Multiplication and modulo - Order of operations












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$begingroup$


Simple question



I have $$acdot bcdot c;mod;p$$



How do I interpret this? Do i read it as this $$(abc);mod;p$$ or is it $$acdot bcdot (c;mod;p)$$



Or do I completely misunderstand modular arithmetic?



Thanks!






modular-arithmetic







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    0












    $begingroup$


    Simple question



    I have $$acdot bcdot c;mod;p$$



    How do I interpret this? Do i read it as this $$(abc);mod;p$$ or is it $$acdot bcdot (c;mod;p)$$



    Or do I completely misunderstand modular arithmetic?



    Thanks!






    modular-arithmetic







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Simple question



      I have $$acdot bcdot c;mod;p$$



      How do I interpret this? Do i read it as this $$(abc);mod;p$$ or is it $$acdot bcdot (c;mod;p)$$



      Or do I completely misunderstand modular arithmetic?



      Thanks!






      modular-arithmetic







      share|cite|improve this question









      $endgroup$




      Simple question



      I have $$acdot bcdot c;mod;p$$



      How do I interpret this? Do i read it as this $$(abc);mod;p$$ or is it $$acdot bcdot (c;mod;p)$$



      Or do I completely misunderstand modular arithmetic?



      Thanks!






      modular-arithmetic




      modular-arithmetic






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 '18 at 7:10









      Jamal RahmanJamal Rahman

      31




      31






















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          0












          $begingroup$

          Yes, it's $(a b c) mod p$. In fact
          $$
          x mod p =(x)mod p text{ for any expression } x
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh brilliant explanation by generalising x as any expression
            $endgroup$
            – Jamal Rahman
            Dec 10 '18 at 7:30



















          -1












          $begingroup$

          You can see this as $a*b*c(mod p)$ or $a(mod p) *b(mod p)* c(mod p)$.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
            2






            active

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            active

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            0












            $begingroup$

            Yes, it's $(a b c) mod p$. In fact
            $$
            x mod p =(x)mod p text{ for any expression } x
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Oh brilliant explanation by generalising x as any expression
              $endgroup$
              – Jamal Rahman
              Dec 10 '18 at 7:30
















            0












            $begingroup$

            Yes, it's $(a b c) mod p$. In fact
            $$
            x mod p =(x)mod p text{ for any expression } x
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Oh brilliant explanation by generalising x as any expression
              $endgroup$
              – Jamal Rahman
              Dec 10 '18 at 7:30














            0












            0








            0





            $begingroup$

            Yes, it's $(a b c) mod p$. In fact
            $$
            x mod p =(x)mod p text{ for any expression } x
            $$






            share|cite|improve this answer









            $endgroup$



            Yes, it's $(a b c) mod p$. In fact
            $$
            x mod p =(x)mod p text{ for any expression } x
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 10 '18 at 7:25









            plus1plus1

            3911




            3911












            • $begingroup$
              Oh brilliant explanation by generalising x as any expression
              $endgroup$
              – Jamal Rahman
              Dec 10 '18 at 7:30


















            • $begingroup$
              Oh brilliant explanation by generalising x as any expression
              $endgroup$
              – Jamal Rahman
              Dec 10 '18 at 7:30
















            $begingroup$
            Oh brilliant explanation by generalising x as any expression
            $endgroup$
            – Jamal Rahman
            Dec 10 '18 at 7:30




            $begingroup$
            Oh brilliant explanation by generalising x as any expression
            $endgroup$
            – Jamal Rahman
            Dec 10 '18 at 7:30











            -1












            $begingroup$

            You can see this as $a*b*c(mod p)$ or $a(mod p) *b(mod p)* c(mod p)$.






            share|cite|improve this answer









            $endgroup$


















              -1












              $begingroup$

              You can see this as $a*b*c(mod p)$ or $a(mod p) *b(mod p)* c(mod p)$.






              share|cite|improve this answer









              $endgroup$
















                -1












                -1








                -1





                $begingroup$

                You can see this as $a*b*c(mod p)$ or $a(mod p) *b(mod p)* c(mod p)$.






                share|cite|improve this answer









                $endgroup$



                You can see this as $a*b*c(mod p)$ or $a(mod p) *b(mod p)* c(mod p)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 10 '18 at 7:18









                nafhgoodnafhgood

                1,805422




                1,805422






























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