How to calculate this congruency?

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Let's say I have this linear congruency: $2x + 1234 = 7 mod 17$.
Without "$+1234$" I would've used the following formulas: $x = x_0 + k(dfracmgcd(a, m))$, whereas $ax_0 + my_0 = b$. But I don't know what to do with $1234$? Thank you in advance.







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    Let's say I have this linear congruency: $2x + 1234 = 7 mod 17$.
    Without "$+1234$" I would've used the following formulas: $x = x_0 + k(dfracmgcd(a, m))$, whereas $ax_0 + my_0 = b$. But I don't know what to do with $1234$? Thank you in advance.







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      up vote
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      favorite









      up vote
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      down vote

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      Let's say I have this linear congruency: $2x + 1234 = 7 mod 17$.
      Without "$+1234$" I would've used the following formulas: $x = x_0 + k(dfracmgcd(a, m))$, whereas $ax_0 + my_0 = b$. But I don't know what to do with $1234$? Thank you in advance.







      share|cite|improve this question














      Let's say I have this linear congruency: $2x + 1234 = 7 mod 17$.
      Without "$+1234$" I would've used the following formulas: $x = x_0 + k(dfracmgcd(a, m))$, whereas $ax_0 + my_0 = b$. But I don't know what to do with $1234$? Thank you in advance.









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      edited Aug 17 at 7:53









      Jack M

      17.2k33473




      17.2k33473










      asked Aug 17 at 7:49









      ponikoli

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          I would calculate everything mod. $17$: $;1234equiv 10mod 17$, so the equation becomes
          $$2x+10equiv 7mod17iff 2xequiv-3equiv14mod 17. $$
          Now as $2$ is a unit mod. $17$, we may apply the cancellation law:
          $$2xequiv 14=2cdot 7mod17iff xequiv 7mod 17.$$






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            No worries, just solve it like you would solve a normal equation.



            $$2x+1234=7$$
            $$2x=7-1234$$



            and then apply your usual method.






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              2 Answers
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              2 Answers
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              I would calculate everything mod. $17$: $;1234equiv 10mod 17$, so the equation becomes
              $$2x+10equiv 7mod17iff 2xequiv-3equiv14mod 17. $$
              Now as $2$ is a unit mod. $17$, we may apply the cancellation law:
              $$2xequiv 14=2cdot 7mod17iff xequiv 7mod 17.$$






              share|cite|improve this answer
























                up vote
                0
                down vote



                accepted










                I would calculate everything mod. $17$: $;1234equiv 10mod 17$, so the equation becomes
                $$2x+10equiv 7mod17iff 2xequiv-3equiv14mod 17. $$
                Now as $2$ is a unit mod. $17$, we may apply the cancellation law:
                $$2xequiv 14=2cdot 7mod17iff xequiv 7mod 17.$$






                share|cite|improve this answer






















                  up vote
                  0
                  down vote



                  accepted







                  up vote
                  0
                  down vote



                  accepted






                  I would calculate everything mod. $17$: $;1234equiv 10mod 17$, so the equation becomes
                  $$2x+10equiv 7mod17iff 2xequiv-3equiv14mod 17. $$
                  Now as $2$ is a unit mod. $17$, we may apply the cancellation law:
                  $$2xequiv 14=2cdot 7mod17iff xequiv 7mod 17.$$






                  share|cite|improve this answer












                  I would calculate everything mod. $17$: $;1234equiv 10mod 17$, so the equation becomes
                  $$2x+10equiv 7mod17iff 2xequiv-3equiv14mod 17. $$
                  Now as $2$ is a unit mod. $17$, we may apply the cancellation law:
                  $$2xequiv 14=2cdot 7mod17iff xequiv 7mod 17.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 17 at 8:07









                  Bernard

                  111k635103




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                      up vote
                      1
                      down vote













                      No worries, just solve it like you would solve a normal equation.



                      $$2x+1234=7$$
                      $$2x=7-1234$$



                      and then apply your usual method.






                      share|cite|improve this answer
























                        up vote
                        1
                        down vote













                        No worries, just solve it like you would solve a normal equation.



                        $$2x+1234=7$$
                        $$2x=7-1234$$



                        and then apply your usual method.






                        share|cite|improve this answer






















                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          No worries, just solve it like you would solve a normal equation.



                          $$2x+1234=7$$
                          $$2x=7-1234$$



                          and then apply your usual method.






                          share|cite|improve this answer












                          No worries, just solve it like you would solve a normal equation.



                          $$2x+1234=7$$
                          $$2x=7-1234$$



                          and then apply your usual method.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Aug 17 at 7:52









                          Jack M

                          17.2k33473




                          17.2k33473






















                               

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