Predicting Blink Times in an Arithmetic Series

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SUMMARY

The discussion focuses on predicting the simultaneous blinking times of two lights using arithmetic series. Light 1 blinks at intervals defined by the formula t(n) = 14n - n², while Light 2 follows s(m) = 16m - m². Both lights blink together at 48 seconds, with the corresponding integers m = 4 and n = 6 providing a solution to the Diophantine equation 16m - m² = 14n - n². The conversation highlights the potential for using simple computer programs to find additional solutions.

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wwilliamson
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Need help with the following:

Imagine you have 2 lights and a timer. You start the timer, and the first light blinks at 13 seconds
and the second light blinks at 15 seconds. At 24 seconds the first light blinks again, and the second
light blinks again at 28 seconds. Following this pattern, what formula can I use to predict that both
lights will blink at the same time at 48 seconds?

The intervals at which the lights blink will always get faster over time and adhere a specific
arithmetic series as such:

15 + 13 + 11 + 9 = 48
13 + 11 + 9 + 7 + 5 + 3 = 48

Here is the problem played out for clarity:

Light 1 blinks at 13 seconds.
Light 2 blinks at 15 seconds.
Light 1 blinks at 24 seconds.
Light 2 blinks at 28 seconds.
Light 1 blinks at 33 seconds.
Light 2 blinks at 39 seconds.
Light 1 blinks at 40 seconds.
Light 1 blinks at 45 seconds.
Lights 1 and 2 blink at 48 seconds.
 
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Using the formula for the sum of an arithmetic sequence, the times light one flashes are:

13,24,33,40,45...

are t(n) = 14n - n2.

For light 2 the times are:

15,28,39,48,...

you get s(m) = 16m-m2.

What you are looking for is for t(n) = s(m) for some integers m and n:

16m-m2 = 14n - n2

This is a Diophantine equation. One solution happens when m = 4 and n = 6. I don't know much about the techniques for solving them. A simple computer program could easily check for solutions for small m and n. Or maybe even large ones.
 
Last edited:
Thank you! I will definitely look into that.
 

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