Mixing
Number of odd and even natural solutions [closed]
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I'm trying to prove or disprove the following statement:
Number of even natural solutions of $x_1+x_2+x_3+x _4=14$ is equal to the number of odd natural solutions of $x_1+x_2+x_3+x _4=14$.
I think that they should be equal because of symmetry but it does not feel true. How to approach this issue?
combinatorics
asked Jan 11 at 19:57
kickstartkickstart
1988
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closed as off-topic by amWhy, Leucippus, Abcd, eyeballfrog, Cesareo Jan 12 at 8:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I'm trying to prove or disprove the following statement:
Number of even natural solutions of $x_1+x_2+x_3+x _4=14$ is equal to the number of odd natural solutions of $x_1+x_2+x_3+x _4=14$.
I think that they should be equal because of symmetry but it does not feel true. How to approach this issue?
combinatorics
asked Jan 11 at 19:57
kickstartkickstart
1988
$endgroup$
closed as off-topic by amWhy, Leucippus, Abcd, eyeballfrog, Cesareo Jan 12 at 8:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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Did you already try to compute the actual numbers separately?
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– Dietrich Burde
Jan 11 at 19:59
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With "even natural solutions" do you mean that each x_i is even?
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– user627482
Jan 11 at 20:03
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@user627482 I guess. I just copied from the my book.
$endgroup$
– kickstart
Jan 11 at 20:07
add a comment |
$begingroup$
I'm trying to prove or disprove the following statement:
Number of even natural solutions of $x_1+x_2+x_3+x _4=14$ is equal to the number of odd natural solutions of $x_1+x_2+x_3+x _4=14$.
I think that they should be equal because of symmetry but it does not feel true. How to approach this issue?
combinatorics
asked Jan 11 at 19:57
kickstartkickstart
1988
$endgroup$
I'm trying to prove or disprove the following statement:
Number of even natural solutions of $x_1+x_2+x_3+x _4=14$ is equal to the number of odd natural solutions of $x_1+x_2+x_3+x _4=14$.
I think that they should be equal because of symmetry but it does not feel true. How to approach this issue?
combinatorics
combinatorics
asked Jan 11 at 19:57
kickstartkickstart
1988
asked Jan 11 at 19:57
kickstartkickstart
1988
asked Jan 11 at 19:57
kickstartkickstart
1988
asked Jan 11 at 19:57
kickstartkickstart
1988
asked Jan 11 at 19:57
kickstartkickstart
1988
1988
closed as off-topic by amWhy, Leucippus, Abcd, eyeballfrog, Cesareo Jan 12 at 8:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Leucippus, Abcd, eyeballfrog, Cesareo Jan 12 at 8:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Did you already try to compute the actual numbers separately?
$endgroup$
– Dietrich Burde
Jan 11 at 19:59
$begingroup$
With "even natural solutions" do you mean that each x_i is even?
$endgroup$
– user627482
Jan 11 at 20:03
$begingroup$
@user627482 I guess. I just copied from the my book.
$endgroup$
– kickstart
Jan 11 at 20:07
add a comment |
1
$begingroup$
Did you already try to compute the actual numbers separately?
$endgroup$
– Dietrich Burde
Jan 11 at 19:59
$begingroup$
With "even natural solutions" do you mean that each x_i is even?
$endgroup$
– user627482
Jan 11 at 20:03
$begingroup$
@user627482 I guess. I just copied from the my book.
$endgroup$
– kickstart
Jan 11 at 20:07
1
1
$begingroup$
Did you already try to compute the actual numbers separately?
$endgroup$
– Dietrich Burde
Jan 11 at 19:59
$begingroup$
Did you already try to compute the actual numbers separately?
$endgroup$
– Dietrich Burde
Jan 11 at 19:59
$begingroup$
With "even natural solutions" do you mean that each x_i is even?
$endgroup$
– user627482
Jan 11 at 20:03
$begingroup$
With "even natural solutions" do you mean that each x_i is even?
$endgroup$
– user627482
Jan 11 at 20:03
$begingroup$
@user627482 I guess. I just copied from the my book.
$endgroup$
– kickstart
Jan 11 at 20:07
$begingroup$
@user627482 I guess. I just copied from the my book.
