$Ax + By = C$ where $A, B, C$ are positive whole numbers. Is it possible to have an infinite number of solutions to this?
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$Ax + By = C$ where $A, B, C$ are positive whole numbers. Is it possible to have an infinite number of solutions to this or will it have a specific number of solutions or just a single solution? (solutions being positive whole numbers too, i.e. both $x$ and $y$ should be positive whole numbers too.)
number-theory
edited Aug 21 at 9:31
tmaths
1,326113
asked Aug 21 at 9:27
Malik Hamza Murtaza
106
add a comment |Â
up vote
1
down vote
favorite
$Ax + By = C$ where $A, B, C$ are positive whole numbers. Is it possible to have an infinite number of solutions to this or will it have a specific number of solutions or just a single solution? (solutions being positive whole numbers too, i.e. both $x$ and $y$ should be positive whole numbers too.)
number-theory
edited Aug 21 at 9:31
tmaths
1,326113
asked Aug 21 at 9:27
Malik Hamza Murtaza
106
What do you think yourself? Hint: just assume $A = B = 1$ and think about it
– Ronald
Aug 21 at 9:29
Obviously it is finite - don't understand the rest of the question.
– metamorphy
Aug 21 at 9:29
2
$0 < x < frac CA$, $0 < y < frac CB$, ...
– Martin R
Aug 21 at 9:36
@ronald putting A=B=1 gives x+y=C where neither x nor y can be negative,will have an more than one solution but not infinite. that's what i think.
– Malik Hamza Murtaza
Aug 21 at 9:41
1
Correct, since all numbers are positive by requirement, you'll have a finite number of solutions for $x$ and $y$. And depending on $A$ and $B$ there will be more or less solutions (e.g. if $A>C$ you'll have no solutions at all). For me this approach to assume some simple values and to look how the equation behaves is a way to learn to understand the question. And from there it'll be a small step to find the solution Martin R posted.
– Ronald
Aug 21 at 9:53
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$Ax + By = C$ where $A, B, C$ are positive whole numbers. Is it possible to have an infinite number of solutions to this or will it have a specific number of solutions or just a single solution? (solutions being positive whole numbers too, i.e. both $x$ and $y$ should be positive whole numbers too.)
number-theory
edited Aug 21 at 9:31
tmaths
1,326113
asked Aug 21 at 9:27
Malik Hamza Murtaza
106
$Ax + By = C$ where $A, B, C$ are positive whole numbers. Is it possible to have an infinite number of solutions to this or will it have a specific number of solutions or just a single solution? (solutions being positive whole numbers too, i.e. both $x$ and $y$ should be positive whole numbers too.)
number-theory
edited Aug 21 at 9:31
tmaths
1,326113
asked Aug 21 at 9:27
Malik Hamza Murtaza
106
edited Aug 21 at 9:31
tmaths
1,326113
edited Aug 21 at 9:31
tmaths
1,326113
edited Aug 21 at 9:31
tmaths
1,326113
1,326113
asked Aug 21 at 9:27
Malik Hamza Murtaza
106
asked Aug 21 at 9:27
Malik Hamza Murtaza
106
asked Aug 21 at 9:27
Malik Hamza Murtaza
106
106
What do you think yourself? Hint: just assume $A = B = 1$ and think about it
– Ronald
Aug 21 at 9:29
Obviously it is finite - don't understand the rest of the question.
– metamorphy
Aug 21 at 9:29
2
$0 < x < frac CA$, $0 < y < frac CB$, ...
– Martin R
Aug 21 at 9:36
@ronald putting A=B=1 gives x+y=C where neither x nor y can be negative,will have an more than one solution but not infinite. that's what i think.
– Malik Hamza Murtaza
Aug 21 at 9:41
1
Correct, since all numbers are positive by requirement, you'll have a finite number of solutions for $x$ and $y$. And depending on $A$ and $B$ there will be more or less solutions (e.g. if $A>C$ you'll have no solutions at all). For me this approach to assume some simple values and to look how the equation behaves is a way to learn to understand the question. And from there it'll be a small step to find the solution Martin R posted.
