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Hexagonal Grid Filling


A Trip to Ripple IslandsA spartan skeleton SudokuSkeleton sudoku, the secondA Knight's KuromasuThe Ludicrous Loop: over a thousand cells of circular logic!RPS Trinity - Teddy and Her RibbonHow does one go about creating a new grid-deduction puzzle?The Terrible Twos of Slitherlink Part DeuxFixing a Binairo MistakeIntersection Puzzle













6












$begingroup$


Fill in each blank white hexagon with a digit from 1 to 9. Each of the grey hexagons is the sum of the white hexagons around it, and all of the white hexagons that are around the same grey hexagon must be distinct.





The answer's already available, but I would like to see how people would logically solve this.



Source: USAMTS










share|improve this question









$endgroup$
















    6












    $begingroup$


    Fill in each blank white hexagon with a digit from 1 to 9. Each of the grey hexagons is the sum of the white hexagons around it, and all of the white hexagons that are around the same grey hexagon must be distinct.





    The answer's already available, but I would like to see how people would logically solve this.



    Source: USAMTS










    share|improve this question









    $endgroup$














      6












      6








      6


      0



      $begingroup$


      Fill in each blank white hexagon with a digit from 1 to 9. Each of the grey hexagons is the sum of the white hexagons around it, and all of the white hexagons that are around the same grey hexagon must be distinct.





      The answer's already available, but I would like to see how people would logically solve this.



      Source: USAMTS










      share|improve this question









      $endgroup$




      Fill in each blank white hexagon with a digit from 1 to 9. Each of the grey hexagons is the sum of the white hexagons around it, and all of the white hexagons that are around the same grey hexagon must be distinct.





      The answer's already available, but I would like to see how people would logically solve this.



      Source: USAMTS







      grid-deduction






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 2 hours ago









      Jason KimJason Kim

      3149




      3149




















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          To get it started:




          • $21$ has to be $1+2+3+4+5+6$


          • $39$ has to be $9+8+7+6+5+4$


          • $21$ already has a $4$, so the overlap between $21$ and $39$ has to be $5$ and $6$, but $29$ already has a $5$, so it must get the $6$, while $30$ shares the $5$. (The hex shared by $21$, $30$, and $39$ is $5$, and the hex shared by $21$, $29$, and $39$ is $6$.)





          share|improve this answer









          $endgroup$













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            1 Answer
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            To get it started:




            • $21$ has to be $1+2+3+4+5+6$


            • $39$ has to be $9+8+7+6+5+4$


            • $21$ already has a $4$, so the overlap between $21$ and $39$ has to be $5$ and $6$, but $29$ already has a $5$, so it must get the $6$, while $30$ shares the $5$. (The hex shared by $21$, $30$, and $39$ is $5$, and the hex shared by $21$, $29$, and $39$ is $6$.)





            share|improve this answer









            $endgroup$

















              4












              $begingroup$

              To get it started:




              • $21$ has to be $1+2+3+4+5+6$


              • $39$ has to be $9+8+7+6+5+4$


              • $21$ already has a $4$, so the overlap between $21$ and $39$ has to be $5$ and $6$, but $29$ already has a $5$, so it must get the $6$, while $30$ shares the $5$. (The hex shared by $21$, $30$, and $39$ is $5$, and the hex shared by $21$, $29$, and $39$ is $6$.)





              share|improve this answer









              $endgroup$















                4












                4








                4





                $begingroup$

                To get it started:




                • $21$ has to be $1+2+3+4+5+6$


                • $39$ has to be $9+8+7+6+5+4$


                • $21$ already has a $4$, so the overlap between $21$ and $39$ has to be $5$ and $6$, but $29$ already has a $5$, so it must get the $6$, while $30$ shares the $5$. (The hex shared by $21$, $30$, and $39$ is $5$, and the hex shared by $21$, $29$, and $39$ is $6$.)





                share|improve this answer









                $endgroup$



                To get it started:




                • $21$ has to be $1+2+3+4+5+6$


                • $39$ has to be $9+8+7+6+5+4$


                • $21$ already has a $4$, so the overlap between $21$ and $39$ has to be $5$ and $6$, but $29$ already has a $5$, so it must get the $6$, while $30$ shares the $5$. (The hex shared by $21$, $30$, and $39$ is $5$, and the hex shared by $21$, $29$, and $39$ is $6$.)






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 1 hour ago









                shoovershoover

                2,090614




                2,090614



























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