Find all values of $n$ with $0 ≤ n ≤ 35$ such that the congruence $24x$ $≡$ $n$ $(mod$ $36)$ has a...












0












$begingroup$


I was checking the following number theory excercise:




Find all values of $n$ with $0 ≤ n ≤ 35$ such that the congruence $24x$ $≡$ $n$ $(mod$ $36)$ has a solution.




I made the $gcd$ between $12$ and $36$ and my conclusion was that the numbers are $12$ and $24$ but I think is not correct.



Any help will be really appreciated.






number-theory modular-arithmetic







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    0












    $begingroup$


    I was checking the following number theory excercise:




    Find all values of $n$ with $0 ≤ n ≤ 35$ such that the congruence $24x$ $≡$ $n$ $(mod$ $36)$ has a solution.




    I made the $gcd$ between $12$ and $36$ and my conclusion was that the numbers are $12$ and $24$ but I think is not correct.



    Any help will be really appreciated.






    number-theory modular-arithmetic







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I was checking the following number theory excercise:




      Find all values of $n$ with $0 ≤ n ≤ 35$ such that the congruence $24x$ $≡$ $n$ $(mod$ $36)$ has a solution.




      I made the $gcd$ between $12$ and $36$ and my conclusion was that the numbers are $12$ and $24$ but I think is not correct.



      Any help will be really appreciated.






      number-theory modular-arithmetic







      share|cite|improve this question











      $endgroup$




      I was checking the following number theory excercise:




      Find all values of $n$ with $0 ≤ n ≤ 35$ such that the congruence $24x$ $≡$ $n$ $(mod$ $36)$ has a solution.




      I made the $gcd$ between $12$ and $36$ and my conclusion was that the numbers are $12$ and $24$ but I think is not correct.



      Any help will be really appreciated.






      number-theory modular-arithmetic




      number-theory modular-arithmetic






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 21 at 5:15







      mraz

















      asked Sep 26 '18 at 5:10









      mrazmraz

      39118




      39118






















          2 Answers
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          $begingroup$

          Take a look at the expression:



          $c equiv a pmod b$



          If $gcd(a,b)=d rightarrow gcd(c,d) = d$



          Since $24 = 2^3cdot 3$ and $36 = 2^2cdot 3^2$



          Then $gcd(24, 36) = 2^2cdot 3 = 12$.



          So the remainder has the form $12y$. How many $y's$ are there until $35$?



          Right: $12cdot 0, 12cdot1$ and $12cdot2=24$






          share|cite|improve this answer









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            0












            $begingroup$

            $n$ has to be a multiple of $operatorname{gcd}(24,36)=12$. So you left out $0$.






            share|cite|improve this answer









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              2 Answers
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              0












              $begingroup$

              Take a look at the expression:



              $c equiv a pmod b$



              If $gcd(a,b)=d rightarrow gcd(c,d) = d$



              Since $24 = 2^3cdot 3$ and $36 = 2^2cdot 3^2$



              Then $gcd(24, 36) = 2^2cdot 3 = 12$.



              So the remainder has the form $12y$. How many $y's$ are there until $35$?



              Right: $12cdot 0, 12cdot1$ and $12cdot2=24$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Take a look at the expression:



                $c equiv a pmod b$



                If $gcd(a,b)=d rightarrow gcd(c,d) = d$



                Since $24 = 2^3cdot 3$ and $36 = 2^2cdot 3^2$



                Then $gcd(24, 36) = 2^2cdot 3 = 12$.



                So the remainder has the form $12y$. How many $y's$ are there until $35$?



                Right: $12cdot 0, 12cdot1$ and $12cdot2=24$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Take a look at the expression:



                  $c equiv a pmod b$



                  If $gcd(a,b)=d rightarrow gcd(c,d) = d$



                  Since $24 = 2^3cdot 3$ and $36 = 2^2cdot 3^2$



                  Then $gcd(24, 36) = 2^2cdot 3 = 12$.



                  So the remainder has the form $12y$. How many $y's$ are there until $35$?



                  Right: $12cdot 0, 12cdot1$ and $12cdot2=24$






                  share|cite|improve this answer









                  $endgroup$



                  Take a look at the expression:



                  $c equiv a pmod b$



                  If $gcd(a,b)=d rightarrow gcd(c,d) = d$



                  Since $24 = 2^3cdot 3$ and $36 = 2^2cdot 3^2$



                  Then $gcd(24, 36) = 2^2cdot 3 = 12$.



                  So the remainder has the form $12y$. How many $y's$ are there until $35$?



                  Right: $12cdot 0, 12cdot1$ and $12cdot2=24$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 26 '18 at 6:05









                  kub0xkub0x

                  1,7411921




                  1,7411921























                      0












                      $begingroup$

                      $n$ has to be a multiple of $operatorname{gcd}(24,36)=12$. So you left out $0$.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        $n$ has to be a multiple of $operatorname{gcd}(24,36)=12$. So you left out $0$.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          $n$ has to be a multiple of $operatorname{gcd}(24,36)=12$. So you left out $0$.






                          share|cite|improve this answer









                          $endgroup$



                          $n$ has to be a multiple of $operatorname{gcd}(24,36)=12$. So you left out $0$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Sep 26 '18 at 6:14









                          Chris CusterChris Custer

                          14.1k3827




                          14.1k3827






























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