Solve the congruence system $ xequiv m_i-1 pmod{m_i},$ for $,i = 1,ldots, k$












1














Find natural number $x$ so that
$$xequiv 9pmod{10},quad xequiv8pmod9,quad ...,quad xequiv 1pmod2$$






elementary-number-theory







share|cite|improve this question





























    1














    Find natural number $x$ so that
    $$xequiv 9pmod{10},quad xequiv8pmod9,quad ...,quad xequiv 1pmod2$$






    elementary-number-theory







    share|cite|improve this question



























      1












      1








      1







      Find natural number $x$ so that
      $$xequiv 9pmod{10},quad xequiv8pmod9,quad ...,quad xequiv 1pmod2$$






      elementary-number-theory







      share|cite|improve this question















      Find natural number $x$ so that
      $$xequiv 9pmod{10},quad xequiv8pmod9,quad ...,quad xequiv 1pmod2$$






      elementary-number-theory




      elementary-number-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Sep 4 '17 at 21:18









      Bill Dubuque

      208k29190626




      208k29190626










      asked Mar 16 '14 at 20:37









      user119081

      373




      373






















          3 Answers
          3






          active

          oldest

          votes


















          1














          Hint: The unnatural number $-1$ works.






          share|cite|improve this answer





















          • I ask for natural number not integer
            – user119081
            Mar 16 '14 at 20:44










          • @user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural?
            – Charles
            Mar 16 '14 at 20:45










          • Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you.
            – André Nicolas
            Mar 16 '14 at 20:46












          • I believe this should be exist a natural solution by extended Euclidean algorithm
            – user119081
            Mar 16 '14 at 20:48






          • 1




            While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution?
            – André Nicolas
            Mar 16 '14 at 20:52



















          1














          Since $,m_i-1equiv color{#c00}{-1}pmod{!m_i},$ we can apply $ $ CCRT = $rmcolor{#c00}{constant}$ case optimization of CRT



          $$begin{align} xequiv color{#c00}{-1}!!pmod{!m_i}&iff xequiv -1!!pmod{{rm lcm}{m_i}}\[.4em]
          text{or, without using CRT:} {rm all} m_i mid x+1 &iff {rm lcm}{m_i}mid x+1
          end{align}qquadqquad$$
          The latter equivalence is by the Universal Property of LCM (= definition of LCM in general)






          share|cite|improve this answer























          • It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution.
            – Bill Dubuque
            Mar 16 '14 at 21:11





















          0














          begin{align}
          x &equiv 9 pmod{10} \
          x &equiv 8 pmod 9 \
          x &equiv 7 pmod 8 \
          x &equiv 6 pmod 7 \
          x &equiv 5 pmod 6 \
          x &equiv 4 pmod 5 \
          x &equiv 3 pmod 4 \
          x &equiv 2 pmod 3 \
          x &equiv 1 pmod 2 \
          end{align}



          Is equivalent to



          begin{align}
          x &equiv -1 pmod{10} \
          x &equiv -1 pmod 9 \
          x &equiv -1 pmod 8 \
          x &equiv -1 pmod 7 \
          x &equiv -1 pmod 6 \
          x &equiv -1 pmod 5 \
          x &equiv -1 pmod 4 \
          x &equiv -1 pmod 3 \
          x &equiv -1 pmod 2 \
          end{align}



          which is equivalent to



          $$x equiv -1 mod{operatorname{lcm}{2,3,4,5,6,7,8,9,10}}$$



          $$x equiv -1 pmod{2520}$$



          $$x equiv 2519 pmod{2520}$$






          share|cite|improve this answer























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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            Hint: The unnatural number $-1$ works.






            share|cite|improve this answer





















            • I ask for natural number not integer
              – user119081
              Mar 16 '14 at 20:44










            • @user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural?
              – Charles
              Mar 16 '14 at 20:45










            • Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you.
              – André Nicolas
              Mar 16 '14 at 20:46












            • I believe this should be exist a natural solution by extended Euclidean algorithm
              – user119081
              Mar 16 '14 at 20:48






            • 1




              While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution?
              – André Nicolas
              Mar 16 '14 at 20:52
















            1














            Hint: The unnatural number $-1$ works.






            share|cite|improve this answer





















            • I ask for natural number not integer
              – user119081
              Mar 16 '14 at 20:44










            • @user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural?
              – Charles
              Mar 16 '14 at 20:45










            • Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you.
              – André Nicolas
              Mar 16 '14 at 20:46












            • I believe this should be exist a natural solution by extended Euclidean algorithm
              – user119081
              Mar 16 '14 at 20:48






            • 1




              While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution?
              – André Nicolas
              Mar 16 '14 at 20:52














            1












            1








            1






            Hint: The unnatural number $-1$ works.






            share|cite|improve this answer












            Hint: The unnatural number $-1$ works.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 16 '14 at 20:41









