Scheduling based problemDeducing Two Numbers based on their Difference and RatioThe 2 million, er, 20 dollar...

Is there any official lore on the Far Realm?

What are the steps to solving this definite integral?

Could the terminal length of components like resistors be reduced?

How to prevent z-fighting in OpenSCAD?

Don’t seats that recline flat defeat the purpose of having seatbelts?

Apply MapThread to all but one variable

Why do games have consumables?

Why didn't the Space Shuttle bounce back into space as many times as possible so as to lose a lot of kinetic energy up there?

Can an Area of Effect spell cast outside a Prismatic Wall extend inside it?

Map of water taps to fill bottles

How to fry ground beef so it is well-browned

Does tea made with boiling water cool faster than tea made with boiled (but still hot) water?

"Whatever a Russian does, they end up making the Kalashnikov gun"? Are there any similar proverbs in English?

Initiative: Do I lose my attack/action if my target moves or dies before my turn in combat?

a sore throat vs a strep throat vs strep throat

555 timer FM transmitter

Why did C use the -> operator instead of reusing the . operator?

Which big number is bigger?

What term is being referred to with "reflected-sound-of-underground-spirits"?

Can we say “you can pay when the order gets ready”?

How to have a sharp product image?

What happens to Mjolnir (Thor's hammer) at the end of Endgame?

What is the philosophical significance of speech acts/implicature?

Checks user level and limit the data before saving it to mongoDB



Scheduling based problem


Deducing Two Numbers based on their Difference and RatioThe 2 million, er, 20 dollar problemAdd a number to each vertex of a triangle such that each edge adds to a perfect squareHexagonal sum fillingEight distinct numbers in the tableIncreasing rows and columnsA simple grid puzzle123456789=1 problemFilling a chessboard with -1, 0 and 1Input/Output Problem #7













4












$begingroup$


Can you place these numbers into 5 rows of 4 such that each row totals 20?




1, 2, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 7, 8, 8, 8











share|improve this question









$endgroup$

















    4












    $begingroup$


    Can you place these numbers into 5 rows of 4 such that each row totals 20?




    1, 2, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 7, 8, 8, 8











    share|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      Can you place these numbers into 5 rows of 4 such that each row totals 20?




      1, 2, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 7, 8, 8, 8











      share|improve this question









      $endgroup$




      Can you place these numbers into 5 rows of 4 such that each row totals 20?




      1, 2, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 7, 8, 8, 8








      mathematics packing






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked yesterday









      JonMark PerryJonMark Perry

      20.9k64199




      20.9k64199






















          3 Answers
          3






          active

          oldest

          votes


















          10












          $begingroup$

          Sure.




          1,5,6,8

          2,4,6,8

          3,4,5,8

          4,4,5,7

          5,5,5,5




          Method:




          Started from the bigger numbers, and partitioned into 5 parts of 20: {8,8,4}, {8,7,5} and so on.


          Then swapped a big number with two smaller ones (with the same sum) on another row until I had 4 numbers on each row.




          With some fiddling, it's also possible to get all the columns to add up to 25:




          1,8,6,5

          5,4,7,4

          5,5,5,5

          6,4,2,8

          8,4,5,3




          And here's a magic square (with duplicates, unavoidably) followed by a row of fives:




          1,6,8,5

          5,7,4,4

          8,5,4,3

          6,2,4,8

          5,5,5,5




          And finally:






          4 4 8 4
          8 5 4 3
          1 5 6 8
          7 6 2 5
          5 5 5 5



          This has

          * 20 on all 5 rows

          * 20 on all 4 long diagonals

          * 25 in all 4 columns

          * a magic square on the first 4 rows





          share|improve this answer











          $endgroup$













          • $begingroup$
            I'm late ! But I have differents rows :D
            $endgroup$
            – Narlore
            yesterday










          • $begingroup$
            I like the way you went on to challenge yourself by finding solutions to your own proposed questions.
            $endgroup$
            – 5202456
            yesterday










          • $begingroup$
            @5202456 Thanks! The original solution seemed to leave an awful lot of "wiggle room" in the pattern, so I wanted to see what I could do with it. I added one more "extra magical" solution after your comment; hopefully I didn't make any mistakes, as the numbers are starting to bounce around in my eyes :-)
            $endgroup$
            – Bass
            yesterday



















          5












          $begingroup$

          A bit late to the party and cannot beat the excellent answer from @Bass.



