$c in mathbb{N}$, so that $c cdot 11 = 23 mod 103$











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0
down vote

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How can one find $c in mathbb{N}$, so that $c cdot 11 = 23 mod 103$?



I know that $a cdot b mod n = (a mod n cdot b mod n) mod n$.



Furthermore,
Fermat's little theorem says that for all prime $p$ and if $p$ does not divide $a$



$a^{p-1} equiv 1 pmod p$



And I know that here all numbers $11, 23, text { and } 103$ are prime numbers.



That still doesn't help me though...






modular-arithmetic







share|cite|improve this question






















  • $103$ is a prime. Also, $11$ is a prime, more specifically, it's relatively prime to $103$. Do you know how to compute multiplicative modular inverses? We have $cequiv 23times 11^{-1}pmod{103}$ and $11^{-1}$ is computed by solving $11zequiv 1pmod{103}$
    – Prasun Biswas
    Nov 22 at 10:52










  • My computations give me: $$11zequiv 1iff 7zequiv 110zequiv 10iff 105zequiv 2zequiv 150iff zequiv 75pmod{103}$$ which gives $cequiv 77pmod{103}$
    – Prasun Biswas
    Nov 22 at 10:56

















up vote
0
down vote

favorite












How can one find $c in mathbb{N}$, so that $c cdot 11 = 23 mod 103$?



I know that $a cdot b mod n = (a mod n cdot b mod n) mod n$.



Furthermore,
Fermat's little theorem says that for all prime $p$ and if $p$ does not divide $a$



$a^{p-1} equiv 1 pmod p$



And I know that here all numbers $11, 23, text { and } 103$ are prime numbers.



That still doesn't help me though...






modular-arithmetic







share|cite|improve this question






















  • $103$ is a prime. Also, $11$ is a prime, more specifically, it's relatively prime to $103$. Do you know how to compute multiplicative modular inverses? We have $cequiv 23times 11^{-1}pmod{103}$ and $11^{-1}$ is computed by solving $11zequiv 1pmod{103}$
    – Prasun Biswas
    Nov 22 at 10:52










  • My computations give me: $$11zequiv 1iff 7zequiv 110zequiv 10iff 105zequiv 2zequiv 150iff zequiv 75pmod{103}$$ which gives $cequiv 77pmod{103}$
    – Prasun Biswas
    Nov 22 at 10:56















up vote
0
down vote

favorite









up vote
0
down vote

favorite











How can one find $c in mathbb{N}$, so that $c cdot 11 = 23 mod 103$?



I know that $a cdot b mod n = (a mod n cdot b mod n) mod n$.



Furthermore,
Fermat's little theorem says that for all prime $p$ and if $p$ does not divide $a$



$a^{p-1} equiv 1 pmod p$



And I know that here all numbers $11, 23, text { and } 103$ are prime numbers.



That still doesn't help me though...






modular-arithmetic







share|cite|improve this question













How can one find $c in mathbb{N}$, so that $c cdot 11 = 23 mod 103$?



I know that $a cdot b mod n = (a mod n cdot b mod n) mod n$.



Furthermore,
Fermat's little theorem says that for all prime $p$ and if $p$ does not divide $a$



$a^{p-1} equiv 1 pmod p$



And I know that here all numbers $11, 23, text { and } 103$ are prime numbers.



That still doesn't help me though...






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 Nov 22 at 10:49









Math Dummy

276




276












  • $103$ is a prime. Also, $11$ is a prime, more specifically, it's relatively prime to $103$. Do you know how to compute multiplicative modular inverses? We have $cequiv 23times 11^{-1}pmod{103}$ and $11^{-1}$ is computed by solving $11zequiv 1pmod{103}$
    – Prasun Biswas
    Nov 22 at 10:52










  • My computations give me: $$11zequiv 1iff 7zequiv 110zequiv 10iff 105zequiv 2zequiv 150iff zequiv 75pmod{103}$$ which gives $cequiv 77pmod{103}$
    – Prasun Biswas
    Nov 22 at 10:56




















  • $103$ is a prime. Also, $11$ is a prime, more specifically, it's relatively prime to $103$. Do you know how to compute multiplicative modular inverses? We have $cequiv 23times 11^{-1}pmod{103}$ and $11^{-1}$ is computed by solving $11zequiv 1pmod{103}$
    – Prasun Biswas
    Nov 22 at 10:52










  • My computations give me: $$11zequiv 1iff 7zequiv 110zequiv 10iff 105zequiv 2zequiv 150iff zequiv 75pmod{103}$$ which gives $cequiv 77pmod{103}$
    – Prasun Biswas
    Nov 22 at 10:56


















