Zero, One, Two, Three, etc
$begingroup$
Find a solution to the following system of simultaneous equations. All letters stand for integers (positive, zero, or negative), and different letters are different integers.
The preferred solution will be the one in which the difference between the largest and smallest of the values is the least.
C+E+R+O = 0
U+N+O = 1
D+O+S = 2
T+R+E+S = 3
C+U+A+T+R+O = 4
C+I+N+C+O = 5
S+E+I+S = 6
S+I+E+T+E = 7
O+C+H+O = 8
N+U+E+V+E = 9
D+I+E+Z = 10
mathematics arithmetic
asked Mar 19 at 0:30
Bernardo Recamán SantosBernardo Recamán Santos
2,7561349
$endgroup$
add a comment |
$begingroup$
Find a solution to the following system of simultaneous equations. All letters stand for integers (positive, zero, or negative), and different letters are different integers.
The preferred solution will be the one in which the difference between the largest and smallest of the values is the least.
C+E+R+O = 0
U+N+O = 1
D+O+S = 2
T+R+E+S = 3
C+U+A+T+R+O = 4
C+I+N+C+O = 5
S+E+I+S = 6
S+I+E+T+E = 7
O+C+H+O = 8
N+U+E+V+E = 9
D+I+E+Z = 10
mathematics arithmetic
asked Mar 19 at 0:30
Bernardo Recamán SantosBernardo Recamán Santos
2,7561349
$endgroup$
add a comment |
$begingroup$
Find a solution to the following system of simultaneous equations. All letters stand for integers (positive, zero, or negative), and different letters are different integers.
The preferred solution will be the one in which the difference between the largest and smallest of the values is the least.
C+E+R+O = 0
U+N+O = 1
D+O+S = 2
T+R+E+S = 3
C+U+A+T+R+O = 4
C+I+N+C+O = 5
S+E+I+S = 6
S+I+E+T+E = 7
O+C+H+O = 8
N+U+E+V+E = 9
D+I+E+Z = 10
mathematics arithmetic
asked Mar 19 at 0:30
Bernardo Recamán SantosBernardo Recamán Santos
2,7561349
$endgroup$
Find a solution to the following system of simultaneous equations. All letters stand for integers (positive, zero, or negative), and different letters are different integers.
The preferred solution will be the one in which the difference between the largest and smallest of the values is the least.
C+E+R+O = 0
U+N+O = 1
D+O+S = 2
T+R+E+S = 3
C+U+A+T+R+O = 4
C+I+N+C+O = 5
S+E+I+S = 6
S+I+E+T+E = 7
O+C+H+O = 8
N+U+E+V+E = 9
D+I+E+Z = 10
mathematics arithmetic
mathematics arithmetic
asked Mar 19 at 0:30
Bernardo Recamán SantosBernardo Recamán Santos
2,7561349
asked Mar 19 at 0:30
Bernardo Recamán SantosBernardo Recamán Santos
2,7561349
edited Mar 19 at 1:10
Bernardo Recamán Santos
asked Mar 19 at 0:30
Bernardo Recamán SantosBernardo Recamán Santos
2,7561349
asked Mar 19 at 0:30
Bernardo Recamán SantosBernardo Recamán Santos
2,7561349
asked Mar 19 at 0:30
Bernardo Recamán SantosBernardo Recamán Santos
2,7561349
2,7561349
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This system of equations is underdetermined, and it has 3 degrees of freedom.
Denoting these with $x$, $y$ and $z$, a possible substitution of the original letters is:
$text{A}=19-5x-3y-2z$
$text{C}=z$
$text{D}=10-5x-y+z$
$text{E}=6-2x-y$
$text{H}=24-8x-2y+z$
$text{I}=y$
$text{N}=13-4x-2y-z$
$text{O}=-8+4x+y-z$
$text{R}=2-2x$
$text{S}=x$
$text{T}=-5+3x+y$
$text{U}=-4+y+2z$
$text{V}=-12+8x+3y-z$
$text{Z}=-6+7x+y-z$
At this point I don't have any better idea than quasi-randomly trying $x$, $y$ and $z$ values to see if these expressions indeed produce distinct values, and if they do, calculate the span of them.
