# Blue light/green light problem

I got this question out of the book "Rapid Calculations" by A.H. Russell ($1956$) and I want to see if my approaches are correct. (Forgive me if it's longer than usual.)

A blue light flashes 5 times, with an interval of 3 seconds, and then has a period of 23 seconds' darkness. A green light flashes 7 times, with an interval of 2 seconds, and then has a period of 15 seconds' darkness. If they both start together, at what time will they end their dark periods simultaneously?

The answer is 15 minutes and 45 seconds. Here are my approaches:

First, consider the flashes. The blue light flashes every 15 seconds, while the green light flashes every 14 seconds ($5 \times 3$ and $7 \times 2$). To find the time they flash together, find the least common multiple of 15 and 14, which is 210. Thus, there are 210 seconds for the two lights to flash together.

Second, consider the dark periods. Again, find the least common multiple of 23 and 15, which is 345. Thus there are 345 seconds for both lights to enter their dark period.

Third, consider the cycles. The blue light has a cycle of $5 \times 3 + 23$ or 38 seconds, so the cycle begins again in the 39th second. The green light has a cycle of $7 \times 2 + 15$ or 29 seconds, so the cycle begins again in the 30th second. They enter their complete cycles together in 390 seconds (the least common multiple of 39 and 30).

The time they enter their dark periods simultaneously is the sum of 210, 345 and 390, or 945 seconds. Thus it takes 15 minutes and 45 seconds for both to enter their dark periods.

Approach #2: Consider the cycles only. Three complete cycles is 1170 seconds. If we subtract the flashes (210 seconds) we have 960 seconds; however, since the green light flashes less and goes into darkness earlier, we subtract an additional 15 seconds to arrive at 945 seconds or 15 min 45 sec as before.

Approach #3: Consider their dark periods only. Three dark periods are 1035 seconds; however, the green light enters its dark period earlier, so we subtract 90 seconds (3 periods where the green light enters its new period) to arrive at 945 seconds or 15 min 45 sec as before.

Are any (or all) of these attempts correct, or is there one I'm missing?

EDIT: This was done for my own recreation. See Ross Milliken's simple (as in, "why didn't I think of that?") answer below.

• If I moved too fast away from these lights, I would only see yellow and red. Also, they would never flick simultaneously. Did I do something wrong? – Batominovski Jun 24 '17 at 15:53
• Christian Blatter - you are correct! I have rewritten the question as it was in the book. (The original question was mine and was likely hastily written!) I've also included the author of the book - he's a British author. – bjcolby15 Jun 24 '17 at 21:10

Presumably the flashes are instantaneous, so five flashes at three seconds take $12$ seconds. There are four gaps between the flashes, each lasting three seconds. In the spirit of the book, you should notice that seven flashes at two seconds also take $12$ seconds and are done. Both lights start a dark period at $12$ seconds.

Added with the changed question. The period of the blue light is $12$ seconds for the flashing (four gaps at three seconds) and $23$ off for $35$ seconds. The period of the green light is $27$ seconds. We want the first time these periods end together, which is $\operatorname{LCM}(35,27)=945$ seconds. We have no reason to consider LCMs of other parts of the periods.

• I cleaned up the question...the time does begin at zero seconds. – bjcolby15 Jun 24 '17 at 21:11
• That clears it up quite nicely. I was figuring everything piece by piece but never considered it your way - much, much simpler! Thank you very much! – bjcolby15 Jun 25 '17 at 0:10