$endgroup$
– kickstart
Jan 11 at 20:07
add a comment |
2 Answers
2
active
oldest
votes
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By letting $x_1'=x_1+1$, $x_2'=x_2+1$, $x_3'=x_1-1$ and $x_4'=x_4-1$, we can produce an odd solution $(x_1',x_2',x_3',x_4')$ from an even solution $(x_1,x_2,x_3,x_4)$. Since this mapping is injective, we know that there are more or equal number of odd solutions than that of even solutions. But this mapping is not surjective since e.g. $(3,1,5,5)$ is not in the image. Thus we conclude there are more odd solutions than even solutions.
answered Jan 11 at 20:09
SongSong
12.2k630
$endgroup$
add a comment |
$begingroup$
Assuming 'natural' means 'positive integer':
The total number of solutions is:$$binom{14-1}{4-1}=286$$
Even solutions are positive integer solutions to $y_1+y_2+y_3+y_4=7$, where $y_i=frac{x_i}{2}$
The number of even solutions is:$$binom{7-1}{4-1}=20$$
Odd solutions are positive integer solutions to $z_1+z_2+z_3+z_4=9$, where $z_i=frac{x_i+1}{2}$
The number of odd solutions is:$$binom{9-1}{4-1}=56$$
answered Jan 11 at 21:03
Daniel MathiasDaniel Mathias
1,10418
$endgroup$
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For the odd part, why is it $9$ and not $7$?
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– kickstart
Jan 12 at 11:45
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Because it is $(14+4)/2$
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– Daniel Mathias
Jan 12 at 11:50
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Why do you add $4$?
$endgroup$
– kickstart
Jan 12 at 14:29
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@kickstart Starting with $x_1+x_2+x_3+x _4=14$ with all $x_i$ odd, we add $1$ to each $x_i$ to get $(x_1+1)+(x_2+1)+(x_3+1)+(x _4+1)=14+4$. Divide by $2$ and let $z_i=frac{x_i}{2}$ to get $z_1+z_2+z_3+z_4=9$
$endgroup$
– Daniel Mathias
Jan 12 at 15:56
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By letting $x_1'=x_1+1$, $x_2'=x_2+1$, $x_3'=x_1-1$ and $x_4'=x_4-1$, we can produce an odd solution $(x_1',x_2',x_3',x_4')$ from an even solution $(x_1,x_2,x_3,x_4)$. Since this mapping is injective, we know that there are more or equal number of odd solutions than that of even solutions. But this mapping is not surjective since e.g. $(3,1,5,5)$ is not in the image. Thus we conclude there are more odd solutions than even solutions.
answered Jan 11 at 20:09
SongSong
12.2k630
$endgroup$
add a comment |
$begingroup$
By letting $x_1'=x_1+1$, $x_2'=x_2+1$, $x_3'=x_1-1$ and $x_4'=x_4-1$, we can produce an odd solution $(x_1',x_2',x_3',x_4')$ from an even solution $(x_1,x_2,x_3,x_4)$. Since this mapping is injective, we know that there are more or equal number of odd solutions than that of even solutions. But this mapping is not surjective since e.g. $(3,1,5,5)$ is not in the image. Thus we conclude there are more odd solutions than even solutions.
answered Jan 11 at 20:09
SongSong
12.2k630
$endgroup$
add a comment |
$begingroup$
By letting $x_1'=x_1+1$, $x_2'=x_2+1$, $x_3'=x_1-1$ and $x_4'=x_4-1$, we can produce an odd solution $(x_1',x_2',x_3',x_4')$ from an even solution $(x_1,x_2,x_3,x_4)$. Since this mapping is injective, we know that there are more or equal number of odd solutions than that of even solutions. But this mapping is not surjective since e.g. $(3,1,5,5)$ is not in the image. Thus we conclude there are more odd solutions than even solutions.
answered Jan 11 at 20:09
SongSong
12.2k630
$endgroup$
By letting $x_1'=x_1+1$, $x_2'=x_2+1$, $x_3'=x_1-1$ and $x_4'=x_4-1$, we can produce an odd solution $(x_1',x_2',x_3',x_4')$ from an even solution $(x_1,x_2,x_3,x_4)$. Since this mapping is injective, we know that there are more or equal number of odd solutions than that of even solutions. But this mapping is not surjective since e.g. $(3,1,5,5)$ is not in the image. Thus we conclude there are more odd solutions than even solutions.
