Use of diophantine equation (money problems, 3 variables)
My friend has in his wallet some notes of $displaystyle{ 20 }$ , $displaystyle{ 50 }$ and $displaystyle{ 100 }$ euros.
He has $displaystyle{ 15 }$ notes and the total value of them is $displaystyle{ 690 }$ euros. Tell me the amount of $displaystyle{ 20 }$ , $displaystyle{ 50 }$ and $displaystyle{ 100 }$ euros he has in his wallet.
My solution:
$displaystyle{ 20x+50y+100z=690 Leftrightarrow 2x+5y+10z=69 (1) }$
$displaystyle{ x+y+z=15 (2) }$
$displaystyle{ x,y,z> 0 }$
$displaystyle{ yequiv 1 mod 2 }$ and $displaystyle{ 5cdot14=70 > 69 }$
so $displaystyle{ y< 14 }$.
Then, $displaystyle{ y=1,3,5,7,9,11,13 }$.
Also $displaystyle{ xequiv 2 mod 5 }$.
Then, $displaystyle{ x=2,7,12 }$.
We also have that $displaystyle{ 10cdot7=70 >69 }$, so $displaystyle{ z<7 }$.
From the equation $displaystyle{ (2) }$ we get that $displaystyle{ (x,y) = (7,9) , (7,11) , (7,13) , (12,5) , (12,7), (12,9), (12,11), (12,13) }$ are rejected.
We know that $displaystyle{ z<7 }$, so $displaystyle{(x,y) = (2,1), (2,3), (2,5), (7,1) }$ are rejected, too.
So I check every $displaystyle{ (x,y) = (2,7) , (2,9), (2,11), (2,13)}$ (rejected because $displaystyle{z=0}$ ) ,$displaystyle{(7,3), (7,5), (7,7), (12,1), (12,3)}$ (rejected because $displaystyle{z=0}$ ) to find out which ones are ok with equations $displaystyle{ (1) , (2) }$.
The final answer is : $displaystyle{ x=7, y=5, z=3 }$.
My question:
How can I find the solution using parameters? Thanks!
number-theory diophantine-equations linear-diophantine-equations
asked Nov 29 at 15:27
Dr.Mathematics
21
add a comment |
My friend has in his wallet some notes of $displaystyle{ 20 }$ , $displaystyle{ 50 }$ and $displaystyle{ 100 }$ euros.
He has $displaystyle{ 15 }$ notes and the total value of them is $displaystyle{ 690 }$ euros. Tell me the amount of $displaystyle{ 20 }$ , $displaystyle{ 50 }$ and $displaystyle{ 100 }$ euros he has in his wallet.
My solution:
$displaystyle{ 20x+50y+100z=690 Leftrightarrow 2x+5y+10z=69 (1) }$
$displaystyle{ x+y+z=15 (2) }$
$displaystyle{ x,y,z> 0 }$
$displaystyle{ yequiv 1 mod 2 }$ and $displaystyle{ 5cdot14=70 > 69 }$
so $displaystyle{ y< 14 }$.
Then, $displaystyle{ y=1,3,5,7,9,11,13 }$.
Also $displaystyle{ xequiv 2 mod 5 }$.
Then, $displaystyle{ x=2,7,12 }$.
We also have that $displaystyle{ 10cdot7=70 >69 }$, so $displaystyle{ z<7 }$.
From the equation $displaystyle{ (2) }$ we get that $displaystyle{ (x,y) = (7,9) , (7,11) , (7,13) , (12,5) , (12,7), (12,9), (12,11), (12,13) }$ are rejected.
We know that $displaystyle{ z<7 }$, so $displaystyle{(x,y) = (2,1), (2,3), (2,5), (7,1) }$ are rejected, too.
