The question is: There is a race on a circular track of length $0.5$ km. The race is $4$ km. Two people start at the same point with speeds $20$ m/s and $10$ m/s respectively. Find the distance covered by the first person when they meet the second person for the third time.

My formulation of the question was, $20t\equiv15t \pmod{0.5}$, where $t$ is time elapsed. However, I have no idea how to solve these type of equations.

  • $\begingroup$ The track length is measured in km, and their speeds in mph. Is that intentional? $\endgroup$ – Arthur Dec 18 '18 at 15:34
  • $\begingroup$ @Arthur Yup, I just checked. Can't we just convert 4Km to 4000m? Their speeds are in meters per second. I accidentally wrote hour $\endgroup$ – Ryder Rude Dec 18 '18 at 15:37
  • $\begingroup$ m/s makes much more sense than miles per hour. That being said, I don't think much conversion is needed. The only really important numbers here are that one runner is twice as fast as the other, and the track is 500m long (which is a very odd length for a running track, now that I think about it; the standard is 400m). And 20m/s is insanely fast (twice the speed of any world class 100m dasher). Math problems aren't always thought through, I guess. $\endgroup$ – Arthur Dec 18 '18 at 15:56
  • $\begingroup$ @RyderRude $20t \equiv 10t( \mod{500})$ is easier to look at right? $\endgroup$ – 1.414212 Dec 18 '18 at 17:31

Once you correct $.5 km = 500m$

We have $20t \equiv 15t \pmod {500}$

So $5t \equiv 0 \pmod {500}$

When confused. Go back to something less confusing.

$5t\equiv 0 \pmod {500}$

so there is an integer $k$ so that

$5t = 500k$

$t = 100k$

$t$ being any multiple of $100$ seconds will do.

For example: After $100$ seconds, runner 1 will have ran $2000 m$ or or $2 km$ or $4$ laps. Will runner 2 will have run $1500 m$ or $1.5 km$ or $3$ laps.


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