Series of binary operations in $(mathbb{N}, *)$












2












$begingroup$



Consider monoid $(mathbb{N},*)$, where operation $*$ is defined as $x*y = xy + x + y$.



What is the result of $1*2*3*text{...}*25 text{ (mod 29)}$ ?




$xy + x + y$ can be written as $(x + 1) y + x$ or $y + x (1 + y)$, but that does not seem to help in any way.



One thing I noticed is that $n * (n + 1) = text{prime number}$. Ex: $12*13=181$; $4*5=29$.



I am still not sure how can I express $1*2*3*text{...}*25$ in such a way that I could find the remainder of it divided by $29$.






abstract-algebra elementary-number-theory modular-arithmetic binary-operations monoid







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    Consider monoid $(mathbb{N},*)$, where operation $*$ is defined as $x*y = xy + x + y$.



    What is the result of $1*2*3*text{...}*25 text{ (mod 29)}$ ?




    $xy + x + y$ can be written as $(x + 1) y + x$ or $y + x (1 + y)$, but that does not seem to help in any way.



    One thing I noticed is that $n * (n + 1) = text{prime number}$. Ex: $12*13=181$; $4*5=29$.



    I am still not sure how can I express $1*2*3*text{...}*25$ in such a way that I could find the remainder of it divided by $29$.






    abstract-algebra elementary-number-theory modular-arithmetic binary-operations monoid







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      Consider monoid $(mathbb{N},*)$, where operation $*$ is defined as $x*y = xy + x + y$.



      What is the result of $1*2*3*text{...}*25 text{ (mod 29)}$ ?




      $xy + x + y$ can be written as $(x + 1) y + x$ or $y + x (1 + y)$, but that does not seem to help in any way.



      One thing I noticed is that $n * (n + 1) = text{prime number}$. Ex: $12*13=181$; $4*5=29$.



      I am still not sure how can I express $1*2*3*text{...}*25$ in such a way that I could find the remainder of it divided by $29$.






      abstract-algebra elementary-number-theory modular-arithmetic binary-operations monoid







      share|cite|improve this question











      $endgroup$





      Consider monoid $(mathbb{N},*)$, where operation $*$ is defined as $x*y = xy + x + y$.



      What is the result of $1*2*3*text{...}*25 text{ (mod 29)}$ ?




      $xy + x + y$ can be written as $(x + 1) y + x$ or $y + x (1 + y)$, but that does not seem to help in any way.



      One thing I noticed is that $n * (n + 1) = text{prime number}$. Ex: $12*13=181$; $4*5=29$.



      I am still not sure how can I express $1*2*3*text{...}*25$ in such a way that I could find the remainder of it divided by $29$.






      abstract-algebra elementary-number-theory modular-arithmetic binary-operations monoid




      abstract-algebra elementary-number-theory modular-arithmetic binary-operations monoid






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 26 at 12:08









      Maria Mazur

      47.9k1260120




      47.9k1260120










      asked Jan 26 at 11:39









      bitadeptbitadept

      475




      475






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          We can rewrite this operation like this:



          $$x*y = (x+1)(y+1)-1 $$




          It is asociative:



          $$(x*y)*z = (x*y+1)(z+1)-1 = (x+1)(y+1)(z+1)-1$$



          also



          $$x*(y*z) = (x+1)(y*z+1)-1 = (x+1)(y+1)(z+1)-1$$




          So $$1*2*3*text{...}*25 = 2cdot 3cdot text{...}cdot 25cdot 26 -1text{ (mod 29)}$$
          $$= 28^{-1}cdot 27^{-1}cdot 28!-1text{ (mod 29)}$$
          $$= (-1)cdot (14)cdot (-1)-1text{ (mod 29)}$$
          $$= 13text{ (mod 29)}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            It took me awhile to figure out how you wrote $x*y$ as $(x+1)(y+1)-1$. Thanks.
            $endgroup$
            – bitadept
            Jan 26 at 12:07





















          2












          $begingroup$

          Hint: Rename your elements: Instead of $1$ say $2'$, instead of $2$ say $3'$, and so on. Try to figure out what the rule for $x'*y'$ is. For instance,
          $$
          2'*2'=1*1=3=4'\
          2'*3'=1*2=5=6'\
          3'*3'=2*2=8=9'
          $$






          share|cite|improve this answer











          $endgroup$













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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            We can rewrite this operation like this:



            $$x*y = (x+1)(y+1)-1 $$




            It is asociative:



            $$(x*y)*z = (x*y+1)(z+1)-1 = (x+1)(y+1)(z+1)-1$$



            also



            $$x*(y*z) = (x+1)(y*z+1)-1 = (x+1)(y+1)(z+1)-1$$




            So $$1*2*3*text{...}*25 = 2cdot 3cdot text{...}cdot 25cdot 26 -1text{ (mod 29)}$$
            $$= 28^{-1}cdot 27^{-1}cdot 28!-1text{ (mod 29)}$$
            $$= (-1)cdot (14)cdot (-1)-1text{ (mod 29)}$$
            $$= 13text{ (mod 29)}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              It took me awhile to figure out how you wrote $x*y$ as $(x+1)(y+1)-1$. Thanks.
              $endgroup$
              – bitadept
              Jan 26 at 12:07


















