Diophantine Equations (another)












0












$begingroup$


I have the following equation



$756x+630y = 2394$



and i am trying to find the general solution to it. I have done these type of questions numerous times and had no difficulty but I cant seem to do this.



What I have so far is



$756x+630y=2394$



$756-1(630)=126$



$630-5(126)=0$



I tried using the technique roll back but it dosent work.






elementary-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    The usual method is to apply the extended Euclidean algorithm to $756$ and $630$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 15 '18 at 15:17
















0












$begingroup$


I have the following equation



$756x+630y = 2394$



and i am trying to find the general solution to it. I have done these type of questions numerous times and had no difficulty but I cant seem to do this.



What I have so far is



$756x+630y=2394$



$756-1(630)=126$



$630-5(126)=0$



I tried using the technique roll back but it dosent work.






elementary-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    The usual method is to apply the extended Euclidean algorithm to $756$ and $630$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 15 '18 at 15:17














0












0








0





$begingroup$


I have the following equation



$756x+630y = 2394$



and i am trying to find the general solution to it. I have done these type of questions numerous times and had no difficulty but I cant seem to do this.



What I have so far is



$756x+630y=2394$



$756-1(630)=126$



$630-5(126)=0$



I tried using the technique roll back but it dosent work.






elementary-number-theory







share|cite|improve this question











$endgroup$




I have the following equation



$756x+630y = 2394$



and i am trying to find the general solution to it. I have done these type of questions numerous times and had no difficulty but I cant seem to do this.



What I have so far is



$756x+630y=2394$



$756-1(630)=126$



$630-5(126)=0$



I tried using the technique roll back but it dosent work.






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 Dec 15 '18 at 15:56









Ethan Bolker

44.5k553120




44.5k553120










asked Dec 15 '18 at 15:15









user147825user147825

758




758












  • $begingroup$
    The usual method is to apply the extended Euclidean algorithm to $756$ and $630$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 15 '18 at 15:17


















  • $begingroup$
    The usual method is to apply the extended Euclidean algorithm to $756$ and $630$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 15 '18 at 15:17
















$begingroup$
The usual method is to apply the extended Euclidean algorithm to $756$ and $630$.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 15:17




$begingroup$
The usual method is to apply the extended Euclidean algorithm to $756$ and $630$.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 15:17










3 Answers
3






active

oldest

votes


















1












$begingroup$

What your calculation shows is that $126$ is the greatest common divisor of $756$ and $630$ and that $756-630=126$ and $6times630-5times756=0$. Since $2394/126=19$ is not fractional, we can solve the equation. $19(756-630)+k(6times630-5times756)=2394$.



Note that $6times630-5times756=0$ comes from substituting $756-630=126$ into $630-5times126=0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    5×630−4×756=126 and 6×630−5×756=0 sorry im not sure where you are getting these. I can see they are right and i was wondering where they came from
    $endgroup$
    – user147825
    Dec 15 '18 at 15:38










  • $begingroup$
    Edited to make it more clear.
    $endgroup$
    – SmileyCraft
    Dec 15 '18 at 15:54










  • $begingroup$
    Note that this is basically the extended Euclidean algorithm, which solves this in general.
    $endgroup$
    – SmileyCraft
    Dec 15 '18 at 15:55



















1












$begingroup$

You can divide all by $$126$$ and you have to solve $$6x+5y=19$$
and then you can write $$y=4-x-frac{1+x}{5}$$
Substitute $$frac{1+x}{5}=t$$ we get $$x=5t-1,y=5-6t$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is the solution! +1
    $endgroup$
    – greedoid
    Dec 15 '18 at 17:41



















0












$begingroup$

If x= 1 and y= 1 then 6x+ 5y= 1. So if x= 19 and y= -19, 6x+ 5y= 19. But x= 19- 5k, y= -19+ 6k is also a solution: 6(19- 5k)+ 5(-19+ 6k)= 6(19)- 6(5)k- 5(19)+ 5(6)k= 114-95= 19 for all k.






share|cite|improve this answer









$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    What your calculation shows is that $126$ is the greatest common divisor of $756$ and $630$ and that $756-630=126$ and $6times630-5times756=0$. Since $2394/126=19$ is not fractional, we can solve the equation. $19(756-630)+k(6times630-5times756)=2394$.



