The incongruent solutions of a linear congruence












2














My question is to do with the incongruent solutions of a linear congruence.
This is the problem:




Find all integer solutions to the linear congruence $15x equiv 36 mod 57$.




I'm able to use Euclid's algorithm, the gcd etc to solve the linear Diophantine equation and get a general solution for $x$.
I get $x=48+19t$ with $tinmathbb{Z}$.



Now I am required to express my answer as a linear congruence:
So from the above it follows that $x equiv 48 mod 19 $.



However, I don't understand the next steps and would appreciate an explanation.



Notes then go on to say "now express your answer in the same modulus as the question (i.e., $57$). If we vary $t (=-2,-1,0,1,2)$ we find solutions $10,29,48,67$. But $67equiv10 mod 57$ and thus after $10,29,48$ we get no new solutions mod 57."



My questions are to do with the statement in bold:



Why does $67equiv 10 mod 57$ imply that we would get no new solutions?
Also, why are there only $3$ incongruent solutions?



(I cooked up a sort of rough explanation, but it doesn't exactly satisfy me: $x=10,29,48,67,86$ etc depending on the value if $t$ we choose.
But as $19(3)$ every $3$ solutions from $10,29,48$ will be equivalent to adding $3(19)=57$ (or a multiple of $57$) to one of $10,29,48$ and thus all the 'new' solutions will be equivalent to the original three solutions mod $57$.)






modular-arithmetic linear-diophantine-equations







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  • 2




    The residue class $48pmod{19}$ (which BTW is the same as $10pmod{19}$) splits into $57/19=3$ residue classes modulo $57$: namely, $10pmod{57}$, $29pmod{57}$, and $48pmod{57}$. The residue class $67pmod{57}$ is identical to the residue class $10pmod{57}$.
    – W-t-P
    2 days ago
















2














My question is to do with the incongruent solutions of a linear congruence.
This is the problem:




Find all integer solutions to the linear congruence $15x equiv 36 mod 57$.




I'm able to use Euclid's algorithm, the gcd etc to solve the linear Diophantine equation and get a general solution for $x$.
I get $x=48+19t$ with $tinmathbb{Z}$.



Now I am required to express my answer as a linear congruence:
So from the above it follows that $x equiv 48 mod 19 $.



However, I don't understand the next steps and would appreciate an explanation.



Notes then go on to say "now express your answer in the same modulus as the question (i.e., $57$). If we vary $t (=-2,-1,0,1,2)$ we find solutions $10,29,48,67$. But $67equiv10 mod 57$ and thus after $10,29,48$ we get no new solutions mod 57."



My questions are to do with the statement in bold:



Why does $67equiv 10 mod 57$ imply that we would get no new solutions?
Also, why are there only $3$ incongruent solutions?



(I cooked up a sort of rough explanation, but it doesn't exactly satisfy me: $x=10,29,48,67,86$ etc depending on the value if $t$ we choose.
But as $19(3)$ every $3$ solutions from $10,29,48$ will be equivalent to adding $3(19)=57$ (or a multiple of $57$) to one of $10,29,48$ and thus all the 'new' solutions will be equivalent to the original three solutions mod $57$.)






modular-arithmetic linear-diophantine-equations







share|cite|improve this question




















  • 2




    The residue class $48pmod{19}$ (which BTW is the same as $10pmod{19}$) splits into $57/19=3$ residue classes modulo $57$: namely, $10pmod{57}$, $29pmod{57}$, and $48pmod{57}$. The residue class $67pmod{57}$ is identical to the residue class $10pmod{57}$.
    – W-t-P
    2 days ago














2












2








2







My question is to do with the incongruent solutions of a linear congruence.
This is the problem:




Find all integer solutions to the linear congruence $15x equiv 36 mod 57$.




I'm able to use Euclid's algorithm, the gcd etc to solve the linear Diophantine equation and get a general solution for $x$.
I get $x=48+19t$ with $tinmathbb{Z}$.



Now I am required to express my answer as a linear congruence:
So from the above it follows that $x equiv 48 mod 19 $.



However, I don't understand the next steps and would appreciate an explanation.



Notes then go on to say "now express your answer in the same modulus as the question (i.e., $57$). If we vary $t (=-2,-1,0,1,2)$ we find solutions $10,29,48,67$. But $67equiv10 mod 57$ and thus after $10,29,48$ we get no new solutions mod 57."



My questions are to do with the statement in bold:



Why does $67equiv 10 mod 57$ imply that we would get no new solutions?
Also, why are there only $3$ incongruent solutions?



(I cooked up a sort of rough explanation, but it doesn't exactly satisfy me: $x=10,29,48,67,86$ etc depending on the value if $t$ we choose.
But as $19(3)$ every $3$ solutions from $10,29,48$ will be equivalent to adding $3(19)=57$ (or a multiple of $57$) to one of $10,29,48$ and thus all the 'new' solutions will be equivalent to the original three solutions mod $57$.)






modular-arithmetic linear-diophantine-equations







share|cite|improve this question















My question is to do with the incongruent solutions of a linear congruence.
This is the problem:




Find all integer solutions to the linear congruence $15x equiv 36 mod 57$.