answered Jan 11 at 20:09
SongSong
12.2k630
edited Jan 11 at 21:46
answered Jan 11 at 20:09
SongSong
12.2k630
answered Jan 11 at 20:09
SongSong
12.2k630
answered Jan 11 at 20:09
SongSong
12.2k630
12.2k630
add a comment |
add a comment |
$begingroup$
Assuming 'natural' means 'positive integer':
The total number of solutions is:$$binom{14-1}{4-1}=286$$
Even solutions are positive integer solutions to $y_1+y_2+y_3+y_4=7$, where $y_i=frac{x_i}{2}$
The number of even solutions is:$$binom{7-1}{4-1}=20$$
Odd solutions are positive integer solutions to $z_1+z_2+z_3+z_4=9$, where $z_i=frac{x_i+1}{2}$
The number of odd solutions is:$$binom{9-1}{4-1}=56$$
answered Jan 11 at 21:03
Daniel MathiasDaniel Mathias
1,10418
$endgroup$
$begingroup$
For the odd part, why is it $9$ and not $7$?
$endgroup$
– kickstart
Jan 12 at 11:45
$begingroup$
Because it is $(14+4)/2$
$endgroup$
– Daniel Mathias
Jan 12 at 11:50
$begingroup$
Why do you add $4$?
$endgroup$
– kickstart
Jan 12 at 14:29
$begingroup$
@kickstart Starting with $x_1+x_2+x_3+x _4=14$ with all $x_i$ odd, we add $1$ to each $x_i$ to get $(x_1+1)+(x_2+1)+(x_3+1)+(x _4+1)=14+4$. Divide by $2$ and let $z_i=frac{x_i}{2}$ to get $z_1+z_2+z_3+z_4=9$
$endgroup$
– Daniel Mathias
Jan 12 at 15:56
add a comment |
$begingroup$
Assuming 'natural' means 'positive integer':
The total number of solutions is:$$binom{14-1}{4-1}=286$$
Even solutions are positive integer solutions to $y_1+y_2+y_3+y_4=7$, where $y_i=frac{x_i}{2}$
The number of even solutions is:$$binom{7-1}{4-1}=20$$
Odd solutions are positive integer solutions to $z_1+z_2+z_3+z_4=9$, where $z_i=frac{x_i+1}{2}$
The number of odd solutions is:$$binom{9-1}{4-1}=56$$
answered Jan 11 at 21:03
Daniel MathiasDaniel Mathias
1,10418
$endgroup$
$begingroup$
For the odd part, why is it $9$ and not $7$?
$endgroup$
– kickstart
Jan 12 at 11:45
$begingroup$
Because it is $(14+4)/2$
$endgroup$
– Daniel Mathias
Jan 12 at 11:50
$begingroup$
Why do you add $4$?
$endgroup$
– kickstart
Jan 12 at 14:29
$begingroup$
@kickstart Starting with $x_1+x_2+x_3+x _4=14$ with all $x_i$ odd, we add $1$ to each $x_i$ to get $(x_1+1)+(x_2+1)+(x_3+1)+(x _4+1)=14+4$. Divide by $2$ and let $z_i=frac{x_i}{2}$ to get $z_1+z_2+z_3+z_4=9$
$endgroup$
– Daniel Mathias
Jan 12 at 15:56
add a comment |
$begingroup$
Assuming 'natural' means 'positive integer':
The total number of solutions is:$$binom{14-1}{4-1}=286$$
Even solutions are positive integer solutions to $y_1+y_2+y_3+y_4=7$, where $y_i=frac{x_i}{2}$
The number of even solutions is:$$binom{7-1}{4-1}=20$$
Odd solutions are positive integer solutions to $z_1+z_2+z_3+z_4=9$, where $z_i=frac{x_i+1}{2}$
The number of odd solutions is:$$binom{9-1}{4-1}=56$$
answered Jan 11 at 21:03
Daniel MathiasDaniel Mathias
1,10418
$endgroup$
Assuming 'natural' means 'positive integer':
The total number of solutions is:$$binom{14-1}{4-1}=286$$
Even solutions are positive integer solutions to $y_1+y_2+y_3+y_4=7$, where $y_i=frac{x_i}{2}$
The number of even solutions is:$$binom{7-1}{4-1}=20$$
Odd solutions are positive integer solutions to $z_1+z_2+z_3+z_4=9$, where $z_i=frac{x_i+1}{2}$
The number of odd solutions is:$$binom{9-1}{4-1}=56$$
answered Jan 11 at 21:03
Daniel MathiasDaniel Mathias
1,10418
answered Jan 11 at 21:03
Daniel MathiasDaniel Mathias
1,10418
answered Jan 11 at 21:03
Daniel MathiasDaniel Mathias
1,10418
answered Jan 11 at 21:03
Daniel MathiasDaniel Mathias
1,10418
1,10418
$begingroup$
For the odd part, why is it $9$ and not $7$?