          I worked out the number of distinct solutions bearing in mind they can be further permuted by ordering each row and the row sequence. I found




          16 distinct solutions




          I did this by first finding all the sets of four digits which sum to 20.




          There are 17 sets of digits


            1  3  8  8
          1 4 7 8
          1 5 6 8
          1 6 6 7
          2 3 7 8
          2 4 6 8
          2 5 5 8
          2 5 6 7
          3 4 5 8
          3 4 6 7
          3 5 5 7
          3 5 6 6
          4 4 4 8
          4 4 5 7
          4 4 6 6
          4 5 5 6
          5 5 5 5



          I then permuted them for each row so that each digit is used the right number of times.




          These are the 16 solutions


           1  3  8  8     1  3  8  8     1  3  8  8     1  3  8  8
          2 4 6 8 2 5 5 8 2 5 5 8 2 5 6 7
          4 4 5 7 4 4 5 7 4 4 5 7 4 4 4 8
          4 5 5 6 4 4 6 6 4 5 5 6 4 5 5 6
          5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5

          1 4 7 8 1 4 7 8 1 4 7 8 1 4 7 8
          2 4 6 8 2 5 5 8 2 5 5 8 2 5 5 8
          3 4 5 8 3 4 5 8 3 4 5 8 3 5 6 6
          4 5 5 6 4 4 6 6 4 5 5 6 4 4 4 8
          5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5

          1 5 6 8 1 5 6 8 1 5 6 8 1 5 6 8
          2 3 7 8 2 4 6 8 2 4 6 8 2 5 5 8
          4 4 4 8 3 4 5 8 3 5 5 7 3 4 5 8
          4 5 5 6 4 4 5 7 4 4 4 8 4 4 5 7
          5 5 5 5 5 5 5 5 5 5 5 5 4 5 5 6

          1 5 6 8 1 5 6 8 1 5 6 8 1 6 6 7
          2 5 5 8 2 5 5 8 2 5 6 7 2 5 5 8
          3 4 6 7 3 5 5 7 3 4 5 8 3 4 5 8
          4 4 4 8 4 4 4 8 4 4 4 8 4 4 4 8
          5 5 5 5 4 5 5 6 5 5 5 5 5 5 5 5



          Method:




          A computer program written in C.







          share|improve this answer









          $endgroup$













          • $begingroup$
            Nice! FYI I generated the puzzle from row 2 column 4.
            $endgroup$
            – JonMark Perry
            yesterday



















          3












          $begingroup$

          This one works




          7 3 5 5

          8 2 5 5

          6 4 5 5

          8 6 1 5

          8 4 4 4




          Method :




          Write randomly the numbers on a piece of paper during 5 min until the solution appears magically.







          share|improve this answer











          $endgroup$














            Your Answer








            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "559"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: false,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: null,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f83173%2fscheduling-based-problem%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            10












            $begingroup$

            Sure.




            1,5,6,8

            2,4,6,8

            3,4,5,8

            4,4,5,7

            5,5,5,5




            Method:




            Started from the bigger numbers, and partitioned into 5 parts of 20: {8,8,4}, {8,7,5} and so on.


            Then swapped a big number with two smaller ones (with the same sum) on another row until I had 4 numbers on each row.