– Ronald
Aug 21 at 9:53
add a comment |Â
What do you think yourself? Hint: just assume $A = B = 1$ and think about it
– Ronald
Aug 21 at 9:29
Obviously it is finite - don't understand the rest of the question.
– metamorphy
Aug 21 at 9:29
2
$0 < x < frac CA$, $0 < y < frac CB$, ...
– Martin R
Aug 21 at 9:36
@ronald putting A=B=1 gives x+y=C where neither x nor y can be negative,will have an more than one solution but not infinite. that's what i think.
– Malik Hamza Murtaza
Aug 21 at 9:41
1
Correct, since all numbers are positive by requirement, you'll have a finite number of solutions for $x$ and $y$. And depending on $A$ and $B$ there will be more or less solutions (e.g. if $A>C$ you'll have no solutions at all). For me this approach to assume some simple values and to look how the equation behaves is a way to learn to understand the question. And from there it'll be a small step to find the solution Martin R posted.
– Ronald
Aug 21 at 9:53
What do you think yourself? Hint: just assume $A = B = 1$ and think about it
– Ronald
Aug 21 at 9:29
What do you think yourself? Hint: just assume $A = B = 1$ and think about it
– Ronald
Aug 21 at 9:29
Obviously it is finite - don't understand the rest of the question.
– metamorphy
Aug 21 at 9:29
Obviously it is finite - don't understand the rest of the question.
– metamorphy
Aug 21 at 9:29
2
2
$0 < x < frac CA$, $0 < y < frac CB$, ...
– Martin R
Aug 21 at 9:36
$0 < x < frac CA$, $0 < y < frac CB$, ...
– Martin R
Aug 21 at 9:36
@ronald putting A=B=1 gives x+y=C where neither x nor y can be negative,will have an more than one solution but not infinite. that's what i think.
– Malik Hamza Murtaza
Aug 21 at 9:41
@ronald putting A=B=1 gives x+y=C where neither x nor y can be negative,will have an more than one solution but not infinite. that's what i think.
– Malik Hamza Murtaza
Aug 21 at 9:41
1
1
Correct, since all numbers are positive by requirement, you'll have a finite number of solutions for $x$ and $y$. And depending on $A$ and $B$ there will be more or less solutions (e.g. if $A>C$ you'll have no solutions at all). For me this approach to assume some simple values and to look how the equation behaves is a way to learn to understand the question. And from there it'll be a small step to find the solution Martin R posted.
– Ronald
Aug 21 at 9:53
Correct, since all numbers are positive by requirement, you'll have a finite number of solutions for $x$ and $y$. And depending on $A$ and $B$ there will be more or less solutions (e.g. if $A>C$ you'll have no solutions at all). For me this approach to assume some simple values and to look how the equation behaves is a way to learn to understand the question. And from there it'll be a small step to find the solution Martin R posted.
– Ronald
Aug 21 at 9:53
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
There can be no solution (e.g. $x+y=1$), exactly one solution (e.g. $x+y=2$),
or multiple solutions (e.g. $x + y = 3$).
In any case, the number of solutions is finite, since
$0 < x < frac CA$ and $0 < y < frac CB$.
answered Aug 21 at 9:41
Martin R
23.9k32743
add a comment |Â
up vote
0
down vote
Since all numbers are positive, we have
$$1le x,y <Ax+By,quadtexthencequad 1le x,yle C-1,$$
so there can't be more than $(C-1)^2$ solutions.
answered Aug 21 at 9:41
Bernard
111k635103
could you please show me how did you come up with that limit??
– Malik Hamza Murtaza
Aug 21 at 9:45
@Malik: Just order the solutions you've found yourself while assuming $A = B = 1$ in a square and count the number of entries.
– Ronald
Aug 21 at 9:56
Well, both $Ax$ and $By$ are $ge 1$, and since their sum is $C$, each has to be $le C-1)$. Note however this is not an optimal bound – it's only the simplest to obtain.
– Bernard
Aug 21 at 9:57
@Ronald And Bernard, Understood. Am i correct to assume that the actual number of solutions will be dependent on the values of A and B?
– Malik Hamza Murtaza
Aug 21 at 10:02
Just imagine $A$ and $B$ are even (e.g. 2) and $C$ is odd. It won't be hard to count all solutions, even if $C$ is incredibly large. And as I pointed out before, if either $A>C$ or $B>C$ you also won't find any solutions. So yes, the number of solutions definitely depends on the values of $A$ and $B$.
