Solving Problems After 55: "I Saw a Problem

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Discussion Overview

The discussion revolves around a problem involving the scheduling of trains that leave a station at different intervals. Participants explore methods to determine when the trains next leave together after their initial departure, considering both specific examples and general approaches to similar problems.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant describes a specific problem involving trains leaving every 15 and 20 minutes and attempts to find a general solution using sequences.
  • Another participant suggests the concept of Diophantine equations as a potential approach to the problem, breaking it down into prime factors.
  • A third participant proposes using the least common multiple (LCM) of the two intervals as a straightforward solution, calculating it to find that the trains leave together every 60 minutes.
  • A later reply expresses gratitude for the clarification provided regarding the least common multiple approach.

Areas of Agreement / Disagreement

Participants present multiple approaches to the problem, including sequences, Diophantine equations, and the least common multiple. There is no consensus on a single method, as different participants highlight various techniques without resolving which is superior.

Contextual Notes

Some assumptions about the nature of the problem and the methods used remain implicit, and there may be limitations in the generalizability of the approaches discussed.

CWatters
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I saw a problem the other day that belongs to a class of problems that I have forgotten how to solve (I'm 55). It went like this..

Trains leave a station every 15mins and every 20 mins. The first two leave together at 9.10am. Do any others leave at the same time and when is the next time?

The simple way to solve it is just to write down the two sequences...

15mins: 9.10 9.25 9.40 9.55 10.10 etc
20mins: 9.10 9.30 9.50 10.10 etc

and note that they match again at 10.10am

Clearly that works for this problem but there must be a better way. I just can't seem to figure out how to do the general case. For example you could assume there is a match at some time ΔT after 9.10 then write...

ΔT = n*15 = m*20

where n and m are unknown integers. But that's one equation with two unknowns.

I half remember that ΔT must be a multiple of |n-m| or something like that but I think that just adds another unknown integer to the mix.
 
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Are you thinking of Diophantine equations?

In this case, you can break it into prime factors:

3*5*n = 2*2*5*m

3*n = 4*m. Hence n=4 and m=3

Just a guess
 
The simplest way to deal with this is "least common multiple" of 15 and 20. 15= 3*5 and 20= 4*5 so the least common multiple is 3*4*5= 60 minutes. Two trains leave together 10 minutes after every hour.
 
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Thanks HallsofIvy. That's exactly what I was trying to remember.
 

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