            With some fiddling, it's also possible to get all the columns to add up to 25:




            1,8,6,5

            5,4,7,4

            5,5,5,5

            6,4,2,8

            8,4,5,3




            And here's a magic square (with duplicates, unavoidably) followed by a row of fives:




            1,6,8,5

            5,7,4,4

            8,5,4,3

            6,2,4,8

            5,5,5,5




            And finally:






            4 4 8 4
            8 5 4 3
            1 5 6 8
            7 6 2 5
            5 5 5 5



            This has

            * 20 on all 5 rows

            * 20 on all 4 long diagonals

            * 25 in all 4 columns

            * a magic square on the first 4 rows





            share|improve this answer











            $endgroup$













            • $begingroup$
              I'm late ! But I have differents rows :D
              $endgroup$
              – Narlore
              yesterday










            • $begingroup$
              I like the way you went on to challenge yourself by finding solutions to your own proposed questions.
              $endgroup$
              – 5202456
              yesterday










            • $begingroup$
              @5202456 Thanks! The original solution seemed to leave an awful lot of "wiggle room" in the pattern, so I wanted to see what I could do with it. I added one more "extra magical" solution after your comment; hopefully I didn't make any mistakes, as the numbers are starting to bounce around in my eyes :-)
              $endgroup$
              – Bass
              yesterday
















            10












            $begingroup$

            Sure.




            1,5,6,8

            2,4,6,8

            3,4,5,8

            4,4,5,7

            5,5,5,5




            Method:




            Started from the bigger numbers, and partitioned into 5 parts of 20: {8,8,4}, {8,7,5} and so on.


            Then swapped a big number with two smaller ones (with the same sum) on another row until I had 4 numbers on each row.




            With some fiddling, it's also possible to get all the columns to add up to 25:




            1,8,6,5

            5,4,7,4

            5,5,5,5

            6,4,2,8

            8,4,5,3




            And here's a magic square (with duplicates, unavoidably) followed by a row of fives:




            1,6,8,5

            5,7,4,4

            8,5,4,3

            6,2,4,8

            5,5,5,5




            And finally:






            4 4 8 4
            8 5 4 3
            1 5 6 8
            7 6 2 5
            5 5 5 5



            This has

            * 20 on all 5 rows

            * 20 on all 4 long diagonals

            * 25 in all 4 columns

            * a magic square on the first 4 rows





            share|improve this answer











            $endgroup$













            • $begingroup$
              I'm late ! But I have differents rows :D
              $endgroup$
              – Narlore
              yesterday










            • $begingroup$
              I like the way you went on to challenge yourself by finding solutions to your own proposed questions.
              $endgroup$
              – 5202456
              yesterday










            • $begingroup$
              @5202456 Thanks! The original solution seemed to leave an awful lot of "wiggle room" in the pattern, so I wanted to see what I could do with it. I added one more "extra magical" solution after your comment; hopefully I didn't make any mistakes, as the numbers are starting to bounce around in my eyes :-)
              $endgroup$
              – Bass
              yesterday














            10












            10








            10





            $begingroup$

            Sure.




            1,5,6,8

            2,4,6,8

            3,4,5,8

            4,4,5,7

            5,5,5,5




            Method:




            Started from the bigger numbers, and partitioned into 5 parts of 20: {8,8,4}, {8,7,5} and so on.


            Then swapped a big number with two smaller ones (with the same sum) on another row until I had 4 numbers on each row.




            With some fiddling, it's also possible to get all the columns to add up to 25:




            1,8,6,5

            5,4,7,4

            5,5,5,5

            6,4,2,8

            8,4,5,3




            And here's a magic square (with duplicates, unavoidably) followed by a row of fives:




            1,6,8,5

            5,7,4,4

            8,5,4,3

            6,2,4,8

            5,5,5,5




            And finally:






            4 4 8 4
            8 5 4 3
            1 5 6 8
            7 6 2 5
            5 5 5 5



            This has

            * 20 on all 5 rows

            * 20 on all 4 long diagonals

            * 25 in all 4 columns

            * a magic square on the first 4 rows





            share|improve this answer











            $endgroup$



            Sure.




