Six real numbers so that product of any five is the sixth one












2












$begingroup$


enter image description here



I believe this is not so hard problem but I got no clue to proceed. The work I did till now.



Say the numbers be $a,b,c,d,e, abcde$. Then, $b.c.d.e.abcde=a$ hence $bcde= +-1$.



Basically I couldn't even find a single occasion where such things occcur except all of these numbers being either $1$ or $-1$. Are these all cases?






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    They could all be $0$ (unless the image, which I can't access right now, has a rule forbidding that).
    $endgroup$
    – J.G.
    16 mins ago
















2












$begingroup$


enter image description here



I believe this is not so hard problem but I got no clue to proceed. The work I did till now.



Say the numbers be $a,b,c,d,e, abcde$. Then, $b.c.d.e.abcde=a$ hence $bcde= +-1$.



Basically I couldn't even find a single occasion where such things occcur except all of these numbers being either $1$ or $-1$. Are these all cases?






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    They could all be $0$ (unless the image, which I can't access right now, has a rule forbidding that).
    $endgroup$
    – J.G.
    16 mins ago














2












2








2


2



$begingroup$


enter image description here



I believe this is not so hard problem but I got no clue to proceed. The work I did till now.



Say the numbers be $a,b,c,d,e, abcde$. Then, $b.c.d.e.abcde=a$ hence $bcde= +-1$.



Basically I couldn't even find a single occasion where such things occcur except all of these numbers being either $1$ or $-1$. Are these all cases?






combinatorics







share|cite|improve this question











$endgroup$




enter image description here



I believe this is not so hard problem but I got no clue to proceed. The work I did till now.



Say the numbers be $a,b,c,d,e, abcde$. Then, $b.c.d.e.abcde=a$ hence $bcde= +-1$.



Basically I couldn't even find a single occasion where such things occcur except all of these numbers being either $1$ or $-1$. Are these all cases?






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 39 mins ago









Vinyl_coat_jawa

3,14711233




3,14711233










asked 41 mins ago









ChakSayantanChakSayantan

15416




15416












  • $begingroup$
    They could all be $0$ (unless the image, which I can't access right now, has a rule forbidding that).
    $endgroup$
    – J.G.
    16 mins ago


















  • $begingroup$
    They could all be $0$ (unless the image, which I can't access right now, has a rule forbidding that).
    $endgroup$
    – J.G.
    16 mins ago
















$begingroup$
They could all be $0$ (unless the image, which I can't access right now, has a rule forbidding that).
$endgroup$
– J.G.
16 mins ago




$begingroup$
They could all be $0$ (unless the image, which I can't access right now, has a rule forbidding that).
$endgroup$
– J.G.
16 mins ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

Well, if there are an even number of “1”s and “-1”s, then the property holds, so that’s $binom{6}{0}+ binom{6}{2}+ binom{6}{4}+ binom{6}{6}=32$. And then there’s the case that they’re all zero. Thus, there are 33 total cases.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I was in right direction then. These two last options just confused me. I was thinking there could be much more elements.
    $endgroup$
    – ChakSayantan
    7 mins ago



















1












$begingroup$

Here's an argument which extends to $2n$ real numbers quite nicely! We can start by noting that
$$x_1 (x_2x_3x_4x_5x_6) = x_1^2$$
And by commutativity, we get that in fact all the magnitudes are equal.



Now if $L$ is the magnitude, we also must have $L = L^5$, and from this we can conclude that $L=0$ or $1$.



Now we just have to go through the possibilities. If $L = 0$, then we have all $0$'s



If $L = 1$, then we have all $-1$'s and $1$'s. This configuration works iff the number of $-1$'s is even, as this would imply that
$$frac{prod_{i=1}^6 x_i}{1} = 1 text{ and } frac{prod_{i=1}^6 x_i}{-1} = -1$$
Now we can count configurations. There will be
$$sum_{i=0}^3 binom{6}{2i} = 2^5$$
possibilities. And finally, we have $1+32 = 33$.