– Ronald
Aug 21 at 10:06
 |Â
show 1 more comment
up vote
0
down vote
Since $Bneq0$ any $x$ there is at most one $y$ that satisfies $Ax+By=C$.
Also for $x$ large enough we have $Ax>C$ so that $Ax+text any positive number>C$
Since $x$ is demanded to positive we conclude that there are only a finite number of $x$'s for wich there is an $y$ that satisfies $Ax+By=C$.
answered Aug 21 at 9:42
drhab
87.8k541119
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There can be no solution (e.g. $x+y=1$), exactly one solution (e.g. $x+y=2$),
or multiple solutions (e.g. $x + y = 3$).
In any case, the number of solutions is finite, since
$0 < x < frac CA$ and $0 < y < frac CB$.
answered Aug 21 at 9:41
Martin R
23.9k32743
add a comment |Â
up vote
1
down vote
accepted
There can be no solution (e.g. $x+y=1$), exactly one solution (e.g. $x+y=2$),
or multiple solutions (e.g. $x + y = 3$).
In any case, the number of solutions is finite, since
$0 < x < frac CA$ and $0 < y < frac CB$.
answered Aug 21 at 9:41
Martin R
23.9k32743
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There can be no solution (e.g. $x+y=1$), exactly one solution (e.g. $x+y=2$),
or multiple solutions (e.g. $x + y = 3$).
In any case, the number of solutions is finite, since
$0 < x < frac CA$ and $0 < y < frac CB$.
answered Aug 21 at 9:41
Martin R
23.9k32743
There can be no solution (e.g. $x+y=1$), exactly one solution (e.g. $x+y=2$),
or multiple solutions (e.g. $x + y = 3$).
In any case, the number of solutions is finite, since
$0 < x < frac CA$ and $0 < y < frac CB$.
answered Aug 21 at 9:41
Martin R
23.9k32743
answered Aug 21 at 9:41
Martin R
23.9k32743
answered Aug 21 at 9:41
Martin R
23.9k32743
answered Aug 21 at 9:41
Martin R
23.9k32743
23.9k32743
add a comment |Â
add a comment |Â
up vote
0
down vote
Since all numbers are positive, we have
$$1le x,y <Ax+By,quadtexthencequad 1le x,yle C-1,$$
so there can't be more than $(C-1)^2$ solutions.
answered Aug 21 at 9:41
Bernard
111k635103
could you please show me how did you come up with that limit??
– Malik Hamza Murtaza
Aug 21 at 9:45
@Malik: Just order the solutions you've found yourself while assuming $A = B = 1$ in a square and count the number of entries.
– Ronald
Aug 21 at 9:56
Well, both $Ax$ and $By$ are $ge 1$, and since their sum is $C$, each has to be $le C-1)$. Note however this is not an optimal bound – it's only the simplest to obtain.
– Bernard
Aug 21 at 9:57
@Ronald And Bernard, Understood. Am i correct to assume that the actual number of solutions will be dependent on the values of A and B?
– Malik Hamza Murtaza
Aug 21 at 10:02
Just imagine $A$ and $B$ are even (e.g. 2) and $C$ is odd. It won't be hard to count all solutions, even if $C$ is incredibly large. And as I pointed out before, if either $A>C$ or $B>C$ you also won't find any solutions. So yes, the number of solutions definitely depends on the values of $A$ and $B$.
– Ronald
Aug 21 at 10:06
 |Â
show 1 more comment
up vote
0
down vote
Since all numbers are positive, we have
$$1le x,y <Ax+By,quadtexthencequad 1le x,yle C-1,$$
so there can't be more than $(C-1)^2$ solutions.
answered Aug 21 at 9:41
Bernard
111k635103
could you please show me how did you come up with that limit??
– Malik Hamza Murtaza
Aug 21 at 9:45
@Malik: Just order the solutions you've found yourself while assuming $A = B = 1$ in a square and count the number of entries.
– Ronald
Aug 21 at 9:56
Well, both $Ax$ and $By$ are $ge 1$, and since their sum is $C$, each has to be $le C-1)$. Note however this is not an optimal bound – it's only the simplest to obtain.