            1,5,6,8

            2,4,6,8

            3,4,5,8

            4,4,5,7

            5,5,5,5




            Method:




            Started from the bigger numbers, and partitioned into 5 parts of 20: {8,8,4}, {8,7,5} and so on.


            Then swapped a big number with two smaller ones (with the same sum) on another row until I had 4 numbers on each row.




            With some fiddling, it's also possible to get all the columns to add up to 25:




            1,8,6,5

            5,4,7,4

            5,5,5,5

            6,4,2,8

            8,4,5,3




            And here's a magic square (with duplicates, unavoidably) followed by a row of fives:




            1,6,8,5

            5,7,4,4

            8,5,4,3

            6,2,4,8

            5,5,5,5




            And finally:






            4 4 8 4
            8 5 4 3
            1 5 6 8
            7 6 2 5
            5 5 5 5



            This has

            * 20 on all 5 rows

            * 20 on all 4 long diagonals

            * 25 in all 4 columns

            * a magic square on the first 4 rows






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited yesterday

























            answered yesterday









            BassBass

            31.7k475194




            31.7k475194












            • $begingroup$
              I'm late ! But I have differents rows :D
              $endgroup$
              – Narlore
              yesterday










            • $begingroup$
              I like the way you went on to challenge yourself by finding solutions to your own proposed questions.
              $endgroup$
              – 5202456
              yesterday










            • $begingroup$
              @5202456 Thanks! The original solution seemed to leave an awful lot of "wiggle room" in the pattern, so I wanted to see what I could do with it. I added one more "extra magical" solution after your comment; hopefully I didn't make any mistakes, as the numbers are starting to bounce around in my eyes :-)
              $endgroup$
              – Bass
              yesterday


















            • $begingroup$
              I'm late ! But I have differents rows :D
              $endgroup$
              – Narlore
              yesterday










            • $begingroup$
              I like the way you went on to challenge yourself by finding solutions to your own proposed questions.
              $endgroup$
              – 5202456
              yesterday










            • $begingroup$
              @5202456 Thanks! The original solution seemed to leave an awful lot of "wiggle room" in the pattern, so I wanted to see what I could do with it. I added one more "extra magical" solution after your comment; hopefully I didn't make any mistakes, as the numbers are starting to bounce around in my eyes :-)
              $endgroup$
              – Bass
              yesterday
















            $begingroup$
            I'm late ! But I have differents rows :D
            $endgroup$
            – Narlore
            yesterday




            $begingroup$
            I'm late ! But I have differents rows :D
            $endgroup$
            – Narlore
            yesterday












            $begingroup$
            I like the way you went on to challenge yourself by finding solutions to your own proposed questions.
            $endgroup$
            – 5202456
            yesterday




            $begingroup$
            I like the way you went on to challenge yourself by finding solutions to your own proposed questions.
            $endgroup$
            – 5202456
            yesterday












            $begingroup$
            @5202456 Thanks! The original solution seemed to leave an awful lot of "wiggle room" in the pattern, so I wanted to see what I could do with it. I added one more "extra magical" solution after your comment; hopefully I didn't make any mistakes, as the numbers are starting to bounce around in my eyes :-)
            $endgroup$
            – Bass
            yesterday




            $begingroup$
            @5202456 Thanks! The original solution seemed to leave an awful lot of "wiggle room" in the pattern, so I wanted to see what I could do with it. I added one more "extra magical" solution after your comment; hopefully I didn't make any mistakes, as the numbers are starting to bounce around in my eyes :-)
            $endgroup$
            – Bass
            yesterday











            5












            $begingroup$

            A bit late to the party and cannot beat the excellent answer from @Bass.



            I worked out the number of distinct solutions bearing in mind they can be further permuted by ordering each row and the row sequence. I found




            16 distinct solutions




            I did this by first finding all the sets of four digits which sum to 20.