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%2f3134563%2fsix-real-numbers-so-that-product-of-any-five-is-the-sixth-one%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Well, if there are an even number of “1”s and “-1”s, then the property holds, so that’s $binom{6}{0}+ binom{6}{2}+ binom{6}{4}+ binom{6}{6}=32$. And then there’s the case that they’re all zero. Thus, there are 33 total cases.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I was in right direction then. These two last options just confused me. I was thinking there could be much more elements.
      $endgroup$
      – ChakSayantan
      7 mins ago
















    3












    $begingroup$

    Well, if there are an even number of “1”s and “-1”s, then the property holds, so that’s $binom{6}{0}+ binom{6}{2}+ binom{6}{4}+ binom{6}{6}=32$. And then there’s the case that they’re all zero. Thus, there are 33 total cases.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I was in right direction then. These two last options just confused me. I was thinking there could be much more elements.
      $endgroup$
      – ChakSayantan
      7 mins ago














    3












    3








    3





    $begingroup$

    Well, if there are an even number of “1”s and “-1”s, then the property holds, so that’s $binom{6}{0}+ binom{6}{2}+ binom{6}{4}+ binom{6}{6}=32$. And then there’s the case that they’re all zero. Thus, there are 33 total cases.






    share|cite|improve this answer









    $endgroup$



    Well, if there are an even number of “1”s and “-1”s, then the property holds, so that’s $binom{6}{0}+ binom{6}{2}+ binom{6}{4}+ binom{6}{6}=32$. And then there’s the case that they’re all zero. Thus, there are 33 total cases.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 20 mins ago









    Zachary HunterZachary Hunter

    682111




    682111












    • $begingroup$
      I was in right direction then. These two last options just confused me. I was thinking there could be much more elements.
      $endgroup$
      – ChakSayantan
      7 mins ago


















    • $begingroup$
      I was in right direction then. These two last options just confused me. I was thinking there could be much more elements.
      $endgroup$
      – ChakSayantan
      7 mins ago
















    $begingroup$
    I was in right direction then. These two last options just confused me. I was thinking there could be much more elements.
    $endgroup$
    – ChakSayantan
    7 mins ago




    $begingroup$
    I was in right direction then. These two last options just confused me. I was thinking there could be much more elements.
    $endgroup$
    – ChakSayantan
    7 mins ago











    1












    $begingroup$

    Here's an argument which extends to $2n$ real numbers quite nicely! We can start by noting that
    $$x_1 (x_2x_3x_4x_5x_6) = x_1^2$$
    And by commutativity, we get that in fact all the magnitudes are equal.



    Now if $L$ is the magnitude, we also must have $L = L^5$, and from this we can conclude that $L=0$ or $1$.



    Now we just have to go through the possibilities. If $L = 0$, then we have all $0$'s



    If $L = 1$, then we have all $-1$'s and $1$'s. This configuration works iff the number of $-1$'s is even, as this would imply that
    $$frac{prod_{i=1}^6 x_i}{1} = 1 text{ and } frac{prod_{i=1}^6 x_i}{-1} = -1$$
    Now we can count configurations. There will be
    $$sum_{i=0}^3 binom{6}{2i} = 2^5$$
    possibilities. And finally, we have $1+32 = 33$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Here's an argument which extends to $2n$ real numbers quite nicely! We can start by noting that
      $$x_1 (x_2x_3x_4x_5x_6) = x_1^2$$
      And by commutativity, we get that in fact all the magnitudes are equal.



      Now if $L$ is the magnitude, we also must have $L = L^5$, and from this we can conclude that $L=0$ or $1$.



      Now we just have to go through the possibilities. If $L = 0$, then we have all $0$'s



      If $L = 1$, then we have all $-1$'s and $1$'s. This configuration works iff the number of $-1$'s is even, as this would imply that
      $$frac{prod_{i=1}^6 x_i}{1} = 1 text{ and } frac{prod_{i=1}^6 x_i}{-1} = -1$$
      Now we can count configurations. There will be
      $$sum_{i=0}^3 binom{6}{2i} = 2^5$$
      possibilities. And finally, we have $1+32 = 33$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Here's an argument which extends to $2n$ real numbers quite nicely! We can start by noting that
        $$x_1 (x_2x_3x_4x_5x_6) = x_1^2$$
        And by commutativity, we get that in fact all the magnitudes are equal.



        Now if $L$ is the magnitude, we also must have $L = L^5$, and from this we can conclude that $L=0$ or $1$.



        Now we just have to go through the possibilities. If $L = 0$, then we have all $0$'s



        If $L = 1$, then we have all $-1$'s and $1$'s. This configuration works iff the number of $-1$'s is even, as this would imply that
        $$frac{prod_{i=1}^6 x_i}{1} = 1 text{ and } frac{prod_{i=1}^6 x_i}{-1} = -1$$
        Now we can count configurations. There will be
        $$sum_{i=0}^3 binom{6}{2i} = 2^5$$
        possibilities. And finally, we have $1+32 = 33$.






        share|cite|improve this answer









        $endgroup$



        Here's an argument which extends to $2n$ real numbers quite nicely! We can start by noting that
        $$x_1 (x_2x_3x_4x_5x_6) = x_1^2$$
        And by commutativity, we get that in fact all the magnitudes are equal.



