For what values of t, $20tequiv15t pmod {0.5}$?












0












$begingroup$


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, $20tequiv15t pmod{0.5}$, where $t$ is time elapsed. However, I have no idea how to solve these type of equations.






modular-arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















0












$begingroup$


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, $20tequiv15t pmod{0.5}$, where $t$ is time elapsed. However, I have no idea how to solve these type of equations.






modular-arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$


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, $20tequiv15t pmod{0.5}$, where $t$ is time elapsed. However, I have no idea how to solve these type of equations.






modular-arithmetic







share|cite|improve this question











$endgroup$




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, $20tequiv15t pmod{0.5}$, where $t$ is time elapsed. However, I have no idea how to solve these type of equations.






modular-arithmetic




modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 16:18









Brahadeesh

6,51442364




6,51442364










asked Dec 18 '18 at 15:29









Ryder RudeRyder Rude

447112




447112












  • $begingroup$
    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




















  • $begingroup$
    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


















$begingroup$
The track length is measured in km, and their speeds in mph. Is that intentional?
$endgroup$
– Arthur
Dec 18 '18 at 15:34




$begingroup$
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






$begingroup$
@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






$begingroup$
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






$begingroup$
@RyderRude $20t equiv 10t( mod{500})$ is easier to look at right?
$endgroup$
– 1.414212
Dec 18 '18 at 17:31












1 Answer
1






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3












$begingroup$

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.



$5tequiv 0 pmod {500}$



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.






share|cite|improve this answer









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    $begingroup$

    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.



    $5tequiv 0 pmod {500}$



    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.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      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.



      $5tequiv 0 pmod {500}$



      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.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        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.



        $5tequiv 0 pmod {500}$



        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.






        share|cite|improve this answer









        $endgroup$



        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.



        $5tequiv 0 pmod {500}$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 17:51









        fleabloodfleablood

        73.5k22891




        73.5k22891






























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