Math puzzle - sudoku like











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Sudoku



I am having problems solving this puzzle. The sum of each 3x3 square should be 2019, how to find the number in the bottom right corner? Labelling each field we can gain some information about the numbers by subtracting two neighbouring squares, but is looks like a very cumbersome and long proces. And am not even sure that it would solve the problem. Maybe one can find some invariant to use?










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    up vote
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    Sudoku



    I am having problems solving this puzzle. The sum of each 3x3 square should be 2019, how to find the number in the bottom right corner? Labelling each field we can gain some information about the numbers by subtracting two neighbouring squares, but is looks like a very cumbersome and long proces. And am not even sure that it would solve the problem. Maybe one can find some invariant to use?










    share|cite|improve this question
























      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      Sudoku



      I am having problems solving this puzzle. The sum of each 3x3 square should be 2019, how to find the number in the bottom right corner? Labelling each field we can gain some information about the numbers by subtracting two neighbouring squares, but is looks like a very cumbersome and long proces. And am not even sure that it would solve the problem. Maybe one can find some invariant to use?










      share|cite|improve this question













      Sudoku



      I am having problems solving this puzzle. The sum of each 3x3 square should be 2019, how to find the number in the bottom right corner? Labelling each field we can gain some information about the numbers by subtracting two neighbouring squares, but is looks like a very cumbersome and long proces. And am not even sure that it would solve the problem. Maybe one can find some invariant to use?







      combinatorics number-theory






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      asked 9 hours ago









      Nikolaj K

      475




      475






















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          Note that the sum of the top six elements in the first column is the same as the sum of the top six elements in the last column because you can make four $3 times 3$ squares up against the top left and get a sum of $4cdot 2019$ and four $3 times 3$ squares up agains the top right and also get a sum of $4 cdot 2019$. A similar argument says the sum of the bottom six elements in the first column is the same as the sum of the bottom six elements of the last column.



          Now adding the top six of the first and the bottom six of the last gives the same sum as the top six of the last and the bottom six of the first. The center five squares on each edge cancel out and we are left with $10+?=11+6, ?=7$






          share|cite|improve this answer





















          • There is just one top element in the first column; its value is $10$.
            – Christian Blatter
            8 hours ago










          • Thank you! That was fast!
            – Nikolaj K
            8 hours ago










          • @ChristianBlatter The usage of "Top N" to mean "Elements 1 through N" was first popularized by Professor D. Letterman.
            – MartianInvader
            3 hours ago


















          up vote
          2
          down vote













          Please consider this as a supplement to Ross Millikan excellent answer...



          A sudoku like puzzle



          $$(10 + ?) - (6+11) = 4(2019) + 4(2019) - 4(2019) - 4(2019) = 0\
          implies ? = 7$$






          share|cite|improve this answer





















          • Thanks. I couldn't think of an easy way to draw the figure.
            – Ross Millikan
            7 hours ago


















          up vote
          1
          down vote













          enter image description here



          Notice that because every $3times 3$ square has the same sum that



          $10 + A+ B = 8+E+F$ and that $A+B + 7 = E+F + X$.



          Which means $X = 5$.



          Likewise $7+C+D= X + G+H = 5+G+H$ and $C+D+6 = G+H+Z$



          Which means $Z = 4$.



          And so on.



          $8 + E+F = 11 + J+K$ while $E+F+5 = J+K + Y$ so $Y= 8$.



          And $5+G+H = 8 + L+M$ while $G+H+4 = L + M +?$ so $?=7$.



          And that's that.



          Or simply, by $3times 3$ squares adding to the same, the difference between any two squares $3$ terms apart, must be equal to the corresponding difference of two squares $3$ terms apart in a row or column three rows or columns away.



          So $10 -8 = 7 - X= 6-Z$ and 11-8 = Y-X= ? - Z$ and so.....$X,Y,Z, ?$ are easily solved.






          share|cite|improve this answer























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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            Note that the sum of the top six elements in the first column is the same as the sum of the top six elements in the last column because you can make four $3 times 3$ squares up against the top left and get a sum of $4cdot 2019$ and four $3 times 3$ squares up agains the top right and also get a sum of $4 cdot 2019$. A similar argument says the sum of the bottom six elements in the first column is the same as the sum of the bottom six elements of the last column.



            Now adding the top six of the first and the bottom six of the last gives the same sum as the top six of the last and the bottom six of the first. The center five squares on each edge cancel out and we are left with $10+?=11+6, ?=7$






            share|cite|improve this answer





















            • There is just one top element in the first column; its value is $10$.
              – Christian Blatter
              8 hours ago










            • Thank you! That was fast!
              – Nikolaj K
              8 hours ago










            • @ChristianBlatter The usage of "Top N" to mean "Elements 1 through N" was first popularized by Professor D. Letterman.
              – MartianInvader
              3 hours ago















            up vote
            1
            down vote



            accepted










            Note that the sum of the top six elements in the first column is the same as the sum of the top six elements in the last column because you can make four $3 times 3$ squares up against the top left and get a sum of $4cdot 2019$ and four $3 times 3$ squares up agains the top right and also get a sum of $4 cdot 2019$. A similar argument says the sum of the bottom six elements in the first column is the same as the sum of the bottom six elements of the last column.