        Now if $L$ is the magnitude, we also must have $L = L^5$, and from this we can conclude that $L=0$ or $1$.



        Now we just have to go through the possibilities. If $L = 0$, then we have all $0$'s



        If $L = 1$, then we have all $-1$'s and $1$'s. This configuration works iff the number of $-1$'s is even, as this would imply that
        $$frac{prod_{i=1}^6 x_i}{1} = 1 text{ and } frac{prod_{i=1}^6 x_i}{-1} = -1$$
        Now we can count configurations. There will be
        $$sum_{i=0}^3 binom{6}{2i} = 2^5$$
        possibilities. And finally, we have $1+32 = 33$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 15 mins ago









        Isaac BrowneIsaac Browne

        4,67731132




        4,67731132






























            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%2f3134563%2fsix-real-numbers-so-that-product-of-any-five-is-the-sixth-one%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

            Olav Thon

            Image
            Olav Thon Fødd 29. juni 1923 Ål Yrke forretningsdrivande Olav Thon på Commons Olav Thon (fødd 29. juni 1923) er ein norsk, stort sett sjølvlært forretningsmann, fødd i Ål i Hallingdal, som har gjort store pengar på særleg kjøp, bygging og utvikling av fast eigedom, mellom anna kjøpesentre, hotell og restaurantar, i både Noreg og somme andre land. Han vart Ridder av 1. klasse av St. Olavs Orden i 2003, og er som ivrig friluftsmann æresmedlem i Den Norske Turistforening etter å ha ytt store økonomiske bidrag til foreininga sine hytter og ruter i fjellet. Han er kjend for alltid å gå med ei karakteristisk strikkelue. Olav Thon i samtale med Trond Helleland på sentralbanksjefens årstale i februar 2016 I ung alder reiste Olav Thon til Newcastle for å lære seg engelsk, og han var seinare ei tid i Sveits, der han gjorde forretningar i import og eksport. Deretter tok han til å kjøpe faste eigedomar i Noreg med stor økonomisk suksess, verksemda voks snøgt og v...
            Read more

            Waikiki

            Image
            Denne artikkelen kan ha godt av ein språkvask , som reinskar opp målføringa og/eller innfører same språkstilen overalt. Denne artikkelen kan ha godt av ei opprydding for å nå ein høgare standard og/eller for å verta i tråd med standardoppsettet. Sjå korleis du redigerer ei side og stilmanualen for hjelp. Waikiki Land   USA Stat Hawaii Stad Honolulu Wikimedia Commons: Waikiki Utsyn over Waikiki Waikīkī er ein bydel i Honolulu, sør på øya Oahu, som er ei av Hawaii-øyane. Waikīkīstranda er ein av dei meste berømte strendene i verda. Waikīkī strekkjer seg langs stillehavskysten og har ei rekkje velrenommerte hotell. Bydelen er Hawaii-turismens høyborg. På Waikīkī Beach står ei statue av den lokale sportsstjerna Duka Kahanamoku. Kjelder | Denne artikkelen bygger på «Waikiki» frå Wikipedia på bokmål, den 4. november 2018. Denne artikkelen manglar kjelder eller referansar . Hjelp Wikipedia med å finna truverdige kjelder...
            Read more

            Tårekanal

            Image
            Tåreapparatet ( a ) Tårekjertel ( b ) Øvre tårepunkt ( c ) Øvre tårekanalikkel ( d ) Tåresekk ( e ) Nedre tårepunkt ( f ) Nedre tårekanalikkel ( g ) Tårekanal Tårekanalen eller nase-tårekanalen (av latin ductus nasolacrimalis ) er ein ductus som frakter overflødige tårer frå tåresekken til nasehola. Opninga inn mot nedre nasegang er delvis tildekt av ei ufullstendig slimhinnefold kalla Hasners ventil. Overflødige tårer strøymer gjennom tårekanalen og trekker inn i nedre nasegang. Dette er grunnen til at ein kan få rennande nase når ein gret, og at ein somme gonger kan kjenne saltsmaken av tårer i munnen. Både tårekanalen og tåresekken er tekt med fleirlaga sylinderepitel som inneheld slimproduserande begerceller, og er begge omgjevne av bindevev. Dakryostenose (tårekanalstenose) er ei innsnevring av tårekanalen som fører til epifora, ei overflod av tårer. Medfødd dakryostenose kan føra til cystisk utviding av tårekanalen ved at det dannast ein dakryocystocele. Personar m...
            Read more