            André Nicolas

            451k36421805




            451k36421805












            • I ask for natural number not integer
              – user119081
              Mar 16 '14 at 20:44










            • @user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural?
              – Charles
              Mar 16 '14 at 20:45










            • Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you.
              – André Nicolas
              Mar 16 '14 at 20:46












            • I believe this should be exist a natural solution by extended Euclidean algorithm
              – user119081
              Mar 16 '14 at 20:48






            • 1




              While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution?
              – André Nicolas
              Mar 16 '14 at 20:52


















            • I ask for natural number not integer
              – user119081
              Mar 16 '14 at 20:44










            • @user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural?
              – Charles
              Mar 16 '14 at 20:45










            • Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you.
              – André Nicolas
              Mar 16 '14 at 20:46












            • I believe this should be exist a natural solution by extended Euclidean algorithm
              – user119081
              Mar 16 '14 at 20:48






            • 1




              While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution?
              – André Nicolas
              Mar 16 '14 at 20:52
















            I ask for natural number not integer
            – user119081
            Mar 16 '14 at 20:44




            I ask for natural number not integer
            – user119081
            Mar 16 '14 at 20:44












            @user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural?
            – Charles
            Mar 16 '14 at 20:45




            @user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural?
            – Charles
            Mar 16 '14 at 20:45












            Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you.
            – André Nicolas
            Mar 16 '14 at 20:46






            Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you.
            – André Nicolas
            Mar 16 '14 at 20:46














            I believe this should be exist a natural solution by extended Euclidean algorithm
            – user119081
            Mar 16 '14 at 20:48




            I believe this should be exist a natural solution by extended Euclidean algorithm
            – user119081
            Mar 16 '14 at 20:48




            1




            1




            While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution?
            – André Nicolas
            Mar 16 '14 at 20:52




            While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution?
            – André Nicolas
            Mar 16 '14 at 20:52











            1














            Since $,m_i-1equiv color{#c00}{-1}pmod{!m_i},$ we can apply $ $ CCRT = $rmcolor{#c00}{constant}$ case optimization of CRT



            $$begin{align} xequiv color{#c00}{-1}!!pmod{!m_i}&iff xequiv -1!!pmod{{rm lcm}{m_i}}\[.4em]
            text{or, without using CRT:} {rm all} m_i mid x+1 &iff {rm lcm}{m_i}mid x+1
            end{align}qquadqquad$$
            The latter equivalence is by the Universal Property of LCM (= definition of LCM in general)






            share|cite|improve this answer























            • It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution.
              – Bill Dubuque
              Mar 16 '14 at 21:11


















            1














            Since $,m_i-1equiv color{#c00}{-1}pmod{!m_i},$ we can apply $ $ CCRT = $rmcolor{#c00}{constant}$ case optimization of CRT



            $$begin{align} xequiv color{#c00}{-1}!!pmod{!m_i}&iff xequiv -1!!pmod{{rm lcm}{m_i}}\[.4em]
            text{or, without using CRT:} {rm all} m_i mid x+1 &iff {rm lcm}{m_i}mid x+1
            end{align}qquadqquad$$
            The latter equivalence is by the Universal Property of LCM (= definition of LCM in general)






            share|cite|improve this answer























            • It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution.
              – Bill Dubuque
              Mar 16 '14 at 21:11
















            1












            1








            1






            Since $,m_i-1equiv color{#c00}{-1}pmod{!m_i},$ we can apply $ $ CCRT = $rmcolor{#c00}{constant}$ case optimization of CRT



            $$begin{align} xequiv color{#c00}{-1}!!pmod{!m_i}&iff xequiv -1!!pmod{{rm lcm}{m_i}}\[.4em]
            text{or, without using CRT:} {rm all} m_i mid x+1 &iff {rm lcm}{m_i}mid x+1
            end{align}qquadqquad$$
            The latter equivalence is by the Universal Property of LCM (= definition of LCM in general)






            share|cite|improve this answer














            Since $,m_i-1equiv color{#c00}{-1}pmod{!m_i},$ we can apply $ $ CCRT = $rmcolor{#c00}{constant}$ case optimization of CRT



            $$begin{align} xequiv color{#c00}{-1}!!pmod{!m_i}&iff xequiv -1!!pmod{{rm lcm}{m_i}}\[.4em]
            text{or, without using CRT:} {rm all} m_i mid x+1 &iff {rm lcm}{m_i}mid x+1
            end{align}qquadqquad$$
            The latter equivalence is by the Universal Property of LCM (= definition of LCM in general)







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Sep 4 '17 at 21:20

























            answered Mar 16 '14 at 20:52









            Bill Dubuque

            208k29190626




            208k29190626












            • It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution.
              – Bill Dubuque
              Mar 16 '14 at 21:11




















            • It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution.
              – Bill Dubuque
              Mar 16 '14 at 21:11


















