Number of solutions in linear equation with 3 variables
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Is there a way how to determine number of solutions in linear equation like this: $ax + by + cz = d$, where $a,b,c,x,y,z,d$ are non-negative integers and $a,b,c,d$ are known?
diophantine-equations
edited Nov 14 at 20:10
Alex D
494218
asked Nov 14 at 19:47
Andreisk
61
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up vote
1
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Is there a way how to determine number of solutions in linear equation like this: $ax + by + cz = d$, where $a,b,c,x,y,z,d$ are non-negative integers and $a,b,c,d$ are known?
diophantine-equations
edited Nov 14 at 20:10
Alex D
494218
asked Nov 14 at 19:47
Andreisk
61
2
You can use generating functions and look at the coefficient of $X^d$ in the expansion of $(1+X^a+X^{2a}+X^{3a}+dots)(1+X^b+X^{2b}+dots)(1+X^c+X^{2c}+dots)$
– JMoravitz
Nov 14 at 19:50
See also this question and similar ones.
– Dietrich Burde
Nov 14 at 19:57
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is there a way how to determine number of solutions in linear equation like this: $ax + by + cz = d$, where $a,b,c,x,y,z,d$ are non-negative integers and $a,b,c,d$ are known?
diophantine-equations
edited Nov 14 at 20:10
Alex D
494218
asked Nov 14 at 19:47
Andreisk
61
Is there a way how to determine number of solutions in linear equation like this: $ax + by + cz = d$, where $a,b,c,x,y,z,d$ are non-negative integers and $a,b,c,d$ are known?
diophantine-equations
diophantine-equations
edited Nov 14 at 20:10
Alex D
494218
asked Nov 14 at 19:47
Andreisk
61
edited Nov 14 at 20:10
Alex D
494218
asked Nov 14 at 19:47
Andreisk
61
edited Nov 14 at 20:10
Alex D
494218
edited Nov 14 at 20:10
Alex D
494218
edited Nov 14 at 20:10
Alex D
494218
494218
asked Nov 14 at 19:47
Andreisk
61
asked Nov 14 at 19:47
Andreisk
61
asked Nov 14 at 19:47
Andreisk
61
61
2
You can use generating functions and look at the coefficient of $X^d$ in the expansion of $(1+X^a+X^{2a}+X^{3a}+dots)(1+X^b+X^{2b}+dots)(1+X^c+X^{2c}+dots)$
– JMoravitz
Nov 14 at 19:50
See also this question and similar ones.
– Dietrich Burde
Nov 14 at 19:57
add a comment |
2
You can use generating functions and look at the coefficient of $X^d$ in the expansion of $(1+X^a+X^{2a}+X^{3a}+dots)(1+X^b+X^{2b}+dots)(1+X^c+X^{2c}+dots)$
– JMoravitz
Nov 14 at 19:50
See also this question and similar ones.
– Dietrich Burde
Nov 14 at 19:57
2
2
You can use generating functions and look at the coefficient of $X^d$ in the expansion of $(1+X^a+X^{2a}+X^{3a}+dots)(1+X^b+X^{2b}+dots)(1+X^c+X^{2c}+dots)$
– JMoravitz
Nov 14 at 19:50
You can use generating functions and look at the coefficient of $X^d$ in the expansion of $(1+X^a+X^{2a}+X^{3a}+dots)(1+X^b+X^{2b}+dots)(1+X^c+X^{2c}+dots)$
– JMoravitz
Nov 14 at 19:50
See also this question and similar ones.
– Dietrich Burde
Nov 14 at 19:57
See also this question and similar ones.
– Dietrich Burde
Nov 14 at 19:57
add a comment |
1 Answer
1
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Above equation shown below:
$ax + by + cz = d$ -------$(1)$
For (a, b, c, d)= (3, 2, 5, 32), equation $(1)$ has parametric solution given below:
$x=(1-v)$
$y=(-5u+9v+87)$
$z=(2u-3v-29)$
For $(u,v)$ = $(15, 0)$ we get,
$(x, y, z)$ = $(1, 12, 1)$
answered Nov 15 at 16:15
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Above equation shown below:
$ax + by + cz = d$ -------$(1)$
For (a, b, c, d)= (3, 2, 5, 32), equation $(1)$ has parametric solution given below:
$x=(1-v)$
$y=(-5u+9v+87)$
$z=(2u-3v-29)$
For $(u,v)$ = $(15, 0)$ we get,
$(x, y, z)$ = $(1, 12, 1)$
answered Nov 15 at 16:15
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
Above equation shown below:
$ax + by + cz = d$ -------$(1)$
For (a, b, c, d)= (3, 2, 5, 32), equation $(1)$ has parametric solution given below:
$x=(1-v)$
$y=(-5u+9v+87)$
$z=(2u-3v-29)$
For $(u,v)$ = $(15, 0)$ we get,
$(x, y, z)$ = $(1, 12, 1)$
answered Nov 15 at 16:15
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
Above equation shown below:
$ax + by + cz = d$ -------$(1)$
For (a, b, c, d)= (3, 2, 5, 32), equation $(1)$ has parametric solution given below:
$x=(1-v)$
$y=(-5u+9v+87)$
$z=(2u-3v-29)$
For $(u,v)$ = $(15, 0)$ we get,
$(x, y, z)$ = $(1, 12, 1)$
answered Nov 15 at 16:15
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Above equation shown below:
$ax + by + cz = d$ -------$(1)$
For (a, b, c, d)= (3, 2, 5, 32), equation $(1)$ has parametric solution given below:
$x=(1-v)$
$y=(-5u+9v+87)$
$z=(2u-3v-29)$
For $(u,v)$ = $(15, 0)$ we get,
$(x, y, z)$ = $(1, 12, 1)$
answered Nov 15 at 16:15
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 15 at 16:15
Sam
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 15 at 16:15
Sam
1
answered Nov 15 at 16:15
Sam
1
1
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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You can use generating functions and look at the coefficient of $X^d$ in the expansion of $(1+X^a+X^{2a}+X^{3a}+dots)(1+X^b+X^{2b}+dots)(1+X^c+X^{2c}+dots)$
– JMoravitz
Nov 14 at 19:50
See also this question and similar ones.
– Dietrich Burde
Nov 14 at 19:57