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







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$endgroup$












  • $begingroup$
    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









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    2 Answers
    2






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    2 Answers
    2






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    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






























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