious. It's not that hard to prove it but for a first pass at the subject it might be as well to just accept it. – lulu Sep 02 '17 at 23:00
  • @lulu For reference see the OP's prior question, where the similar questions are posed. – Bill Dubuque Sep 02 '17 at 23:03
  • @BillDubuque Ah, thanks. – lulu Sep 02 '17 at 23:06

3 Answers3

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It seems you need some help in understanding how $\,\rm lcm\,$ plays a role in such problems. Let's count time in seconds, relative to the starting time 8:20 when they all changed. Then light $\#1$ changes every $48$ seconds, i.e. at times $48n = 48, 96,\ldots =$ all multiples of $48,\,$ which are listed in the first row in the table below. $ $ Similarly for lights $\#2$ and $\#3$ in the $2$nd and $3$rd rows.

$$\begin{array}{|r|r|rrrrrrrr|} \hline \#1 & 48n & 48 \!&\! \!&\! 96 \!&\! \!&\! \color{#c00}{144} \!&\! 192 \!&\! \!&\! 240 \!&\! \color{#c00}{288} \!&\! \!&\! 336 \!&\! \!&\! 384 \!&\! \color{#0a0}{432} \!&\! \ldots\, \\ \hline \#2 & 72n & \!&\! 72 \!&\! \!&\! \!&\! \color{#c00}{144} \!&\! \!&\! \color{#c00}{216} \!&\! \!&\! \color{#c00}{288} \!&\! \!&\! \!&\! 360 \!&\! \!&\! \color{#0a0}{432} \!&\! \ldots\,\\ \hline \#3 & 108n \!&\! \!&\! \!&\! \!&\! 108 \!&\! \!&\! \!&\! \color{#c00}{216} \!&\! \!&\! \!&\! 324 \!&\! \!&\! \!&\! \!&\! \color{#0a0}{432} \!&\! \ldots\,\\ \hline \end{array}$$

The times when $2$ of $3$ lights change are shown in $\rm\color{#c00}{red},$ and $\rm\color{#0a0}{green}$ shows the time when all $3$ change. By construction, the number $\color{#0a0}{432}$ is a common multiple of $48, 72,108.\,$ Since no prior common multiple of all three occurs, it is their least common multiple $(\rm lcm).$

Recall, as I explained in your prior question, the $\rm lcm$ is characterized by the universal property

$${ 48,72,108\mid m\iff \overbrace{{\rm lcm}(48,72,108)}^{\large \color{#0a0}{432}}\mid m}$$

This says that the times $m$ when all $3$ change $ $ (i.e. when $m$ is a common multiple of $48,72,108),\,$ are equivalent to the times when $m$ is a multiple of their $\,{\rm lcm} = \color{#0a0}{432}.\,$ Interpreted in terms of the above table, this is true because the displayed pattern repeats if we consider the times $\!\!\pmod{\!\color{#0a0}{432}}$

Bill Dubuque
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here's an expansion on lulu's writing:

  1. 8:20:00 all three are changing
  2. 8:20:48 light 1 changes 48 seconds
  3. 8:21:12 light 2 changes 72 seconds
  4. 8:21:36 light 1 changes 96 seconds
  5. 8:21:48 light 3 changes 108 seconds
  6. 8:22:24 light 1 and light 2 change 144 seconds
  7. 8:23:12 light 1 changes 192 seconds
  8. 8:23:36 light 2 and light 3 change 216 seconds
  9. 8:24:00 light 1 changes 240 seconds
  10. 8:24:48 light 1 and light 2 change 288 seconds
  11. 8:25:24 light 3 changes 324 seconds
  12. 8:25:36 light 1 changes 336 seconds
  13. 8:26:00 light 2 changes 360 seconds
  14. 8:26:24 light 1 changes 384 seconds
  15. ??? all three change ??? seconds

this took me very much more time than before bill's first comment to work out by mind when at least with computer I can compute the final time fairly fast with LCM so maybe they'd think to use it because it took them so much time otherwise.

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    It would probably help the OP if you showed the time in seconds relative to the starting time, so the common multiples can be seen more clearly. – Bill Dubuque Sep 02 '17 at 23:37
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"okay seriously how would someone even guess we have to do this by LCM GCD method ?"

If the first light blinks ever $48$ seconds then any time it blinks will be at a mmultiple of $48$. So the answer when the all blink will have to be a multiple of $48$ because those are the only times the first one blinks.

The second only blinks at a multiple of $72$ seconds. So the answer has to be a common multiple of both $48$ and $72$.

And so on. The answer has to be a common multiple of all $48$, $72$, and $108$. What is a common multiple of $48$, $72$, and $108$.

Well... notice the first light will blink at: $48, 96, 144, 192, 240, 288, 336, 384, 432, 480,......$

The second light at: $72, 144, 216, 288, 360, 432, 504, 576, 648, 720 ....$

The third at: $108, 216, 324, 432, 540, 648, 756, 864, 972, 1080 ....$

What number do they all have in common? Well.... I can see it is $432$.

Dang. That was tedious. Maybe there is an easier way to figure out what is the least common mulitple of $48, 72$ and $108$.

fleablood
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