– Bernard
Aug 21 at 9:57
@Ronald And Bernard, Understood. Am i correct to assume that the actual number of solutions will be dependent on the values of A and B?
– Malik Hamza Murtaza
Aug 21 at 10:02
Just imagine $A$ and $B$ are even (e.g. 2) and $C$ is odd. It won't be hard to count all solutions, even if $C$ is incredibly large. And as I pointed out before, if either $A>C$ or $B>C$ you also won't find any solutions. So yes, the number of solutions definitely depends on the values of $A$ and $B$.
– Ronald
Aug 21 at 10:06
 |Â
show 1 more comment
up vote
0
down vote
up vote
0
down vote
Since all numbers are positive, we have
$$1le x,y <Ax+By,quadtexthencequad 1le x,yle C-1,$$
so there can't be more than $(C-1)^2$ solutions.
answered Aug 21 at 9:41
Bernard
111k635103
Since all numbers are positive, we have
$$1le x,y <Ax+By,quadtexthencequad 1le x,yle C-1,$$
so there can't be more than $(C-1)^2$ solutions.
answered Aug 21 at 9:41
Bernard
111k635103
answered Aug 21 at 9:41
Bernard
111k635103
answered Aug 21 at 9:41
Bernard
111k635103
answered Aug 21 at 9:41
Bernard
111k635103
111k635103
could you please show me how did you come up with that limit??
– Malik Hamza Murtaza
Aug 21 at 9:45
@Malik: Just order the solutions you've found yourself while assuming $A = B = 1$ in a square and count the number of entries.
– Ronald
Aug 21 at 9:56
Well, both $Ax$ and $By$ are $ge 1$, and since their sum is $C$, each has to be $le C-1)$. Note however this is not an optimal bound – it's only the simplest to obtain.
– Bernard
Aug 21 at 9:57
@Ronald And Bernard, Understood. Am i correct to assume that the actual number of solutions will be dependent on the values of A and B?
– Malik Hamza Murtaza
Aug 21 at 10:02
Just imagine $A$ and $B$ are even (e.g. 2) and $C$ is odd. It won't be hard to count all solutions, even if $C$ is incredibly large. And as I pointed out before, if either $A>C$ or $B>C$ you also won't find any solutions. So yes, the number of solutions definitely depends on the values of $A$ and $B$.
– Ronald
Aug 21 at 10:06
 |Â
show 1 more comment
could you please show me how did you come up with that limit??
– Malik Hamza Murtaza
Aug 21 at 9:45
@Malik: Just order the solutions you've found yourself while assuming $A = B = 1$ in a square and count the number of entries.
– Ronald
Aug 21 at 9:56
Well, both $Ax$ and $By$ are $ge 1$, and since their sum is $C$, each has to be $le C-1)$. Note however this is not an optimal bound – it's only the simplest to obtain.
– Bernard
Aug 21 at 9:57
@Ronald And Bernard, Understood. Am i correct to assume that the actual number of solutions will be dependent on the values of A and B?
– Malik Hamza Murtaza
Aug 21 at 10:02
Just imagine $A$ and $B$ are even (e.g. 2) and $C$ is odd. It won't be hard to count all solutions, even if $C$ is incredibly large. And as I pointed out before, if either $A>C$ or $B>C$ you also won't find any solutions. So yes, the number of solutions definitely depends on the values of $A$ and $B$.
– Ronald
Aug 21 at 10:06
could you please show me how did you come up with that limit??
– Malik Hamza Murtaza
Aug 21 at 9:45
could you please show me how did you come up with that limit??
– Malik Hamza Murtaza
Aug 21 at 9:45
@Malik: Just order the solutions you've found yourself while assuming $A = B = 1$ in a square and count the number of entries.
– Ronald
Aug 21 at 9:56
@Malik: Just order the solutions you've found yourself while assuming $A = B = 1$ in a square and count the number of entries.
– Ronald
Aug 21 at 9:56
Well, both $Ax$ and $By$ are $ge 1$, and since their sum is $C$, each has to be $le C-1)$. Note however this is not an optimal bound – it's only the simplest to obtain.