To provide a working example, $x=0$, $y=-1$, $z=1$ seems to produce distinct numbers in the range of $[-16; 27]$, so that produces a span of 43. I doubt this is minimal, feel free to use this input for further finetuning.
Update
$text{span}([0,-1,-2])=39$
$text{span}([2,3,-6])=32$
$text{span}([2,5,1])=26$
$text{span}([-3,14,-9])=24$
$text{span}([6,-11,12])=24$
also, confirmed with programming: $24$ is the minimal span if $max(|x|, |y|, |z|) le 120$
Update 2
Now it can easily be proven, that $24$ is the minimal span in general.
$text{S}-text{R}=3x+2$
If we want the span to be $24$ at most, any letter pairs should have a difference of at most $24$, so $|3x+2|le24$, so we get $|x|le8$.
But also $text{S}-text{I}$ and $text{S}-text{C}$ has to be at most $24$ (in its absolute value), which then gives $|y|le32$ and $|z|le32$, and since every tuplet in this range has been checked, we are done.
So the value of the letters can be any of the following two sets to achieve the minimal span of $24$:
A C D E H I N O R S T U V Z
10 -9 2 -2 11 14 6 3 8 -3 0 -8 15 -4
-2 12 3 5 10 -11 -1 -7 -10 6 2 9 -9 13
answered Mar 19 at 7:40
eliaselias
8,90332455
$endgroup$
$begingroup$
@JonMarkPerry, but if x=-8, |3x+2|=22
$endgroup$
– elias
Mar 19 at 9:16
$begingroup$
For my proof, the bound |x|<=8 is enough, no need to be more strict. I could not use it anyway in the next step, when trying to bound y and z.
$endgroup$
– elias
Mar 19 at 9:22
$begingroup$
I'm not stating that $|3x+2|le24$ is equivalent with $|x|le8$. I'm just saying the latter is a consequence of the former.
$endgroup$
– elias
Mar 19 at 9:27
$begingroup$
So you claim that $|3x+2|le24$ doesn't cause $|x|le8$? Could you provide a counterexample? I mean an $x$ that fulfills the first, but does not fulfill the second?
$endgroup$
– elias
Mar 19 at 9:33
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
This system of equations is underdetermined, and it has 3 degrees of freedom.
Denoting these with $x$, $y$ and $z$, a possible substitution of the original letters is:
$text{A}=19-5x-3y-2z$
$text{C}=z$
$text{D}=10-5x-y+z$
$text{E}=6-2x-y$
$text{H}=24-8x-2y+z$
$text{I}=y$
$text{N}=13-4x-2y-z$
$text{O}=-8+4x+y-z$
$text{R}=2-2x$
$text{S}=x$
$text{T}=-5+3x+y$
$text{U}=-4+y+2z$
$text{V}=-12+8x+3y-z$
$text{Z}=-6+7x+y-z$
At this point I don't have any better idea than quasi-randomly trying $x$, $y$ and $z$ values to see if these expressions indeed produce distinct values, and if they do, calculate the span of them.
To provide a working example, $x=0$, $y=-1$, $z=1$ seems to produce distinct numbers in the range of $[-16; 27]$, so that produces a span of 43. I doubt this is minimal, feel free to use this input for further finetuning.
Update
$text{span}([0,-1,-2])=39$
$text{span}([2,3,-6])=32$
$text{span}([2,5,1])=26$
$text{span}([-3,14,-9])=24$
$text{span}([6,-11,12])=24$
also, confirmed with programming: $24$ is the minimal span if $max(|x|, |y|, |z|) le 120$
Update 2
Now it can easily be proven, that $24$ is the minimal span in general.
$text{S}-text{R}=3x+2$
If we want the span to be $24$ at most, any letter pairs should have a difference of at most $24$, so $|3x+2|le24$, so we get $|x|le8$.
But also $text{S}-text{I}$ and $text{S}-text{C}$ has to be at most $24$ (in its absolute value), which then gives $|y|le32$ and $|z|le32$, and since every tuplet in this range has been checked, we are done.
