Zero, One, Two, Three, etc The Next CEO of Stack OverflowThree secret numbers (Variant 1 and Variant 2)A new six-sided dieSquares in seriesSum of three six digit numbersFind the value of $bigstar$: Puzzle 3 - SubstitutionFind the value of $bigstar$: Puzzle 4 - In BetweenFind the value of $bigstar$: Puzzle 7 - Boss BattleFind the value of $bigstar$: Puzzle 8 - InequalityDistinct digits but Close EnoughA 4 x 4 Magic Square with Pairwise Relatively Prime Entries

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Zero, One, Two, Three, etc



The Next CEO of Stack OverflowThree secret numbers (Variant 1 and Variant 2)A new six-sided dieSquares in seriesSum of three six digit numbersFind the value of $bigstar$: Puzzle 3 - SubstitutionFind the value of $bigstar$: Puzzle 4 - In BetweenFind the value of $bigstar$: Puzzle 7 - Boss BattleFind the value of $bigstar$: Puzzle 8 - InequalityDistinct digits but Close EnoughA 4 x 4 Magic Square with Pairwise Relatively Prime Entries










10












$begingroup$


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










share|improve this question











$endgroup$
















    10












    $begingroup$


    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










    share|improve this question











    $endgroup$














      10












      10








      10





      $begingroup$


      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










      share|improve this question











      $endgroup$




      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 19 at 1:10







      Bernardo Recamán Santos

















      asked Mar 19 at 0:30









      Bernardo Recamán SantosBernardo Recamán Santos

      2,7611349




      2,7611349




















          1 Answer
          1






          active

          oldest

          votes


















          10












          $begingroup$

          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:
          $textA=19-5x-3y-2z$
          $textC=z$
          $textD=10-5x-y+z$
          $textE=6-2x-y$
          $textH=24-8x-2y+z$
          $textI=y$
          $textN=13-4x-2y-z$
          $textO=-8+4x+y-z$
          $textR=2-2x$
          $textS=x$
          $textT=-5+3x+y$
          $textU=-4+y+2z$
          $textV=-12+8x+3y-z$
          $textZ=-6+7x+y-z$



          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
          $textspan([0,-1,-2])=39$
          $textspan([2,3,-6])=32$
          $textspan([2,5,1])=26$
          $textspan([-3,14,-9])=24$
          $textspan([6,-11,12])=24$

          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.
          $textS-textR=3x+2$

          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 $textS-textI$ and $textS-textC$ 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





          share|improve this answer











          $endgroup$












          • $begingroup$
            @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











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          10












          $begingroup$

          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:
          $textA=19-5x-3y-2z$
          $textC=z$
          $textD=10-5x-y+z$
          $textE=6-2x-y$
          $textH=24-8x-2y+z$
          $textI=y$
          $textN=13-4x-2y-z$
          $textO=-8+4x+y-z$
          $textR=2-2x$
          $textS=x$
          $textT=-5+3x+y$
          $textU=-4+y+2z$
          $textV=-12+8x+3y-z$
          $textZ=-6+7x+y-z$



          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
          $textspan([0,-1,-2])=39$
          $textspan([2,3,-6])=32$
          $textspan([2,5,1])=26$
          $textspan([-3,14,-9])=24$
          $textspan([6,-11,12])=24$

          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.
          $textS-textR=3x+2$

          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 $textS-textI$ and $textS-textC$ 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





          share|improve this answer











          $endgroup$












          • $begingroup$
            @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















          10












          $begingroup$

          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:
          $textA=19-5x-3y-2z$
          $textC=z$
          $textD=10-5x-y+z$
          $textE=6-2x-y$
          $textH=24-8x-2y+z$
          $textI=y$
          $textN=13-4x-2y-z$
          $textO=-8+4x+y-z$
          $textR=2-2x$
          $textS=x$
          $textT=-5+3x+y$
          $textU=-4+y+2z$
          $textV=-12+8x+3y-z$
          $textZ=-6+7x+y-z$



          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
          $textspan([0,-1,-2])=39$
          $textspan([2,3,-6])=32$
          $textspan([2,5,1])=26$
          $textspan([-3,14,-9])=24$
          $textspan([6,-11,12])=24$

          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.
          $textS-textR=3x+2$

          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 $textS-textI$ and $textS-textC$ 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





          share|improve this answer











          $endgroup$












          • $begingroup$
            @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













          10












          10








          10





          $begingroup$

          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:
          $textA=19-5x-3y-2z$
          $textC=z$
          $textD=10-5x-y+z$
          $textE=6-2x-y$
          $textH=24-8x-2y+z$
          $textI=y$
          $textN=13-4x-2y-z$
          $textO=-8+4x+y-z$
          $textR=2-2x$
          $textS=x$
          $textT=-5+3x+y$
          $textU=-4+y+2z$
          $textV=-12+8x+3y-z$
          $textZ=-6+7x+y-z$



          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
          $textspan([0,-1,-2])=39$
          $textspan([2,3,-6])=32$
          $textspan([2,5,1])=26$
          $textspan([-3,14,-9])=24$
          $textspan([6,-11,12])=24$

          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.
          $textS-textR=3x+2$

          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 $textS-textI$ and $textS-textC$ 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





          share|improve this answer











          $endgroup$



          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:
          $textA=19-5x-3y-2z$
          $textC=z$
          $textD=10-5x-y+z$
          $textE=6-2x-y$
          $textH=24-8x-2y+z$
          $textI=y$
          $textN=13-4x-2y-z$
          $textO=-8+4x+y-z$
          $textR=2-2x$
          $textS=x$
          $textT=-5+3x+y$
          $textU=-4+y+2z$
          $textV=-12+8x+3y-z$
          $textZ=-6+7x+y-z$



          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
          $textspan([0,-1,-2])=39$
          $textspan([2,3,-6])=32$
          $textspan([2,5,1])=26$
          $textspan([-3,14,-9])=24$
          $textspan([6,-11,12])=24$

          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.
          $textS-textR=3x+2$

          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 $textS-textI$ and $textS-textC$ 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






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Mar 19 at 9:24

























          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
















          • $begingroup$
            @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

















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