– Bernard
Aug 21 at 9:57
Well, both $Ax$ and $By$ are $ge 1$, and since their sum is $C$, each has to be $le C-1)$. Note however this is not an optimal bound – it's only the simplest to obtain.
– Bernard
Aug 21 at 9:57
@Ronald And Bernard, Understood. Am i correct to assume that the actual number of solutions will be dependent on the values of A and B?
– Malik Hamza Murtaza
Aug 21 at 10:02
@Ronald And Bernard, Understood. Am i correct to assume that the actual number of solutions will be dependent on the values of A and B?
– Malik Hamza Murtaza
Aug 21 at 10:02
Just imagine $A$ and $B$ are even (e.g. 2) and $C$ is odd. It won't be hard to count all solutions, even if $C$ is incredibly large. And as I pointed out before, if either $A>C$ or $B>C$ you also won't find any solutions. So yes, the number of solutions definitely depends on the values of $A$ and $B$.
– Ronald
Aug 21 at 10:06
Just imagine $A$ and $B$ are even (e.g. 2) and $C$ is odd. It won't be hard to count all solutions, even if $C$ is incredibly large. And as I pointed out before, if either $A>C$ or $B>C$ you also won't find any solutions. So yes, the number of solutions definitely depends on the values of $A$ and $B$.
– Ronald
Aug 21 at 10:06
 |Â
show 1 more comment
up vote
0
down vote
Since $Bneq0$ any $x$ there is at most one $y$ that satisfies $Ax+By=C$.
Also for $x$ large enough we have $Ax>C$ so that $Ax+text any positive number>C$
Since $x$ is demanded to positive we conclude that there are only a finite number of $x$'s for wich there is an $y$ that satisfies $Ax+By=C$.
answered Aug 21 at 9:42
drhab
87.8k541119
add a comment |Â
up vote
0
down vote
Since $Bneq0$ any $x$ there is at most one $y$ that satisfies $Ax+By=C$.
Also for $x$ large enough we have $Ax>C$ so that $Ax+text any positive number>C$
Since $x$ is demanded to positive we conclude that there are only a finite number of $x$'s for wich there is an $y$ that satisfies $Ax+By=C$.
answered Aug 21 at 9:42
drhab
87.8k541119
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since $Bneq0$ any $x$ there is at most one $y$ that satisfies $Ax+By=C$.
Also for $x$ large enough we have $Ax>C$ so that $Ax+text any positive number>C$
Since $x$ is demanded to positive we conclude that there are only a finite number of $x$'s for wich there is an $y$ that satisfies $Ax+By=C$.
answered Aug 21 at 9:42
drhab
87.8k541119
Since $Bneq0$ any $x$ there is at most one $y$ that satisfies $Ax+By=C$.
Also for $x$ large enough we have $Ax>C$ so that $Ax+text any positive number>C$
Since $x$ is demanded to positive we conclude that there are only a finite number of $x$'s for wich there is an $y$ that satisfies $Ax+By=C$.
answered Aug 21 at 9:42
drhab
87.8k541119
answered Aug 21 at 9:42
drhab
87.8k541119
answered Aug 21 at 9:42
drhab
87.8k541119
answered Aug 21 at 9:42
drhab
87.8k541119
87.8k541119
add a comment |Â
add a comment |Â
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What do you think yourself? Hint: just assume $A = B = 1$ and think about it
– Ronald
Aug 21 at 9:29
Obviously it is finite - don't understand the rest of the question.
– metamorphy
Aug 21 at 9:29
2
$0 < x < frac CA$, $0 < y < frac CB$, ...
– Martin R
Aug 21 at 9:36
@ronald putting A=B=1 gives x+y=C where neither x nor y can be negative,will have an more than one solution but not infinite. that's what i think.
– Malik Hamza Murtaza
Aug 21 at 9:41
1
Correct, since all numbers are positive by requirement, you'll have a finite number of solutions for $x$ and $y$. And depending on $A$ and $B$ there will be more or less solutions (e.g. if $A>C$ you'll have no solutions at all). For me this approach to assume some simple values and to look how the equation behaves is a way to learn to understand the question. And from there it'll be a small step to find the solution Martin R posted.
– Ronald
Aug 21 at 9:53