So the value of the letters can be any of the following two sets to achieve the minimal span of $24$:
A C D E H I N O R S T U V Z
10 -9 2 -2 11 14 6 3 8 -3 0 -8 15 -4
-2 12 3 5 10 -11 -1 -7 -10 6 2 9 -9 13
answered Mar 19 at 7:40
eliaselias
8,90332455
$endgroup$
$begingroup$
@JonMarkPerry, but if x=-8, |3x+2|=22
$endgroup$
– elias
Mar 19 at 9:16
$begingroup$
For my proof, the bound |x|<=8 is enough, no need to be more strict. I could not use it anyway in the next step, when trying to bound y and z.
$endgroup$
– elias
Mar 19 at 9:22
$begingroup$
I'm not stating that $|3x+2|le24$ is equivalent with $|x|le8$. I'm just saying the latter is a consequence of the former.
$endgroup$
– elias
Mar 19 at 9:27
$begingroup$
So you claim that $|3x+2|le24$ doesn't cause $|x|le8$? Could you provide a counterexample? I mean an $x$ that fulfills the first, but does not fulfill the second?
$endgroup$
– elias
Mar 19 at 9:33
add a comment |
$begingroup$
This system of equations is underdetermined, and it has 3 degrees of freedom.
Denoting these with $x$, $y$ and $z$, a possible substitution of the original letters is:
$text{A}=19-5x-3y-2z$
$text{C}=z$
$text{D}=10-5x-y+z$
$text{E}=6-2x-y$
$text{H}=24-8x-2y+z$
$text{I}=y$
$text{N}=13-4x-2y-z$
$text{O}=-8+4x+y-z$
$text{R}=2-2x$
$text{S}=x$
$text{T}=-5+3x+y$
$text{U}=-4+y+2z$
$text{V}=-12+8x+3y-z$
$text{Z}=-6+7x+y-z$
At this point I don't have any better idea than quasi-randomly trying $x$, $y$ and $z$ values to see if these expressions indeed produce distinct values, and if they do, calculate the span of them.
To provide a working example, $x=0$, $y=-1$, $z=1$ seems to produce distinct numbers in the range of $[-16; 27]$, so that produces a span of 43. I doubt this is minimal, feel free to use this input for further finetuning.
Update
$text{span}([0,-1,-2])=39$
$text{span}([2,3,-6])=32$
$text{span}([2,5,1])=26$
$text{span}([-3,14,-9])=24$
$text{span}([6,-11,12])=24$
also, confirmed with programming: $24$ is the minimal span if $max(|x|, |y|, |z|) le 120$
Update 2
Now it can easily be proven, that $24$ is the minimal span in general.
$text{S}-text{R}=3x+2$
If we want the span to be $24$ at most, any letter pairs should have a difference of at most $24$, so $|3x+2|le24$, so we get $|x|le8$.
But also $text{S}-text{I}$ and $text{S}-text{C}$ has to be at most $24$ (in its absolute value), which then gives $|y|le32$ and $|z|le32$, and since every tuplet in this range has been checked, we are done.
So the value of the letters can be any of the following two sets to achieve the minimal span of $24$:
A C D E H I N O R S T U V Z
10 -9 2 -2 11 14 6 3 8 -3 0 -8 15 -4
-2 12 3 5 10 -11 -1 -7 -10 6 2 9 -9 13
answered Mar 19 at 7:40
eliaselias
8,90332455
$endgroup$
$begingroup$
@JonMarkPerry, but if x=-8, |3x+2|=22
$endgroup$
– elias
Mar 19 at 9:16
$begingroup$
For my proof, the bound |x|<=8 is enough, no need to be more strict. I could not use it anyway in the next step, when trying to bound y and z.
$endgroup$
– elias
Mar 19 at 9:22
$begingroup$
I'm not stating that $|3x+2|le24$ is equivalent with $|x|le8$. I'm just saying the latter is a consequence of the former.
