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






active

oldest

votes


















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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045289%2ffor-what-values-of-t-20t-equiv15t-pmod-0-5%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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












      $begingroup$

      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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045289%2ffor-what-values-of-t-20t-equiv15t-pmod-0-5%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
            Read more

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

            Image
            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
            Read more