            There are 17 sets of digits


              1  3  8  8
            1 4 7 8
            1 5 6 8
            1 6 6 7
            2 3 7 8
            2 4 6 8
            2 5 5 8
            2 5 6 7
            3 4 5 8
            3 4 6 7
            3 5 5 7
            3 5 6 6
            4 4 4 8
            4 4 5 7
            4 4 6 6
            4 5 5 6
            5 5 5 5



            I then permuted them for each row so that each digit is used the right number of times.




            These are the 16 solutions


             1  3  8  8     1  3  8  8     1  3  8  8     1  3  8  8
            2 4 6 8 2 5 5 8 2 5 5 8 2 5 6 7
            4 4 5 7 4 4 5 7 4 4 5 7 4 4 4 8
            4 5 5 6 4 4 6 6 4 5 5 6 4 5 5 6
            5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5

            1 4 7 8 1 4 7 8 1 4 7 8 1 4 7 8
            2 4 6 8 2 5 5 8 2 5 5 8 2 5 5 8
            3 4 5 8 3 4 5 8 3 4 5 8 3 5 6 6
            4 5 5 6 4 4 6 6 4 5 5 6 4 4 4 8
            5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5

            1 5 6 8 1 5 6 8 1 5 6 8 1 5 6 8
            2 3 7 8 2 4 6 8 2 4 6 8 2 5 5 8
            4 4 4 8 3 4 5 8 3 5 5 7 3 4 5 8
            4 5 5 6 4 4 5 7 4 4 4 8 4 4 5 7
            5 5 5 5 5 5 5 5 5 5 5 5 4 5 5 6

            1 5 6 8 1 5 6 8 1 5 6 8 1 6 6 7
            2 5 5 8 2 5 5 8 2 5 6 7 2 5 5 8
            3 4 6 7 3 5 5 7 3 4 5 8 3 4 5 8
            4 4 4 8 4 4 4 8 4 4 4 8 4 4 4 8
            5 5 5 5 4 5 5 6 5 5 5 5 5 5 5 5



            Method:




            A computer program written in C.







            share|improve this answer









            $endgroup$













            • $begingroup$
              Nice! FYI I generated the puzzle from row 2 column 4.
              $endgroup$
              – JonMark Perry
              yesterday
















            5












            $begingroup$

            A bit late to the party and cannot beat the excellent answer from @Bass.



            I worked out the number of distinct solutions bearing in mind they can be further permuted by ordering each row and the row sequence. I found




            16 distinct solutions




            I did this by first finding all the sets of four digits which sum to 20.




            There are 17 sets of digits


              1  3  8  8
            1 4 7 8
            1 5 6 8
            1 6 6 7
            2 3 7 8
            2 4 6 8
            2 5 5 8
            2 5 6 7
            3 4 5 8
            3 4 6 7
            3 5 5 7
            3 5 6 6
            4 4 4 8
            4 4 5 7
            4 4 6 6
            4 5 5 6
            5 5 5 5



            I then permuted them for each row so that each digit is used the right number of times.




            These are the 16 solutions


             1  3  8  8     1  3  8  8     1  3  8  8     1  3  8  8
            2 4 6 8 2 5 5 8 2 5 5 8 2 5 6 7
            4 4 5 7 4 4 5 7 4 4 5 7 4 4 4 8
            4 5 5 6 4 4 6 6 4 5 5 6 4 5 5 6
            5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5

            1 4 7 8 1 4 7 8 1 4 7 8 1 4 7 8
            2 4 6 8 2 5 5 8 2 5 5 8 2 5 5 8
            3 4 5 8 3 4 5 8 3 4 5 8 3 5 6 6
            4 5 5 6 4 4 6 6 4 5 5 6 4 4 4 8
            5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5