So I check every $displaystyle{ (x,y) = (2,7) , (2,9), (2,11), (2,13)}$ (rejected because $displaystyle{z=0}$ ) ,$displaystyle{(7,3), (7,5), (7,7), (12,1), (12,3)}$ (rejected because $displaystyle{z=0}$ ) to find out which ones are ok with equations $displaystyle{ (1) , (2) }$.
The final answer is : $displaystyle{ x=7, y=5, z=3 }$.
My question:
How can I find the solution using parameters? Thanks!
number-theory diophantine-equations linear-diophantine-equations
asked Nov 29 at 15:27
Dr.Mathematics
21
I found a much easier way to get the x,y,z values but still not by using parameters.
– Dr.Mathematics
Nov 29 at 15:50
What do you mean by "by using parameters"?
– Servaes
Nov 29 at 16:37
Solving 2x+5y=W and finding x,y with a parameter and then solving W+10z=69 and finding w,z with parameters.. And then taking into account that x+y+z=15 and x,y,z>0 we can find the right parameters. Something like that..
– Dr.Mathematics
Nov 29 at 17:14
That raises the very similar question of what you mean by "...finding $x$, $y$ (or $w$, $z$) with parameters"?
– Servaes
Nov 30 at 0:09
add a comment |
My friend has in his wallet some notes of $displaystyle{ 20 }$ , $displaystyle{ 50 }$ and $displaystyle{ 100 }$ euros.
He has $displaystyle{ 15 }$ notes and the total value of them is $displaystyle{ 690 }$ euros. Tell me the amount of $displaystyle{ 20 }$ , $displaystyle{ 50 }$ and $displaystyle{ 100 }$ euros he has in his wallet.
My solution:
$displaystyle{ 20x+50y+100z=690 Leftrightarrow 2x+5y+10z=69 (1) }$
$displaystyle{ x+y+z=15 (2) }$
$displaystyle{ x,y,z> 0 }$
$displaystyle{ yequiv 1 mod 2 }$ and $displaystyle{ 5cdot14=70 > 69 }$
so $displaystyle{ y< 14 }$.
Then, $displaystyle{ y=1,3,5,7,9,11,13 }$.
Also $displaystyle{ xequiv 2 mod 5 }$.
Then, $displaystyle{ x=2,7,12 }$.
We also have that $displaystyle{ 10cdot7=70 >69 }$, so $displaystyle{ z<7 }$.
From the equation $displaystyle{ (2) }$ we get that $displaystyle{ (x,y) = (7,9) , (7,11) , (7,13) , (12,5) , (12,7), (12,9), (12,11), (12,13) }$ are rejected.
We know that $displaystyle{ z<7 }$, so $displaystyle{(x,y) = (2,1), (2,3), (2,5), (7,1) }$ are rejected, too.
So I check every $displaystyle{ (x,y) = (2,7) , (2,9), (2,11), (2,13)}$ (rejected because $displaystyle{z=0}$ ) ,$displaystyle{(7,3), (7,5), (7,7), (12,1), (12,3)}$ (rejected because $displaystyle{z=0}$ ) to find out which ones are ok with equations $displaystyle{ (1) , (2) }$.
The final answer is : $displaystyle{ x=7, y=5, z=3 }$.
My question:
How can I find the solution using parameters? Thanks!
number-theory diophantine-equations linear-diophantine-equations
asked Nov 29 at 15:27
Dr.Mathematics
21
My friend has in his wallet some notes of $displaystyle{ 20 }$ , $displaystyle{ 50 }$ and $displaystyle{ 100 }$ euros.
He has $displaystyle{ 15 }$ notes and the total value of them is $displaystyle{ 690 }$ euros. Tell me the amount of $displaystyle{ 20 }$ , $displaystyle{ 50 }$ and $displaystyle{ 100 }$ euros he has in his wallet.
