Diophantine Equations (another)
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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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
$begingroup$
The usual method is to apply the extended Euclidean algorithm to $756$ and $630$.
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– Lord Shark the Unknown
Dec 15 '18 at 15:17
$begingroup$
You should first reduce the equation by dividing by 2.
$endgroup$
– dssknj
Dec 15 '18 at 15:18
add a comment |
$begingroup$
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
$endgroup$
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
edited Dec 15 '18 at 15:56
Ethan Bolker
42.1k548111
42.1k548111
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
$begingroup$
You should first reduce the equation by dividing by 2.
$endgroup$
– dssknj
Dec 15 '18 at 15:18
add a comment |
$begingroup$
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
$begingroup$
You should first reduce the equation by dividing by 2.
$endgroup$
– dssknj
Dec 15 '18 at 15:18
$begingroup$
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
$begingroup$
You should first reduce the equation by dividing by 2.
$endgroup$
– dssknj
Dec 15 '18 at 15:18
$begingroup$
You should first reduce the equation by dividing by 2.
$endgroup$
– dssknj
Dec 15 '18 at 15:18
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 $630-5times126=0$.
$endgroup$
$begingroup$
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
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– user147825
Dec 15 '18 at 15:38
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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
add a comment |
$begingroup$
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 $$x=5t-1,y=5-6t$$
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$begingroup$
This is the solution! +1
$endgroup$
– greedoid
Dec 15 '18 at 17:41
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $630-5times126=0$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $630-5times126=0$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $630-5times126=0$.
$endgroup$
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 $630-5times126=0$.
edited Dec 15 '18 at 15:54
answered Dec 15 '18 at 15:22
SmileyCraftSmileyCraft
3,401516
3,401516
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $$x=5t-1,y=5-6t$$
$endgroup$
$begingroup$
This is the solution! +1
$endgroup$
– greedoid
Dec 15 '18 at 17:41
add a comment |
$begingroup$
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 $$x=5t-1,y=5-6t$$
$endgroup$
$begingroup$
This is the solution! +1
$endgroup$
– greedoid
Dec 15 '18 at 17:41
add a comment |
$begingroup$
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 $$x=5t-1,y=5-6t$$
$endgroup$
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 $$x=5t-1,y=5-6t$$
answered Dec 15 '18 at 15:20
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
73.7k42864
$begingroup$
This is the solution! +1
$endgroup$
– greedoid
Dec 15 '18 at 17:41
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Dec 15 '18 at 15:26
user247327user247327
10.6k1515
10.6k1515
add a comment |
add a comment |
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$begingroup$
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
$begingroup$
You should first reduce the equation by dividing by 2.
$endgroup$
– dssknj
Dec 15 '18 at 15:18