            2












            $begingroup$

            We can rewrite this operation like this:



            $$x*y = (x+1)(y+1)-1 $$




            It is asociative:



            $$(x*y)*z = (x*y+1)(z+1)-1 = (x+1)(y+1)(z+1)-1$$



            also



            $$x*(y*z) = (x+1)(y*z+1)-1 = (x+1)(y+1)(z+1)-1$$




            So $$1*2*3*text{...}*25 = 2cdot 3cdot text{...}cdot 25cdot 26 -1text{ (mod 29)}$$
            $$= 28^{-1}cdot 27^{-1}cdot 28!-1text{ (mod 29)}$$
            $$= (-1)cdot (14)cdot (-1)-1text{ (mod 29)}$$
            $$= 13text{ (mod 29)}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              It took me awhile to figure out how you wrote $x*y$ as $(x+1)(y+1)-1$. Thanks.
              $endgroup$
              – bitadept
              Jan 26 at 12:07
















            2












            2








            2





            $begingroup$

            We can rewrite this operation like this:



            $$x*y = (x+1)(y+1)-1 $$




            It is asociative:



            $$(x*y)*z = (x*y+1)(z+1)-1 = (x+1)(y+1)(z+1)-1$$



            also



            $$x*(y*z) = (x+1)(y*z+1)-1 = (x+1)(y+1)(z+1)-1$$




            So $$1*2*3*text{...}*25 = 2cdot 3cdot text{...}cdot 25cdot 26 -1text{ (mod 29)}$$
            $$= 28^{-1}cdot 27^{-1}cdot 28!-1text{ (mod 29)}$$
            $$= (-1)cdot (14)cdot (-1)-1text{ (mod 29)}$$
            $$= 13text{ (mod 29)}$$






            share|cite|improve this answer











            $endgroup$



            We can rewrite this operation like this:



            $$x*y = (x+1)(y+1)-1 $$




            It is asociative:



            $$(x*y)*z = (x*y+1)(z+1)-1 = (x+1)(y+1)(z+1)-1$$



            also



            $$x*(y*z) = (x+1)(y*z+1)-1 = (x+1)(y+1)(z+1)-1$$




            So $$1*2*3*text{...}*25 = 2cdot 3cdot text{...}cdot 25cdot 26 -1text{ (mod 29)}$$
            $$= 28^{-1}cdot 27^{-1}cdot 28!-1text{ (mod 29)}$$
            $$= (-1)cdot (14)cdot (-1)-1text{ (mod 29)}$$
            $$= 13text{ (mod 29)}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 26 at 12:01

























            answered Jan 26 at 11:44









            Maria MazurMaria Mazur

            47.9k1260120




            47.9k1260120












            • $begingroup$
              It took me awhile to figure out how you wrote $x*y$ as $(x+1)(y+1)-1$. Thanks.
              $endgroup$
              – bitadept
              Jan 26 at 12:07




















            • $begingroup$
              It took me awhile to figure out how you wrote $x*y$ as $(x+1)(y+1)-1$. Thanks.
              $endgroup$
              – bitadept
              Jan 26 at 12:07


















            $begingroup$
            It took me awhile to figure out how you wrote $x*y$ as $(x+1)(y+1)-1$. Thanks.
            $endgroup$
            – bitadept
            Jan 26 at 12:07






            $begingroup$
            It took me awhile to figure out how you wrote $x*y$ as $(x+1)(y+1)-1$. Thanks.
            $endgroup$
            – bitadept
            Jan 26 at 12:07













            2












            $begingroup$

            Hint: Rename your elements: Instead of $1$ say $2'$, instead of $2$ say $3'$, and so on. Try to figure out what the rule for $x'*y'$ is. For instance,
            $$
            2'*2'=1*1=3=4'\
            2'*3'=1*2=5=6'\
            3'*3'=2*2=8=9'
            $$






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Hint: Rename your elements: Instead of $1$ say $2'$, instead of $2$ say $3'$, and so on. Try to figure out what the rule for $x'*y'$ is. For instance,
              $$
              2'*2'=1*1=3=4'\
              2'*3'=1*2=5=6'\
              3'*3'=2*2=8=9'
              $$






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Hint: Rename your elements: Instead of $1$ say $2'$, instead of $2$ say $3'$, and so on. Try to figure out what the rule for $x'*y'$ is. For instance,
                $$
                2'*2'=1*1=3=4'\
                2'*3'=1*2=5=6'\
                3'*3'=2*2=8=9'
                $$






                share|cite|improve this answer











                $endgroup$



                Hint: Rename your elements: Instead of $1$ say $2'$, instead of $2$ say $3'$, and so on. Try to figure out what the rule for $x'*y'$ is. For instance,
                $$
                2'*2'=1*1=3=4'\
                2'*3'=1*2=5=6'\
                3'*3'=2*2=8=9'
                $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 26 at 11:59

























                answered Jan 26 at 11:48









                ArthurArthur

                119k7118202




                119k7118202






























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