            Now adding the top six of the first and the bottom six of the last gives the same sum as the top six of the last and the bottom six of the first. The center five squares on each edge cancel out and we are left with $10+?=11+6, ?=7$






            share|cite|improve this answer





















            • There is just one top element in the first column; its value is $10$.
              – Christian Blatter
              8 hours ago










            • Thank you! That was fast!
              – Nikolaj K
              8 hours ago










            • @ChristianBlatter The usage of "Top N" to mean "Elements 1 through N" was first popularized by Professor D. Letterman.
              – MartianInvader
              3 hours ago













            up vote
            1
            down vote



            accepted







            up vote
            1
            down vote



            accepted






            Note that the sum of the top six elements in the first column is the same as the sum of the top six elements in the last column because you can make four $3 times 3$ squares up against the top left and get a sum of $4cdot 2019$ and four $3 times 3$ squares up agains the top right and also get a sum of $4 cdot 2019$. A similar argument says the sum of the bottom six elements in the first column is the same as the sum of the bottom six elements of the last column.



            Now adding the top six of the first and the bottom six of the last gives the same sum as the top six of the last and the bottom six of the first. The center five squares on each edge cancel out and we are left with $10+?=11+6, ?=7$






            share|cite|improve this answer












            Note that the sum of the top six elements in the first column is the same as the sum of the top six elements in the last column because you can make four $3 times 3$ squares up against the top left and get a sum of $4cdot 2019$ and four $3 times 3$ squares up agains the top right and also get a sum of $4 cdot 2019$. A similar argument says the sum of the bottom six elements in the first column is the same as the sum of the bottom six elements of the last column.



            Now adding the top six of the first and the bottom six of the last gives the same sum as the top six of the last and the bottom six of the first. The center five squares on each edge cancel out and we are left with $10+?=11+6, ?=7$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 9 hours ago









            Ross Millikan

            288k23195364




            288k23195364












            • There is just one top element in the first column; its value is $10$.
              – Christian Blatter
              8 hours ago










            • Thank you! That was fast!
              – Nikolaj K
              8 hours ago










            • @ChristianBlatter The usage of "Top N" to mean "Elements 1 through N" was first popularized by Professor D. Letterman.
              – MartianInvader
              3 hours ago


















            • There is just one top element in the first column; its value is $10$.
              – Christian Blatter
              8 hours ago










            • Thank you! That was fast!
              – Nikolaj K
              8 hours ago










            • @ChristianBlatter The usage of "Top N" to mean "Elements 1 through N" was first popularized by Professor D. Letterman.
              – MartianInvader
              3 hours ago
















            There is just one top element in the first column; its value is $10$.
            – Christian Blatter
            8 hours ago




            There is just one top element in the first column; its value is $10$.
            – Christian Blatter
            8 hours ago












            Thank you! That was fast!
            – Nikolaj K
            8 hours ago




            Thank you! That was fast!
            – Nikolaj K
            8 hours ago












            @ChristianBlatter The usage of "Top N" to mean "Elements 1 through N" was first popularized by Professor D. Letterman.
            – MartianInvader
            3 hours ago




            @ChristianBlatter The usage of "Top N" to mean "Elements 1 through N" was first popularized by Professor D. Letterman.
            – MartianInvader
            3 hours ago










            up vote
            2
            down vote













            Please consider this as a supplement to Ross Millikan excellent answer...



            A sudoku like puzzle



            $$(10 + ?) - (6+11) = 4(2019) + 4(2019) - 4(2019) - 4(2019) = 0\
            implies ? = 7$$






            share|cite|improve this answer





















            • Thanks. I couldn't think of an easy way to draw the figure.
              – Ross Millikan
              7 hours ago















            up vote
            2
            down vote













            Please consider this as a supplement to Ross Millikan excellent answer...



            A sudoku like puzzle



            $$(10 + ?) - (6+11) = 4(2019) + 4(2019) - 4(2019) - 4(2019) = 0\
            implies ? = 7$$






            share|cite|improve this answer





















            • Thanks. I couldn't think of an easy way to draw the figure.
              – Ross Millikan
              7 hours ago













            up vote
            2
            down vote










            up vote
            2
            down vote









            Please consider this as a supplement to Ross Millikan excellent answer...



            A sudoku like puzzle



            $$(10 + ?) - (6+11) = 4(2019) + 4(2019) - 4(2019) - 4(2019) = 0\
            implies ? = 7$$






            share|cite|improve this answer












            Please consider this as a supplement to Ross Millikan excellent answer...