My solution:
$displaystyle{ 20x+50y+100z=690 Leftrightarrow 2x+5y+10z=69 (1) }$
$displaystyle{ x+y+z=15 (2) }$
$displaystyle{ x,y,z> 0 }$
$displaystyle{ yequiv 1 mod 2 }$ and $displaystyle{ 5cdot14=70 > 69 }$
so $displaystyle{ y< 14 }$.
Then, $displaystyle{ y=1,3,5,7,9,11,13 }$.
Also $displaystyle{ xequiv 2 mod 5 }$.
Then, $displaystyle{ x=2,7,12 }$.
We also have that $displaystyle{ 10cdot7=70 >69 }$, so $displaystyle{ z<7 }$.
From the equation $displaystyle{ (2) }$ we get that $displaystyle{ (x,y) = (7,9) , (7,11) , (7,13) , (12,5) , (12,7), (12,9), (12,11), (12,13) }$ are rejected.
We know that $displaystyle{ z<7 }$, so $displaystyle{(x,y) = (2,1), (2,3), (2,5), (7,1) }$ are rejected, too.
So I check every $displaystyle{ (x,y) = (2,7) , (2,9), (2,11), (2,13)}$ (rejected because $displaystyle{z=0}$ ) ,$displaystyle{(7,3), (7,5), (7,7), (12,1), (12,3)}$ (rejected because $displaystyle{z=0}$ ) to find out which ones are ok with equations $displaystyle{ (1) , (2) }$.
The final answer is : $displaystyle{ x=7, y=5, z=3 }$.
My question:
How can I find the solution using parameters? Thanks!
number-theory diophantine-equations linear-diophantine-equations
number-theory diophantine-equations linear-diophantine-equations
asked Nov 29 at 15:27
Dr.Mathematics
21
asked Nov 29 at 15:27
Dr.Mathematics
21
asked Nov 29 at 15:27
Dr.Mathematics
21
asked Nov 29 at 15:27
Dr.Mathematics
21
asked Nov 29 at 15:27
Dr.Mathematics
21
21
I found a much easier way to get the x,y,z values but still not by using parameters.
– Dr.Mathematics
Nov 29 at 15:50
What do you mean by "by using parameters"?
– Servaes
Nov 29 at 16:37
Solving 2x+5y=W and finding x,y with a parameter and then solving W+10z=69 and finding w,z with parameters.. And then taking into account that x+y+z=15 and x,y,z>0 we can find the right parameters. Something like that..
– Dr.Mathematics
Nov 29 at 17:14
That raises the very similar question of what you mean by "...finding $x$, $y$ (or $w$, $z$) with parameters"?
– Servaes
Nov 30 at 0:09
add a comment |
I found a much easier way to get the x,y,z values but still not by using parameters.
– Dr.Mathematics
Nov 29 at 15:50
What do you mean by "by using parameters"?
– Servaes
Nov 29 at 16:37
Solving 2x+5y=W and finding x,y with a parameter and then solving W+10z=69 and finding w,z with parameters.. And then taking into account that x+y+z=15 and x,y,z>0 we can find the right parameters. Something like that..
– Dr.Mathematics
Nov 29 at 17:14
That raises the very similar question of what you mean by "...finding $x$, $y$ (or $w$, $z$) with parameters"?
– Servaes
Nov 30 at 0:09
I found a much easier way to get the x,y,z values but still not by using parameters.
– Dr.Mathematics
Nov 29 at 15:50
I found a much easier way to get the x,y,z values but still not by using parameters.
– Dr.Mathematics
Nov 29 at 15:50
What do you mean by "by using parameters"?
– Servaes
Nov 29 at 16:37
What do you mean by "by using parameters"?
– Servaes
Nov 29 at 16:37
Solving 2x+5y=W and finding x,y with a parameter and then solving W+10z=69 and finding w,z with parameters.. And then taking into account that x+y+z=15 and x,y,z>0 we can find the right parameters. Something like that..