            1 5 6 8 1 5 6 8 1 5 6 8 1 5 6 8
            2 3 7 8 2 4 6 8 2 4 6 8 2 5 5 8
            4 4 4 8 3 4 5 8 3 5 5 7 3 4 5 8
            4 5 5 6 4 4 5 7 4 4 4 8 4 4 5 7
            5 5 5 5 5 5 5 5 5 5 5 5 4 5 5 6

            1 5 6 8 1 5 6 8 1 5 6 8 1 6 6 7
            2 5 5 8 2 5 5 8 2 5 6 7 2 5 5 8
            3 4 6 7 3 5 5 7 3 4 5 8 3 4 5 8
            4 4 4 8 4 4 4 8 4 4 4 8 4 4 4 8
            5 5 5 5 4 5 5 6 5 5 5 5 5 5 5 5



            Method:




            A computer program written in C.







            share|improve this answer









            $endgroup$













            • $begingroup$
              Nice! FYI I generated the puzzle from row 2 column 4.
              $endgroup$
              – JonMark Perry
              yesterday














            5












            5








            5





            $begingroup$

            A bit late to the party and cannot beat the excellent answer from @Bass.



            I worked out the number of distinct solutions bearing in mind they can be further permuted by ordering each row and the row sequence. I found




            16 distinct solutions




            I did this by first finding all the sets of four digits which sum to 20.




            There are 17 sets of digits


              1  3  8  8
            1 4 7 8
            1 5 6 8
            1 6 6 7
            2 3 7 8
            2 4 6 8
            2 5 5 8
            2 5 6 7
            3 4 5 8
            3 4 6 7
            3 5 5 7
            3 5 6 6
            4 4 4 8
            4 4 5 7
            4 4 6 6
            4 5 5 6
            5 5 5 5



            I then permuted them for each row so that each digit is used the right number of times.




            These are the 16 solutions


             1  3  8  8     1  3  8  8     1  3  8  8     1  3  8  8
            2 4 6 8 2 5 5 8 2 5 5 8 2 5 6 7
            4 4 5 7 4 4 5 7 4 4 5 7 4 4 4 8
            4 5 5 6 4 4 6 6 4 5 5 6 4 5 5 6
            5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5

            1 4 7 8 1 4 7 8 1 4 7 8 1 4 7 8
            2 4 6 8 2 5 5 8 2 5 5 8 2 5 5 8
            3 4 5 8 3 4 5 8 3 4 5 8 3 5 6 6
            4 5 5 6 4 4 6 6 4 5 5 6 4 4 4 8
            5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5

            1 5 6 8 1 5 6 8 1 5 6 8 1 5 6 8
            2 3 7 8 2 4 6 8 2 4 6 8 2 5 5 8
            4 4 4 8 3 4 5 8 3 5 5 7 3 4 5 8
            4 5 5 6 4 4 5 7 4 4 4 8 4 4 5 7
            5 5 5 5 5 5 5 5 5 5 5 5 4 5 5 6

            1 5 6 8 1 5 6 8 1 5 6 8 1 6 6 7
            2 5 5 8 2 5 5 8 2 5 6 7 2 5 5 8
            3 4 6 7 3 5 5 7 3 4 5 8 3 4 5 8
            4 4 4 8 4 4 4 8 4 4 4 8 4 4 4 8
            5 5 5 5 4 5 5 6 5 5 5 5 5 5 5 5



            Method:




            A computer program written in C.







            share|improve this answer









            $endgroup$



            A bit late to the party and cannot beat the excellent answer from @Bass.



            I worked out the number of distinct solutions bearing in mind they can be further permuted by ordering each row and the row sequence. I found




            16 distinct solutions




            I did this by first finding all the sets of four digits which sum to 20.




            There are 17 sets of digits


              1  3  8  8
            1 4 7 8
            1 5 6 8
            1 6 6 7
            2 3 7 8
            2 4 6 8
            2 5 5 8
            2 5 6 7
            3 4 5 8
            3 4 6 7
            3 5 5 7
            3 5 6 6
            4 4 4 8
            4 4 5 7
            4 4 6 6
            4 5 5 6
            5 5 5 5



            I then permuted them for each row so that each digit is used the right number of times.