$103$ is a prime. Also, $11$ is a prime, more specifically, it's relatively prime to $103$. Do you know how to compute multiplicative modular inverses? We have $cequiv 23times 11^{-1}pmod{103}$ and $11^{-1}$ is computed by solving $11zequiv 1pmod{103}$
– Prasun Biswas
Nov 22 at 10:52




$103$ is a prime. Also, $11$ is a prime, more specifically, it's relatively prime to $103$. Do you know how to compute multiplicative modular inverses? We have $cequiv 23times 11^{-1}pmod{103}$ and $11^{-1}$ is computed by solving $11zequiv 1pmod{103}$
– Prasun Biswas
Nov 22 at 10:52












My computations give me: $$11zequiv 1iff 7zequiv 110zequiv 10iff 105zequiv 2zequiv 150iff zequiv 75pmod{103}$$ which gives $cequiv 77pmod{103}$
– Prasun Biswas
Nov 22 at 10:56






My computations give me: $$11zequiv 1iff 7zequiv 110zequiv 10iff 105zequiv 2zequiv 150iff zequiv 75pmod{103}$$ which gives $cequiv 77pmod{103}$
– Prasun Biswas
Nov 22 at 10:56












2 Answers
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0
down vote



accepted










Fermat's little theorem provides a solution:



$$
c cdot 11 equiv 23 bmod 103
implies
c equiv c cdot 11^{102} equiv 23 cdot 11^{101} bmod 103
$$



Thus, $c = 23 cdot 11^{101}$ works. It is not the smallest solution, but the question does not ask for that...






share|cite|improve this answer




























    up vote
    0
    down vote













    By Gauss's algorithm $!bmod 103!:, cequiv dfrac{23}{11}equiv dfrac{9cdot 23}{9cdot 11}equivdfrac{1}{-4}equiv dfrac{104}{-4}equiv -26equiv 77$






    share|cite|improve this answer





















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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

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      up vote
      0
      down vote



      accepted










      Fermat's little theorem provides a solution:



      $$
      c cdot 11 equiv 23 bmod 103
      implies
      c equiv c cdot 11^{102} equiv 23 cdot 11^{101} bmod 103
      $$



      Thus, $c = 23 cdot 11^{101}$ works. It is not the smallest solution, but the question does not ask for that...






      share|cite|improve this answer

























        up vote
        0
        down vote



        accepted










        Fermat's little theorem provides a solution:



        $$
        c cdot 11 equiv 23 bmod 103
        implies
        c equiv c cdot 11^{102} equiv 23 cdot 11^{101} bmod 103
        $$



        Thus, $c = 23 cdot 11^{101}$ works. It is not the smallest solution, but the question does not ask for that...






        share|cite|improve this answer























          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          Fermat's little theorem provides a solution:



          $$
          c cdot 11 equiv 23 bmod 103
          implies
          c equiv c cdot 11^{102} equiv 23 cdot 11^{101} bmod 103
          $$



          Thus, $c = 23 cdot 11^{101}$ works. It is not the smallest solution, but the question does not ask for that...






          share|cite|improve this answer












          Fermat's little theorem provides a solution:



          $$
          c cdot 11 equiv 23 bmod 103
          implies
          c equiv c cdot 11^{102} equiv 23 cdot 11^{101} bmod 103
          $$



          Thus, $c = 23 cdot 11^{101}$ works. It is not the smallest solution, but the question does not ask for that...







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 11:10









          lhf

          162k9166385




          162k9166385






















              up vote
              0
              down vote













              By Gauss's algorithm $!bmod 103!:, cequiv dfrac{23}{11}equiv dfrac{9cdot 23}{9cdot 11}equivdfrac{1}{-4}equiv dfrac{104}{-4}equiv -26equiv 77$






              share|cite|improve this answer

























                up vote
                0
                down vote













                By Gauss's algorithm $!bmod 103!:, cequiv dfrac{23}{11}equiv dfrac{9cdot 23}{9cdot 11}equivdfrac{1}{-4}equiv dfrac{104}{-4}equiv -26equiv 77$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  By Gauss's algorithm $!bmod 103!:, cequiv dfrac{23}{11}equiv dfrac{9cdot 23}{9cdot 11}equivdfrac{1}{-4}equiv dfrac{104}{-4}equiv -26equiv 77$






                  share|cite|improve this answer












                  By Gauss's algorithm $!bmod 103!:, cequiv dfrac{23}{11}equiv dfrac{9cdot 23}{9cdot 11}equivdfrac{1}{-4}equiv dfrac{104}{-4}equiv -26equiv 77$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 19:50









                  Bill Dubuque

                  207k29189625




                  207k29189625






























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