$endgroup$
– kickstart
Jan 12 at 11:45
$begingroup$
Because it is $(14+4)/2$
$endgroup$
– Daniel Mathias
Jan 12 at 11:50
$begingroup$
Why do you add $4$?
$endgroup$
– kickstart
Jan 12 at 14:29
$begingroup$
@kickstart Starting with $x_1+x_2+x_3+x _4=14$ with all $x_i$ odd, we add $1$ to each $x_i$ to get $(x_1+1)+(x_2+1)+(x_3+1)+(x _4+1)=14+4$. Divide by $2$ and let $z_i=frac{x_i}{2}$ to get $z_1+z_2+z_3+z_4=9$
$endgroup$
– Daniel Mathias
Jan 12 at 15:56
add a comment |
$begingroup$
For the odd part, why is it $9$ and not $7$?
$endgroup$
– kickstart
Jan 12 at 11:45
$begingroup$
Because it is $(14+4)/2$
$endgroup$
– Daniel Mathias
Jan 12 at 11:50
$begingroup$
Why do you add $4$?
$endgroup$
– kickstart
Jan 12 at 14:29
$begingroup$
@kickstart Starting with $x_1+x_2+x_3+x _4=14$ with all $x_i$ odd, we add $1$ to each $x_i$ to get $(x_1+1)+(x_2+1)+(x_3+1)+(x _4+1)=14+4$. Divide by $2$ and let $z_i=frac{x_i}{2}$ to get $z_1+z_2+z_3+z_4=9$
$endgroup$
– Daniel Mathias
Jan 12 at 15:56
$begingroup$
For the odd part, why is it $9$ and not $7$?
$endgroup$
– kickstart
Jan 12 at 11:45
$begingroup$
For the odd part, why is it $9$ and not $7$?
$endgroup$
– kickstart
Jan 12 at 11:45
$begingroup$
Because it is $(14+4)/2$
$endgroup$
– Daniel Mathias
Jan 12 at 11:50
$begingroup$
Because it is $(14+4)/2$
$endgroup$
– Daniel Mathias
Jan 12 at 11:50
$begingroup$
Why do you add $4$?
$endgroup$
– kickstart
Jan 12 at 14:29
$begingroup$
Why do you add $4$?
$endgroup$
– kickstart
Jan 12 at 14:29
$begingroup$
@kickstart Starting with $x_1+x_2+x_3+x _4=14$ with all $x_i$ odd, we add $1$ to each $x_i$ to get $(x_1+1)+(x_2+1)+(x_3+1)+(x _4+1)=14+4$. Divide by $2$ and let $z_i=frac{x_i}{2}$ to get $z_1+z_2+z_3+z_4=9$
$endgroup$
– Daniel Mathias
Jan 12 at 15:56
$begingroup$
@kickstart Starting with $x_1+x_2+x_3+x _4=14$ with all $x_i$ odd, we add $1$ to each $x_i$ to get $(x_1+1)+(x_2+1)+(x_3+1)+(x _4+1)=14+4$. Divide by $2$ and let $z_i=frac{x_i}{2}$ to get $z_1+z_2+z_3+z_4=9$
$endgroup$
– Daniel Mathias
Jan 12 at 15:56
add a comment |
1
$begingroup$
Did you already try to compute the actual numbers separately?
$endgroup$
– Dietrich Burde
Jan 11 at 19:59
$begingroup$
With "even natural solutions" do you mean that each x_i is even?
$endgroup$
– user627482
Jan 11 at 20:03
$begingroup$
@user627482 I guess. I just copied from the my book.
$endgroup$
– kickstart
Jan 11 at 20:07