I'm able to use Euclid's algorithm, the gcd etc to solve the linear Diophantine equation and get a general solution for $x$.
I get $x=48+19t$ with $tinmathbb{Z}$.



Now I am required to express my answer as a linear congruence:
So from the above it follows that $x equiv 48 mod 19 $.



However, I don't understand the next steps and would appreciate an explanation.



Notes then go on to say "now express your answer in the same modulus as the question (i.e., $57$). If we vary $t (=-2,-1,0,1,2)$ we find solutions $10,29,48,67$. But $67equiv10 mod 57$ and thus after $10,29,48$ we get no new solutions mod 57."



My questions are to do with the statement in bold:



Why does $67equiv 10 mod 57$ imply that we would get no new solutions?
Also, why are there only $3$ incongruent solutions?



(I cooked up a sort of rough explanation, but it doesn't exactly satisfy me: $x=10,29,48,67,86$ etc depending on the value if $t$ we choose.
But as $19(3)$ every $3$ solutions from $10,29,48$ will be equivalent to adding $3(19)=57$ (or a multiple of $57$) to one of $10,29,48$ and thus all the 'new' solutions will be equivalent to the original three solutions mod $57$.)






modular-arithmetic linear-diophantine-equations




modular-arithmetic linear-diophantine-equations






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edited 2 days ago









N. F. Taussig

43.6k93355




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asked 2 days ago









stochasticmrfox

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607








  • 2




    The residue class $48pmod{19}$ (which BTW is the same as $10pmod{19}$) splits into $57/19=3$ residue classes modulo $57$: namely, $10pmod{57}$, $29pmod{57}$, and $48pmod{57}$. The residue class $67pmod{57}$ is identical to the residue class $10pmod{57}$.
    – W-t-P
    2 days ago














  • 2




    The residue class $48pmod{19}$ (which BTW is the same as $10pmod{19}$) splits into $57/19=3$ residue classes modulo $57$: namely, $10pmod{57}$, $29pmod{57}$, and $48pmod{57}$. The residue class $67pmod{57}$ is identical to the residue class $10pmod{57}$.
    – W-t-P
    2 days ago








2




2




The residue class $48pmod{19}$ (which BTW is the same as $10pmod{19}$) splits into $57/19=3$ residue classes modulo $57$: namely, $10pmod{57}$, $29pmod{57}$, and $48pmod{57}$. The residue class $67pmod{57}$ is identical to the residue class $10pmod{57}$.
– W-t-P
2 days ago




The residue class $48pmod{19}$ (which BTW is the same as $10pmod{19}$) splits into $57/19=3$ residue classes modulo $57$: namely, $10pmod{57}$, $29pmod{57}$, and $48pmod{57}$. The residue class $67pmod{57}$ is identical to the residue class $10pmod{57}$.
– W-t-P
2 days ago










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$48equiv 10pmod{!19},$ so $,x = 10 + 19color{#c00}k.,$ Division $,kdiv 3,Rightarrow,color{#c00}{k = r+3n} $ for $, rin {0,1,2}$



$begin{align}{rm Substituing, yields} x &= 10+19(color{#c00}{r!+!3n})\
&= 10+19r+57n\
&= 10+19{0,1,2} + 57n\
&= {10,29,48} + 57n
end{align}$






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    $48equiv 10pmod{!19},$ so $,x = 10 + 19color{#c00}k.,$ Division $,kdiv 3,Rightarrow,color{#c00}{k = r+3n} $ for $, rin {0,1,2}$



    $begin{align}{rm Substituing, yields} x &= 10+19(color{#c00}{r!+!3n})\
    &= 10+19r+57n\
    &= 10+19{0,1,2} + 57n\
    &= {10,29,48} + 57n
    end{align}$






    share|cite|improve this answer


























      0














      $48equiv 10pmod{!19},$ so $,x = 10 + 19color{#c00}k.,$ Division $,kdiv 3,Rightarrow,color{#c00}{k = r+3n} $ for $, rin {0,1,2}$



      $begin{align}{rm Substituing, yields} x &= 10+19(color{#c00}{r!+!3n})\
      &= 10+19r+57n\
      &= 10+19{0,1,2} + 57n\
      &= {10,29,48} + 57n
      end{align}$






      share|cite|improve this answer
























        0












        0








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        $48equiv 10pmod{!19},$ so $,x = 10 + 19color{#c00}k.,$ Division $,kdiv 3,Rightarrow,color{#c00}{k = r+3n} $ for $, rin {0,1,2}$



        $begin{align}{rm Substituing, yields} x &= 10+19(color{#c00}{r!+!3n})\
        &= 10+19r+57n\
        &= 10+19{0,1,2} + 57n\
        &= {10,29,48} + 57n
        end{align}$






        share|cite|improve this answer












        $48equiv 10pmod{!19},$ so $,x = 10 + 19color{#c00}k.,$ Division $,kdiv 3,Rightarrow,color{#c00}{k = r+3n} $ for $, rin {0,1,2}$



        $begin{align}{rm Substituing, yields} x &= 10+19(color{#c00}{r!+!3n})\
        &= 10+19r+57n\
        &= 10+19{0,1,2} + 57n\
        &= {10,29,48} + 57n
        end{align}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Bill Dubuque

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