    Note that $6times630-5times756=0$ comes from substituting $756-630=126$ into $630-5times126=0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      5×630−4×756=126 and 6×630−5×756=0 sorry im not sure where you are getting these. I can see they are right and i was wondering where they came from
      $endgroup$
      – user147825
      Dec 15 '18 at 15:38










    • $begingroup$
      Edited to make it more clear.
      $endgroup$
      – SmileyCraft
      Dec 15 '18 at 15:54










    • $begingroup$
      Note that this is basically the extended Euclidean algorithm, which solves this in general.
      $endgroup$
      – SmileyCraft
      Dec 15 '18 at 15:55
















    1












    $begingroup$

    What your calculation shows is that $126$ is the greatest common divisor of $756$ and $630$ and that $756-630=126$ and $6times630-5times756=0$. Since $2394/126=19$ is not fractional, we can solve the equation. $19(756-630)+k(6times630-5times756)=2394$.



    Note that $6times630-5times756=0$ comes from substituting $756-630=126$ into $630-5times126=0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      5×630−4×756=126 and 6×630−5×756=0 sorry im not sure where you are getting these. I can see they are right and i was wondering where they came from
      $endgroup$
      – user147825
      Dec 15 '18 at 15:38










    • $begingroup$
      Edited to make it more clear.
      $endgroup$
      – SmileyCraft
      Dec 15 '18 at 15:54










    • $begingroup$
      Note that this is basically the extended Euclidean algorithm, which solves this in general.
      $endgroup$
      – SmileyCraft
      Dec 15 '18 at 15:55














    1












    1








    1





    $begingroup$

    What your calculation shows is that $126$ is the greatest common divisor of $756$ and $630$ and that $756-630=126$ and $6times630-5times756=0$. Since $2394/126=19$ is not fractional, we can solve the equation. $19(756-630)+k(6times630-5times756)=2394$.



    Note that $6times630-5times756=0$ comes from substituting $756-630=126$ into $630-5times126=0$.






    share|cite|improve this answer











    $endgroup$



    What your calculation shows is that $126$ is the greatest common divisor of $756$ and $630$ and that $756-630=126$ and $6times630-5times756=0$. Since $2394/126=19$ is not fractional, we can solve the equation. $19(756-630)+k(6times630-5times756)=2394$.



    Note that $6times630-5times756=0$ comes from substituting $756-630=126$ into $630-5times126=0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 15 '18 at 15:54

























    answered Dec 15 '18 at 15:22









    SmileyCraftSmileyCraft

    3,571518




    3,571518












    • $begingroup$
      5×630−4×756=126 and 6×630−5×756=0 sorry im not sure where you are getting these. I can see they are right and i was wondering where they came from
      $endgroup$
      – user147825
      Dec 15 '18 at 15:38










    • $begingroup$
      Edited to make it more clear.
      $endgroup$
      – SmileyCraft
      Dec 15 '18 at 15:54










    • $begingroup$
      Note that this is basically the extended Euclidean algorithm, which solves this in general.
      $endgroup$
      – SmileyCraft
      Dec 15 '18 at 15:55


















    • $begingroup$
      5×630−4×756=126 and 6×630−5×756=0 sorry im not sure where you are getting these. I can see they are right and i was wondering where they came from
      $endgroup$
      – user147825
      Dec 15 '18 at 15:38










    • $begingroup$
      Edited to make it more clear.
      $endgroup$
      – SmileyCraft
      Dec 15 '18 at 15:54










    • $begingroup$
      Note that this is basically the extended Euclidean algorithm, which solves this in general.
      $endgroup$
      – SmileyCraft
      Dec 15 '18 at 15:55
















    $begingroup$
    5×630−4×756=126 and 6×630−5×756=0 sorry im not sure where you are getting these. I can see they are right and i was wondering where they came from
    $endgroup$
    – user147825
    Dec 15 '18 at 15:38




    $begingroup$
    5×630−4×756=126 and 6×630−5×756=0 sorry im not sure where you are getting these. I can see they are right and i was wondering where they came from
    $endgroup$
    – user147825
    Dec 15 '18 at 15:38