$endgroup$
– elias
Mar 19 at 9:27
$begingroup$
So you claim that $|3x+2|le24$ doesn't cause $|x|le8$? Could you provide a counterexample? I mean an $x$ that fulfills the first, but does not fulfill the second?
$endgroup$
– elias
Mar 19 at 9:33
add a comment |
$begingroup$
This system of equations is underdetermined, and it has 3 degrees of freedom.
Denoting these with $x$, $y$ and $z$, a possible substitution of the original letters is:
$text{A}=19-5x-3y-2z$
$text{C}=z$
$text{D}=10-5x-y+z$
$text{E}=6-2x-y$
$text{H}=24-8x-2y+z$
$text{I}=y$
$text{N}=13-4x-2y-z$
$text{O}=-8+4x+y-z$
$text{R}=2-2x$
$text{S}=x$
$text{T}=-5+3x+y$
$text{U}=-4+y+2z$
$text{V}=-12+8x+3y-z$
$text{Z}=-6+7x+y-z$
At this point I don't have any better idea than quasi-randomly trying $x$, $y$ and $z$ values to see if these expressions indeed produce distinct values, and if they do, calculate the span of them.
To provide a working example, $x=0$, $y=-1$, $z=1$ seems to produce distinct numbers in the range of $[-16; 27]$, so that produces a span of 43. I doubt this is minimal, feel free to use this input for further finetuning.
Update
$text{span}([0,-1,-2])=39$
$text{span}([2,3,-6])=32$
$text{span}([2,5,1])=26$
$text{span}([-3,14,-9])=24$
$text{span}([6,-11,12])=24$
also, confirmed with programming: $24$ is the minimal span if $max(|x|, |y|, |z|) le 120$
Update 2
Now it can easily be proven, that $24$ is the minimal span in general.
$text{S}-text{R}=3x+2$
If we want the span to be $24$ at most, any letter pairs should have a difference of at most $24$, so $|3x+2|le24$, so we get $|x|le8$.
But also $text{S}-text{I}$ and $text{S}-text{C}$ has to be at most $24$ (in its absolute value), which then gives $|y|le32$ and $|z|le32$, and since every tuplet in this range has been checked, we are done.
So the value of the letters can be any of the following two sets to achieve the minimal span of $24$:
A C D E H I N O R S T U V Z
10 -9 2 -2 11 14 6 3 8 -3 0 -8 15 -4
-2 12 3 5 10 -11 -1 -7 -10 6 2 9 -9 13
answered Mar 19 at 7:40
eliaselias
8,90332455
$endgroup$
This system of equations is underdetermined, and it has 3 degrees of freedom.
Denoting these with $x$, $y$ and $z$, a possible substitution of the original letters is:
$text{A}=19-5x-3y-2z$
$text{C}=z$
$text{D}=10-5x-y+z$
$text{E}=6-2x-y$
$text{H}=24-8x-2y+z$
$text{I}=y$
$text{N}=13-4x-2y-z$
$text{O}=-8+4x+y-z$
$text{R}=2-2x$
$text{S}=x$
$text{T}=-5+3x+y$
$text{U}=-4+y+2z$
$text{V}=-12+8x+3y-z$
$text{Z}=-6+7x+y-z$
At this point I don't have any better idea than quasi-randomly trying $x$, $y$ and $z$ values to see if these expressions indeed produce distinct values, and if they do, calculate the span of them.
To provide a working example, $x=0$, $y=-1$, $z=1$ seems to produce distinct numbers in the range of $[-16; 27]$, so that produces a span of 43. I doubt this is minimal, feel free to use this input for further finetuning.
Update
$text{span}([0,-1,-2])=39$
$text{span}([2,3,-6])=32$
$text{span}([2,5,1])=26$
$text{span}([-3,14,-9])=24$
$text{span}([6,-11,12])=24$
also, confirmed with programming: $24$ is the minimal span if $max(|x|, |y|, |z|) le 120$
Update 2
Now it can easily be proven, that $24$ is the minimal span in general.
$text{S}-text{R}=3x+2$
If we want the span to be $24$ at most, any letter pairs should have a difference of at most $24$, so $|3x+2|le24$, so we get $|x|le8$.