            A sudoku like puzzle



            $$(10 + ?) - (6+11) = 4(2019) + 4(2019) - 4(2019) - 4(2019) = 0\
            implies ? = 7$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 7 hours ago









            achille hui

            94.2k5129252




            94.2k5129252












            • Thanks. I couldn't think of an easy way to draw the figure.
              – Ross Millikan
              7 hours ago


















            • Thanks. I couldn't think of an easy way to draw the figure.
              – Ross Millikan
              7 hours ago
















            Thanks. I couldn't think of an easy way to draw the figure.
            – Ross Millikan
            7 hours ago




            Thanks. I couldn't think of an easy way to draw the figure.
            – Ross Millikan
            7 hours ago










            up vote
            1
            down vote













            enter image description here



            Notice that because every $3times 3$ square has the same sum that



            $10 + A+ B = 8+E+F$ and that $A+B + 7 = E+F + X$.



            Which means $X = 5$.



            Likewise $7+C+D= X + G+H = 5+G+H$ and $C+D+6 = G+H+Z$



            Which means $Z = 4$.



            And so on.



            $8 + E+F = 11 + J+K$ while $E+F+5 = J+K + Y$ so $Y= 8$.



            And $5+G+H = 8 + L+M$ while $G+H+4 = L + M +?$ so $?=7$.



            And that's that.



            Or simply, by $3times 3$ squares adding to the same, the difference between any two squares $3$ terms apart, must be equal to the corresponding difference of two squares $3$ terms apart in a row or column three rows or columns away.



            So $10 -8 = 7 - X= 6-Z$ and 11-8 = Y-X= ? - Z$ and so.....$X,Y,Z, ?$ are easily solved.






            share|cite|improve this answer



























              up vote
              1
              down vote













              enter image description here



              Notice that because every $3times 3$ square has the same sum that



              $10 + A+ B = 8+E+F$ and that $A+B + 7 = E+F + X$.



              Which means $X = 5$.



              Likewise $7+C+D= X + G+H = 5+G+H$ and $C+D+6 = G+H+Z$



              Which means $Z = 4$.



              And so on.



              $8 + E+F = 11 + J+K$ while $E+F+5 = J+K + Y$ so $Y= 8$.



              And $5+G+H = 8 + L+M$ while $G+H+4 = L + M +?$ so $?=7$.



              And that's that.



              Or simply, by $3times 3$ squares adding to the same, the difference between any two squares $3$ terms apart, must be equal to the corresponding difference of two squares $3$ terms apart in a row or column three rows or columns away.



              So $10 -8 = 7 - X= 6-Z$ and 11-8 = Y-X= ? - Z$ and so.....$X,Y,Z, ?$ are easily solved.






              share|cite|improve this answer

























                up vote
                1
                down vote










                up vote
                1
                down vote









                enter image description here



                Notice that because every $3times 3$ square has the same sum that



                $10 + A+ B = 8+E+F$ and that $A+B + 7 = E+F + X$.



                Which means $X = 5$.



                Likewise $7+C+D= X + G+H = 5+G+H$ and $C+D+6 = G+H+Z$



                Which means $Z = 4$.



                And so on.



                $8 + E+F = 11 + J+K$ while $E+F+5 = J+K + Y$ so $Y= 8$.



                And $5+G+H = 8 + L+M$ while $G+H+4 = L + M +?$ so $?=7$.



                And that's that.



                Or simply, by $3times 3$ squares adding to the same, the difference between any two squares $3$ terms apart, must be equal to the corresponding difference of two squares $3$ terms apart in a row or column three rows or columns away.



                So $10 -8 = 7 - X= 6-Z$ and 11-8 = Y-X= ? - Z$ and so.....$X,Y,Z, ?$ are easily solved.






                share|cite|improve this answer














                enter image description here



                Notice that because every $3times 3$ square has the same sum that



                $10 + A+ B = 8+E+F$ and that $A+B + 7 = E+F + X$.



                Which means $X = 5$.



                Likewise $7+C+D= X + G+H = 5+G+H$ and $C+D+6 = G+H+Z$



                Which means $Z = 4$.



                And so on.



                $8 + E+F = 11 + J+K$ while $E+F+5 = J+K + Y$ so $Y= 8$.



                And $5+G+H = 8 + L+M$ while $G+H+4 = L + M +?$ so $?=7$.



                And that's that.



                Or simply, by $3times 3$ squares adding to the same, the difference between any two squares $3$ terms apart, must be equal to the corresponding difference of two squares $3$ terms apart in a row or column three rows or columns away.



                So $10 -8 = 7 - X= 6-Z$ and 11-8 = Y-X= ? - Z$ and so.....$X,Y,Z, ?$ are easily solved.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 7 hours ago

























                answered 7 hours ago









                fleablood

                66.4k22684




                66.4k22684






























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