– Dr.Mathematics
Nov 29 at 17:14
Solving 2x+5y=W and finding x,y with a parameter and then solving W+10z=69 and finding w,z with parameters.. And then taking into account that x+y+z=15 and x,y,z>0 we can find the right parameters. Something like that..
– Dr.Mathematics
Nov 29 at 17:14
That raises the very similar question of what you mean by "...finding $x$, $y$ (or $w$, $z$) with parameters"?
– Servaes
Nov 30 at 0:09
That raises the very similar question of what you mean by "...finding $x$, $y$ (or $w$, $z$) with parameters"?
– Servaes
Nov 30 at 0:09
add a comment |
2 Answers
2
active
oldest
votes
Subtracting equation (2) from equation (1) twice yields
$$3y+7z=39.$$
Since $y$ and $z$ are positive integers, there is precisely one solution which is quickly found.
answered Nov 29 at 16:34
Servaes
22.3k33793
add a comment |
"OP" has asked for Parametric solution to below mentioned simultaneous equations:
$displaystyle{ 2x+5y+10z=69 }$
$displaystyle{ x+y+z=15 }$
The solution is given below:
$(x,y,z)=[(5k+2),(k+4),(k+2) ]$
From the above mentioned solution, it can be seen that $(x,y,z)= (7,5,3)$, for $k=1$,
is the only solution that satisfies both the two simultaneous equations.
answered Dec 1 at 17:38
Sam
1
Parametric solution will ordinarily mean that any choice of a parameter will generate a solution, i.e. a family of solutions is expressed through choices of the parameter. But it isn't obvious how substitution of $x=5k+2,y=k+4,z=k+2$ would satisfy either equation. It would improve your Answer to clarify what you mean by a parametric solution.
– hardmath
Dec 1 at 18:02
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Subtracting equation (2) from equation (1) twice yields
$$3y+7z=39.$$
Since $y$ and $z$ are positive integers, there is precisely one solution which is quickly found.
answered Nov 29 at 16:34
Servaes
22.3k33793
add a comment |
Subtracting equation (2) from equation (1) twice yields
$$3y+7z=39.$$
Since $y$ and $z$ are positive integers, there is precisely one solution which is quickly found.
answered Nov 29 at 16:34
Servaes
22.3k33793
add a comment |
Subtracting equation (2) from equation (1) twice yields
$$3y+7z=39.$$
Since $y$ and $z$ are positive integers, there is precisely one solution which is quickly found.
answered Nov 29 at 16:34
Servaes
22.3k33793
Subtracting equation (2) from equation (1) twice yields
$$3y+7z=39.$$
Since $y$ and $z$ are positive integers, there is precisely one solution which is quickly found.
answered Nov 29 at 16:34
Servaes
22.3k33793
answered Nov 29 at 16:34
Servaes
22.3k33793
answered Nov 29 at 16:34
Servaes
22.3k33793
answered Nov 29 at 16:34
Servaes
22.3k33793
22.3k33793
add a comment |
add a comment |
"OP" has asked for Parametric solution to below mentioned simultaneous equations:
$displaystyle{ 2x+5y+10z=69 }$
$displaystyle{ x+y+z=15 }$
The solution is given below:
$(x,y,z)=[(5k+2),(k+4),(k+2) ]$
From the above mentioned solution, it can be seen that $(x,y,z)= (7,5,3)$, for $k=1$,
is the only solution that satisfies both the two simultaneous equations.
answered Dec 1 at 17:38
Sam
1
Parametric solution will ordinarily mean that any choice of a parameter will generate a solution, i.e. a family of solutions is expressed through choices of the parameter. But it isn't obvious how substitution of $x=5k+2,y=k+4,z=k+2$ would satisfy either equation. It would improve your Answer to clarify what you mean by a parametric solution.