    $begingroup$
    Edited to make it more clear.
    $endgroup$
    – SmileyCraft
    Dec 15 '18 at 15:54




    $begingroup$
    Edited to make it more clear.
    $endgroup$
    – SmileyCraft
    Dec 15 '18 at 15:54












    $begingroup$
    Note that this is basically the extended Euclidean algorithm, which solves this in general.
    $endgroup$
    – SmileyCraft
    Dec 15 '18 at 15:55




    $begingroup$
    Note that this is basically the extended Euclidean algorithm, which solves this in general.
    $endgroup$
    – SmileyCraft
    Dec 15 '18 at 15:55











    1












    $begingroup$

    You can divide all by $$126$$ and you have to solve $$6x+5y=19$$
    and then you can write $$y=4-x-frac{1+x}{5}$$
    Substitute $$frac{1+x}{5}=t$$ we get $$x=5t-1,y=5-6t$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This is the solution! +1
      $endgroup$
      – greedoid
      Dec 15 '18 at 17:41
















    1












    $begingroup$

    You can divide all by $$126$$ and you have to solve $$6x+5y=19$$
    and then you can write $$y=4-x-frac{1+x}{5}$$
    Substitute $$frac{1+x}{5}=t$$ we get $$x=5t-1,y=5-6t$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This is the solution! +1
      $endgroup$
      – greedoid
      Dec 15 '18 at 17:41














    1












    1








    1





    $begingroup$

    You can divide all by $$126$$ and you have to solve $$6x+5y=19$$
    and then you can write $$y=4-x-frac{1+x}{5}$$
    Substitute $$frac{1+x}{5}=t$$ we get $$x=5t-1,y=5-6t$$






    share|cite|improve this answer









    $endgroup$



    You can divide all by $$126$$ and you have to solve $$6x+5y=19$$
    and then you can write $$y=4-x-frac{1+x}{5}$$
    Substitute $$frac{1+x}{5}=t$$ we get $$x=5t-1,y=5-6t$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 15 '18 at 15:20









    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

    76.9k42866




    76.9k42866












    • $begingroup$
      This is the solution! +1
      $endgroup$
      – greedoid
      Dec 15 '18 at 17:41


















    • $begingroup$
      This is the solution! +1
      $endgroup$
      – greedoid
      Dec 15 '18 at 17:41
















    $begingroup$
    This is the solution! +1
    $endgroup$
    – greedoid
    Dec 15 '18 at 17:41




    $begingroup$
    This is the solution! +1
    $endgroup$
    – greedoid
    Dec 15 '18 at 17:41











    0












    $begingroup$

    If x= 1 and y= 1 then 6x+ 5y= 1. So if x= 19 and y= -19, 6x+ 5y= 19. But x= 19- 5k, y= -19+ 6k is also a solution: 6(19- 5k)+ 5(-19+ 6k)= 6(19)- 6(5)k- 5(19)+ 5(6)k= 114-95= 19 for all k.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If x= 1 and y= 1 then 6x+ 5y= 1. So if x= 19 and y= -19, 6x+ 5y= 19. But x= 19- 5k, y= -19+ 6k is also a solution: 6(19- 5k)+ 5(-19+ 6k)= 6(19)- 6(5)k- 5(19)+ 5(6)k= 114-95= 19 for all k.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If x= 1 and y= 1 then 6x+ 5y= 1. So if x= 19 and y= -19, 6x+ 5y= 19. But x= 19- 5k, y= -19+ 6k is also a solution: 6(19- 5k)+ 5(-19+ 6k)= 6(19)- 6(5)k- 5(19)+ 5(6)k= 114-95= 19 for all k.






        share|cite|improve this answer









        $endgroup$



        If x= 1 and y= 1 then 6x+ 5y= 1. So if x= 19 and y= -19, 6x+ 5y= 19. But x= 19- 5k, y= -19+ 6k is also a solution: 6(19- 5k)+ 5(-19+ 6k)= 6(19)- 6(5)k- 5(19)+ 5(6)k= 114-95= 19 for all k.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 15 '18 at 15:26









        user247327user247327

        11.2k1515




        11.2k1515






























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