            These are the 16 solutions


             1  3  8  8     1  3  8  8     1  3  8  8     1  3  8  8
            2 4 6 8 2 5 5 8 2 5 5 8 2 5 6 7
            4 4 5 7 4 4 5 7 4 4 5 7 4 4 4 8
            4 5 5 6 4 4 6 6 4 5 5 6 4 5 5 6
            5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5

            1 4 7 8 1 4 7 8 1 4 7 8 1 4 7 8
            2 4 6 8 2 5 5 8 2 5 5 8 2 5 5 8
            3 4 5 8 3 4 5 8 3 4 5 8 3 5 6 6
            4 5 5 6 4 4 6 6 4 5 5 6 4 4 4 8
            5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5

            1 5 6 8 1 5 6 8 1 5 6 8 1 5 6 8
            2 3 7 8 2 4 6 8 2 4 6 8 2 5 5 8
            4 4 4 8 3 4 5 8 3 5 5 7 3 4 5 8
            4 5 5 6 4 4 5 7 4 4 4 8 4 4 5 7
            5 5 5 5 5 5 5 5 5 5 5 5 4 5 5 6

            1 5 6 8 1 5 6 8 1 5 6 8 1 6 6 7
            2 5 5 8 2 5 5 8 2 5 6 7 2 5 5 8
            3 4 6 7 3 5 5 7 3 4 5 8 3 4 5 8
            4 4 4 8 4 4 4 8 4 4 4 8 4 4 4 8
            5 5 5 5 4 5 5 6 5 5 5 5 5 5 5 5



            Method:




            A computer program written in C.








            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered yesterday









            Weather VaneWeather Vane

            2,738114




            2,738114












            • $begingroup$
              Nice! FYI I generated the puzzle from row 2 column 4.
              $endgroup$
              – JonMark Perry
              yesterday


















            • $begingroup$
              Nice! FYI I generated the puzzle from row 2 column 4.
              $endgroup$
              – JonMark Perry
              yesterday
















            $begingroup$
            Nice! FYI I generated the puzzle from row 2 column 4.
            $endgroup$
            – JonMark Perry
            yesterday




            $begingroup$
            Nice! FYI I generated the puzzle from row 2 column 4.
            $endgroup$
            – JonMark Perry
            yesterday











            3












            $begingroup$

            This one works




            7 3 5 5

            8 2 5 5

            6 4 5 5

            8 6 1 5

            8 4 4 4




            Method :




            Write randomly the numbers on a piece of paper during 5 min until the solution appears magically.







            share|improve this answer











            $endgroup$


















              3












              $begingroup$

              This one works




              7 3 5 5

              8 2 5 5

              6 4 5 5

              8 6 1 5

              8 4 4 4




              Method :




              Write randomly the numbers on a piece of paper during 5 min until the solution appears magically.







              share|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                This one works




                7 3 5 5

                8 2 5 5

                6 4 5 5

                8 6 1 5

                8 4 4 4




                Method :




                Write randomly the numbers on a piece of paper during 5 min until the solution appears magically.







                share|improve this answer











                $endgroup$



                This one works




                7 3 5 5

                8 2 5 5

                6 4 5 5

                8 6 1 5

                8 4 4 4




                Method :




                Write randomly the numbers on a piece of paper during 5 min until the solution appears magically.








                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited yesterday

























                answered yesterday









                NarloreNarlore

                37616




                37616






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Puzzling Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f83173%2fscheduling-based-problem%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Couldn't open a raw socket. Error: Permission denied (13) (nmap)Is it possible to run networking commands...

                    VNC viewer RFB protocol error: bad desktop size 0x0I Cannot Type the Key 'd' (lowercase) in VNC Viewer...

                    Why not use the yoke to control yaw, as well as pitch and roll? Announcing the arrival of...