– hardmath
Dec 1 at 18:02
add a comment |
"OP" has asked for Parametric solution to below mentioned simultaneous equations:
$displaystyle{ 2x+5y+10z=69 }$
$displaystyle{ x+y+z=15 }$
The solution is given below:
$(x,y,z)=[(5k+2),(k+4),(k+2) ]$
From the above mentioned solution, it can be seen that $(x,y,z)= (7,5,3)$, for $k=1$,
is the only solution that satisfies both the two simultaneous equations.
answered Dec 1 at 17:38
Sam
1
Parametric solution will ordinarily mean that any choice of a parameter will generate a solution, i.e. a family of solutions is expressed through choices of the parameter. But it isn't obvious how substitution of $x=5k+2,y=k+4,z=k+2$ would satisfy either equation. It would improve your Answer to clarify what you mean by a parametric solution.
– hardmath
Dec 1 at 18:02
add a comment |
"OP" has asked for Parametric solution to below mentioned simultaneous equations:
$displaystyle{ 2x+5y+10z=69 }$
$displaystyle{ x+y+z=15 }$
The solution is given below:
$(x,y,z)=[(5k+2),(k+4),(k+2) ]$
From the above mentioned solution, it can be seen that $(x,y,z)= (7,5,3)$, for $k=1$,
is the only solution that satisfies both the two simultaneous equations.
answered Dec 1 at 17:38
Sam
1
"OP" has asked for Parametric solution to below mentioned simultaneous equations:
$displaystyle{ 2x+5y+10z=69 }$
$displaystyle{ x+y+z=15 }$
The solution is given below:
$(x,y,z)=[(5k+2),(k+4),(k+2) ]$
From the above mentioned solution, it can be seen that $(x,y,z)= (7,5,3)$, for $k=1$,
is the only solution that satisfies both the two simultaneous equations.
answered Dec 1 at 17:38
Sam
1
answered Dec 1 at 17:38
Sam
1
answered Dec 1 at 17:38
Sam
1
answered Dec 1 at 17:38
Sam
1
1
Parametric solution will ordinarily mean that any choice of a parameter will generate a solution, i.e. a family of solutions is expressed through choices of the parameter. But it isn't obvious how substitution of $x=5k+2,y=k+4,z=k+2$ would satisfy either equation. It would improve your Answer to clarify what you mean by a parametric solution.
– hardmath
Dec 1 at 18:02
add a comment |
Parametric solution will ordinarily mean that any choice of a parameter will generate a solution, i.e. a family of solutions is expressed through choices of the parameter. But it isn't obvious how substitution of $x=5k+2,y=k+4,z=k+2$ would satisfy either equation. It would improve your Answer to clarify what you mean by a parametric solution.
– hardmath
Dec 1 at 18:02
Parametric solution will ordinarily mean that any choice of a parameter will generate a solution, i.e. a family of solutions is expressed through choices of the parameter. But it isn't obvious how substitution of $x=5k+2,y=k+4,z=k+2$ would satisfy either equation. It would improve your Answer to clarify what you mean by a parametric solution.
– hardmath
Dec 1 at 18:02
Parametric solution will ordinarily mean that any choice of a parameter will generate a solution, i.e. a family of solutions is expressed through choices of the parameter. But it isn't obvious how substitution of $x=5k+2,y=k+4,z=k+2$ would satisfy either equation. It would improve your Answer to clarify what you mean by a parametric solution.
– hardmath
Dec 1 at 18:02
add a comment |
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I found a much easier way to get the x,y,z values but still not by using parameters.
– Dr.Mathematics
Nov 29 at 15:50
What do you mean by "by using parameters"?
– Servaes
Nov 29 at 16:37
Solving 2x+5y=W and finding x,y with a parameter and then solving W+10z=69 and finding w,z with parameters.. And then taking into account that x+y+z=15 and x,y,z>0 we can find the right parameters. Something like that..
– Dr.Mathematics
Nov 29 at 17:14
That raises the very similar question of what you mean by "...finding $x$, $y$ (or $w$, $z$) with parameters"?
– Servaes
Nov 30 at 0:09