# For what values of t, $20t\equiv15t \pmod {0.5}$?

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.

• The track length is measured in km, and their speeds in mph. Is that intentional? – Arthur Dec 18 '18 at 15:34
• @Arthur Yup, I just checked. Can't we just convert 4Km to 4000m? Their speeds are in meters per second. I accidentally wrote hour – Ryder Rude Dec 18 '18 at 15:37
• 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. – Arthur Dec 18 '18 at 15:56
• @RyderRude $20t \equiv 10t( \mod{500})$ is easier to look at right? – 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.