            It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution.
            – Bill Dubuque
            Mar 16 '14 at 21:11






            It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution.
            – Bill Dubuque
            Mar 16 '14 at 21:11













            0














            begin{align}
            x &equiv 9 pmod{10} \
            x &equiv 8 pmod 9 \
            x &equiv 7 pmod 8 \
            x &equiv 6 pmod 7 \
            x &equiv 5 pmod 6 \
            x &equiv 4 pmod 5 \
            x &equiv 3 pmod 4 \
            x &equiv 2 pmod 3 \
            x &equiv 1 pmod 2 \
            end{align}



            Is equivalent to



            begin{align}
            x &equiv -1 pmod{10} \
            x &equiv -1 pmod 9 \
            x &equiv -1 pmod 8 \
            x &equiv -1 pmod 7 \
            x &equiv -1 pmod 6 \
            x &equiv -1 pmod 5 \
            x &equiv -1 pmod 4 \
            x &equiv -1 pmod 3 \
            x &equiv -1 pmod 2 \
            end{align}



            which is equivalent to



            $$x equiv -1 mod{operatorname{lcm}{2,3,4,5,6,7,8,9,10}}$$



            $$x equiv -1 pmod{2520}$$



            $$x equiv 2519 pmod{2520}$$






            share|cite|improve this answer




























              0














              begin{align}
              x &equiv 9 pmod{10} \
              x &equiv 8 pmod 9 \
              x &equiv 7 pmod 8 \
              x &equiv 6 pmod 7 \
              x &equiv 5 pmod 6 \
              x &equiv 4 pmod 5 \
              x &equiv 3 pmod 4 \
              x &equiv 2 pmod 3 \
              x &equiv 1 pmod 2 \
              end{align}



              Is equivalent to



              begin{align}
              x &equiv -1 pmod{10} \
              x &equiv -1 pmod 9 \
              x &equiv -1 pmod 8 \
              x &equiv -1 pmod 7 \
              x &equiv -1 pmod 6 \
              x &equiv -1 pmod 5 \
              x &equiv -1 pmod 4 \
              x &equiv -1 pmod 3 \
              x &equiv -1 pmod 2 \
              end{align}



              which is equivalent to



              $$x equiv -1 mod{operatorname{lcm}{2,3,4,5,6,7,8,9,10}}$$



              $$x equiv -1 pmod{2520}$$



              $$x equiv 2519 pmod{2520}$$






              share|cite|improve this answer


























                0












                0








                0






                begin{align}
                x &equiv 9 pmod{10} \
                x &equiv 8 pmod 9 \
                x &equiv 7 pmod 8 \
                x &equiv 6 pmod 7 \
                x &equiv 5 pmod 6 \
                x &equiv 4 pmod 5 \
                x &equiv 3 pmod 4 \
                x &equiv 2 pmod 3 \
                x &equiv 1 pmod 2 \
                end{align}



                Is equivalent to



                begin{align}
                x &equiv -1 pmod{10} \
                x &equiv -1 pmod 9 \
                x &equiv -1 pmod 8 \
                x &equiv -1 pmod 7 \
                x &equiv -1 pmod 6 \
                x &equiv -1 pmod 5 \
                x &equiv -1 pmod 4 \
                x &equiv -1 pmod 3 \
                x &equiv -1 pmod 2 \
                end{align}



                which is equivalent to



                $$x equiv -1 mod{operatorname{lcm}{2,3,4,5,6,7,8,9,10}}$$



                $$x equiv -1 pmod{2520}$$



                $$x equiv 2519 pmod{2520}$$






                share|cite|improve this answer














                begin{align}
                x &equiv 9 pmod{10} \
                x &equiv 8 pmod 9 \
                x &equiv 7 pmod 8 \
                x &equiv 6 pmod 7 \
                x &equiv 5 pmod 6 \
                x &equiv 4 pmod 5 \
                x &equiv 3 pmod 4 \
                x &equiv 2 pmod 3 \
                x &equiv 1 pmod 2 \
                end{align}



                Is equivalent to



                begin{align}
                x &equiv -1 pmod{10} \
                x &equiv -1 pmod 9 \
                x &equiv -1 pmod 8 \
                x &equiv -1 pmod 7 \
                x &equiv -1 pmod 6 \
                x &equiv -1 pmod 5 \
                x &equiv -1 pmod 4 \
                x &equiv -1 pmod 3 \
                x &equiv -1 pmod 2 \
                end{align}



                which is equivalent to



                $$x equiv -1 mod{operatorname{lcm}{2,3,4,5,6,7,8,9,10}}$$



                $$x equiv -1 pmod{2520}$$



                $$x equiv 2519 pmod{2520}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 17 '17 at 7:58

























                answered Feb 16 '17 at 14:40









                steven gregory

                17.7k32257




                17.7k32257






























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