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The incongruent solutions of a linear congruence
$begingroup$
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
edited Jan 5 at 11:16
N. F. Taussig
44k93355
asked Jan 5 at 9:54
stochasticmrfoxstochasticmrfox
837
$endgroup$
add a comment |
$begingroup$
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
edited Jan 5 at 11:16
N. F. Taussig
44k93355
asked Jan 5 at 9:54
stochasticmrfoxstochasticmrfox
837
$endgroup$
2
$begingroup$
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 $10pmod{57}$.
$endgroup$
– W-t-P
Jan 5 at 11:18
add a comment |
$begingroup$
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
edited Jan 5 at 11:16
N. F. Taussig
44k93355
asked Jan 5 at 9:54
stochasticmrfoxstochasticmrfox
837
$endgroup$
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
edited Jan 5 at 11:16
N. F. Taussig
44k93355
asked Jan 5 at 9:54
stochasticmrfoxstochasticmrfox
837
edited Jan 5 at 11:16
N. F. Taussig
44k93355
asked Jan 5 at 9:54
stochasticmrfoxstochasticmrfox
837
edited Jan 5 at 11:16
N. F. Taussig
44k93355
edited Jan 5 at 11:16
N. F. Taussig
44k93355
edited Jan 5 at 11:16
N. F. Taussig
44k93355
44k93355
asked Jan 5 at 9:54
stochasticmrfoxstochasticmrfox
837
asked Jan 5 at 9:54
stochasticmrfoxstochasticmrfox
837
asked Jan 5 at 9:54
stochasticmrfoxstochasticmrfox
837
837
2
$begingroup$
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 $10pmod{57}$.
$endgroup$
– W-t-P
Jan 5 at 11:18
add a comment |
2
$begingroup$
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 $10pmod{57}$.
$endgroup$
– W-t-P
Jan 5 at 11:18
2
2
$begingroup$
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 $10pmod{57}$.
$endgroup$
– W-t-P
Jan 5 at 11:18
$begingroup$
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 $10pmod{57}$.
$endgroup$
– W-t-P
Jan 5 at 11:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$48equiv 10pmod{!19},$ so $,x = 10 + 19,color{#c00}k.,$ Division $,kdiv 3,Rightarrow,color{#c00}{k =color{#0a0} r+3n} $ for $, color{#0a0}{rin {0,1,2}}$
$begin{align}{rm Substituing, we find } x &= 10+19(color{#c00}{color{#0a0}r!+!3n})\
&= 10+19,color{#0a0}r+57n\
&= 10+19color{#0a0}{{0,1,2}}! + 57n\
&= {10,29,48} + 57n
end{align}$
answered Jan 5 at 23:42
Bill DubuqueBill Dubuque
209k29191637
$endgroup$
add a comment |
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$begingroup$
$48equiv 10pmod{!19},$ so $,x = 10 + 19,color{#c00}k.,$ Division $,kdiv 3,Rightarrow,color{#c00}{k =color{#0a0} r+3n} $ for $, color{#0a0}{rin {0,1,2}}$
$begin{align}{rm Substituing, we find } x &= 10+19(color{#c00}{color{#0a0}r!+!3n})\
&= 10+19,color{#0a0}r+57n\
&= 10+19color{#0a0}{{0,1,2}}! + 57n\
&= {10,29,48} + 57n
end{align}$
answered Jan 5 at 23:42
Bill DubuqueBill Dubuque
209k29191637
$endgroup$
add a comment |
$begingroup$
$48equiv 10pmod{!19},$ so $,x = 10 + 19,color{#c00}k.,$ Division $,kdiv 3,Rightarrow,color{#c00}{k =color{#0a0} r+3n} $ for $, color{#0a0}{rin {0,1,2}}$
$begin{align}{rm Substituing, we find } x &= 10+19(color{#c00}{color{#0a0}r!+!3n})\
&= 10+19,color{#0a0}r+57n\
&= 10+19color{#0a0}{{0,1,2}}! + 57n\
&= {10,29,48} + 57n
end{align}$
answered Jan 5 at 23:42
Bill DubuqueBill Dubuque
209k29191637
$endgroup$
add a comment |
$begingroup$
$48equiv 10pmod{!19},$ so $,x = 10 + 19,color{#c00}k.,$ Division $,kdiv 3,Rightarrow,color{#c00}{k =color{#0a0} r+3n} $ for $, color{#0a0}{rin {0,1,2}}$
$begin{align}{rm Substituing, we find } x &= 10+19(color{#c00}{color{#0a0}r!+!3n})\
&= 10+19,color{#0a0}r+57n\
&= 10+19color{#0a0}{{0,1,2}}! + 57n\
&= {10,29,48} + 57n
end{align}$
answered Jan 5 at 23:42
Bill DubuqueBill Dubuque
209k29191637
$endgroup$
$48equiv 10pmod{!19},$ so $,x = 10 + 19,color{#c00}k.,$ Division $,kdiv 3,Rightarrow,color{#c00}{k =color{#0a0} r+3n} $ for $, color{#0a0}{rin {0,1,2}}$
$begin{align}{rm Substituing, we find } x &= 10+19(color{#c00}{color{#0a0}r!+!3n})\
&= 10+19,color{#0a0}r+57n\
&= 10+19color{#0a0}{{0,1,2}}! + 57n\
&= {10,29,48} + 57n
end{align}$
answered Jan 5 at 23:42
Bill DubuqueBill Dubuque
209k29191637
edited 2 days ago
answered Jan 5 at 23:42
Bill DubuqueBill Dubuque
209k29191637
answered Jan 5 at 23:42
Bill DubuqueBill Dubuque
209k29191637
answered Jan 5 at 23:42
Bill DubuqueBill Dubuque
209k29191637
209k29191637
add a comment |
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$begingroup$
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 $10pmod{57}$.
$endgroup$
– W-t-P
Jan 5 at 11:18