But also $text{S}-text{I}$ and $text{S}-text{C}$ has to be at most $24$ (in its absolute value), which then gives $|y|le32$ and $|z|le32$, and since every tuplet in this range has been checked, we are done.
So the value of the letters can be any of the following two sets to achieve the minimal span of $24$:
A C D E H I N O R S T U V Z
10 -9 2 -2 11 14 6 3 8 -3 0 -8 15 -4
-2 12 3 5 10 -11 -1 -7 -10 6 2 9 -9 13
answered Mar 19 at 7:40
eliaselias
8,90332455
edited Mar 19 at 9:24
answered Mar 19 at 7:40
eliaselias
8,90332455
answered Mar 19 at 7:40
eliaselias
8,90332455
answered Mar 19 at 7:40
eliaselias
8,90332455
8,90332455
$begingroup$
@JonMarkPerry, but if x=-8, |3x+2|=22
$endgroup$
– elias
Mar 19 at 9:16
$begingroup$
For my proof, the bound |x|<=8 is enough, no need to be more strict. I could not use it anyway in the next step, when trying to bound y and z.
$endgroup$
– elias
Mar 19 at 9:22
$begingroup$
I'm not stating that $|3x+2|le24$ is equivalent with $|x|le8$. I'm just saying the latter is a consequence of the former.
$endgroup$
– elias
Mar 19 at 9:27
$begingroup$
So you claim that $|3x+2|le24$ doesn't cause $|x|le8$? Could you provide a counterexample? I mean an $x$ that fulfills the first, but does not fulfill the second?
$endgroup$
– elias
Mar 19 at 9:33
add a comment |
$begingroup$
@JonMarkPerry, but if x=-8, |3x+2|=22
$endgroup$
– elias
Mar 19 at 9:16
$begingroup$
For my proof, the bound |x|<=8 is enough, no need to be more strict. I could not use it anyway in the next step, when trying to bound y and z.
$endgroup$
– elias
Mar 19 at 9:22
$begingroup$
I'm not stating that $|3x+2|le24$ is equivalent with $|x|le8$. I'm just saying the latter is a consequence of the former.
$endgroup$
– elias
Mar 19 at 9:27
$begingroup$
So you claim that $|3x+2|le24$ doesn't cause $|x|le8$? Could you provide a counterexample? I mean an $x$ that fulfills the first, but does not fulfill the second?
$endgroup$
– elias
Mar 19 at 9:33
$begingroup$
@JonMarkPerry, but if x=-8, |3x+2|=22
$endgroup$
– elias
Mar 19 at 9:16
$begingroup$
@JonMarkPerry, but if x=-8, |3x+2|=22
$endgroup$
– elias
Mar 19 at 9:16
$begingroup$
For my proof, the bound |x|<=8 is enough, no need to be more strict. I could not use it anyway in the next step, when trying to bound y and z.
$endgroup$
– elias
Mar 19 at 9:22
$begingroup$
For my proof, the bound |x|<=8 is enough, no need to be more strict. I could not use it anyway in the next step, when trying to bound y and z.
$endgroup$
– elias
Mar 19 at 9:22
$begingroup$
I'm not stating that $|3x+2|le24$ is equivalent with $|x|le8$. I'm just saying the latter is a consequence of the former.
$endgroup$
– elias
Mar 19 at 9:27
$begingroup$
I'm not stating that $|3x+2|le24$ is equivalent with $|x|le8$. I'm just saying the latter is a consequence of the former.
$endgroup$
– elias
Mar 19 at 9:27
$begingroup$
So you claim that $|3x+2|le24$ doesn't cause $|x|le8$? Could you provide a counterexample? I mean an $x$ that fulfills the first, but does not fulfill the second?
$endgroup$
– elias
Mar 19 at 9:33
$begingroup$
So you claim that $|3x+2|le24$ doesn't cause $|x|le8$? Could you provide a counterexample? I mean an $x$ that fulfills the first, but does not fulfill the second?
$endgroup$
– elias
Mar 19 at 9:33
add a comment |
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