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One-digit products in a row of numbers


Can you fill a 3x3 grid with these numbers so the products of the rows and columns are the same?Increasing rows and columnsWhat is my four digit car number?Four-by-four table with equal row and column products90s Number PuzzleIn a square, arrange the binary numbers such that no $n$:th digit is the same along a row or columnThree-digit multiplication puzzle, part III: Return of the HexHow do I make numbers 50-100 using only the numbers 2, 0, 1, 9?Squares inside a square






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








8












$begingroup$


The digits from 1 to 9 can be arranged in a row, such that any two
neighbouring digits in this row is the product of two one-digit numbers.
Arrangement:




$$728163549$$




Is it possible to do such an arrangement using hexadecimal digits 1-9 and A-F?
Here the row has 15 digits and all numbers are treated as hexadecimal numbers.
Example: 123456789ABCDEF
12=2*9, 23=5*7, 34=4*D, 45 does not work, etc.










share|improve this question











$endgroup$




















    8












    $begingroup$


    The digits from 1 to 9 can be arranged in a row, such that any two
    neighbouring digits in this row is the product of two one-digit numbers.
    Arrangement:




    $$728163549$$




    Is it possible to do such an arrangement using hexadecimal digits 1-9 and A-F?
    Here the row has 15 digits and all numbers are treated as hexadecimal numbers.
    Example: 123456789ABCDEF
    12=2*9, 23=5*7, 34=4*D, 45 does not work, etc.










    share|improve this question











    $endgroup$
















      8












      8








      8


      1



      $begingroup$


      The digits from 1 to 9 can be arranged in a row, such that any two
      neighbouring digits in this row is the product of two one-digit numbers.
      Arrangement:




      $$728163549$$




      Is it possible to do such an arrangement using hexadecimal digits 1-9 and A-F?
      Here the row has 15 digits and all numbers are treated as hexadecimal numbers.
      Example: 123456789ABCDEF
      12=2*9, 23=5*7, 34=4*D, 45 does not work, etc.










      share|improve this question











      $endgroup$




      The digits from 1 to 9 can be arranged in a row, such that any two
      neighbouring digits in this row is the product of two one-digit numbers.
      Arrangement:




      $$728163549$$




      Is it possible to do such an arrangement using hexadecimal digits 1-9 and A-F?
      Here the row has 15 digits and all numbers are treated as hexadecimal numbers.
      Example: 123456789ABCDEF
      12=2*9, 23=5*7, 34=4*D, 45 does not work, etc.







      mathematics calculation-puzzle






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 9 hours ago









      JMP

      25.8k6 gold badges49 silver badges111 bronze badges




      25.8k6 gold badges49 silver badges111 bronze badges










      asked 9 hours ago









      ThomasLThomasL

      7902 silver badges19 bronze badges




      7902 silver badges19 bronze badges























          2 Answers
          2






          active

          oldest

          votes


















          6














          $begingroup$

          One solution is




          $$D2379A5B6C4E18F$$
          enter image description here




          Thought process:




          No product starts with $F$, so $F$ must be at the end, and the only options are $3F$ and $8F$.

          The only $2$-digit numbers that are products of $1$-digit number and start with digits $A, B, C, D, E$ are $$A5, A8, A9, B4, B6, C3, C4, D2, E1.$$

          Therefore we must have subsequences $E1$ and $D2$.

          From this, you quickly get an answer by looking at the above table. I don't know if I was lucky, but apart from the observations above, I guessed all the rest right. Just for the sake of it, here is another one: $$D24E1879A5B6C3F$$







          share|improve this answer











          $endgroup$






















            2














            $begingroup$

            As an addendum to the answer from @Arnaud:




            The smallest such number is 375B6E19C4D2A8F.
            According to the brute-force program I made, there are just $787$ solutions.







            share|improve this answer











            $endgroup$

















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              2 Answers
              2






              active

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              6














              $begingroup$

              One solution is




              $$D2379A5B6C4E18F$$
              enter image description here




              Thought process:




              No product starts with $F$, so $F$ must be at the end, and the only options are $3F$ and $8F$.

              The only $2$-digit numbers that are products of $1$-digit number and start with digits $A, B, C, D, E$ are $$A5, A8, A9, B4, B6, C3, C4, D2, E1.$$

              Therefore we must have subsequences $E1$ and $D2$.

              From this, you quickly get an answer by looking at the above table. I don't know if I was lucky, but apart from the observations above, I guessed all the rest right. Just for the sake of it, here is another one: $$D24E1879A5B6C3F$$







              share|improve this answer











              $endgroup$



















                6














                $begingroup$

                One solution is




                $$D2379A5B6C4E18F$$
                enter image description here




                Thought process:




                No product starts with $F$, so $F$ must be at the end, and the only options are $3F$ and $8F$.

                The only $2$-digit numbers that are products of $1$-digit number and start with digits $A, B, C, D, E$ are $$A5, A8, A9, B4, B6, C3, C4, D2, E1.$$

                Therefore we must have subsequences $E1$ and $D2$.

                From this, you quickly get an answer by looking at the above table. I don't know if I was lucky, but apart from the observations above, I guessed all the rest right. Just for the sake of it, here is another one: $$D24E1879A5B6C3F$$







                share|improve this answer











                $endgroup$

















                  6














                  6










                  6







                  $begingroup$

                  One solution is




                  $$D2379A5B6C4E18F$$
                  enter image description here




                  Thought process:




                  No product starts with $F$, so $F$ must be at the end, and the only options are $3F$ and $8F$.

                  The only $2$-digit numbers that are products of $1$-digit number and start with digits $A, B, C, D, E$ are $$A5, A8, A9, B4, B6, C3, C4, D2, E1.$$

                  Therefore we must have subsequences $E1$ and $D2$.

                  From this, you quickly get an answer by looking at the above table. I don't know if I was lucky, but apart from the observations above, I guessed all the rest right. Just for the sake of it, here is another one: $$D24E1879A5B6C3F$$







                  share|improve this answer











                  $endgroup$



                  One solution is




                  $$D2379A5B6C4E18F$$
                  enter image description here




                  Thought process:




                  No product starts with $F$, so $F$ must be at the end, and the only options are $3F$ and $8F$.

                  The only $2$-digit numbers that are products of $1$-digit number and start with digits $A, B, C, D, E$ are $$A5, A8, A9, B4, B6, C3, C4, D2, E1.$$

                  Therefore we must have subsequences $E1$ and $D2$.

                  From this, you quickly get an answer by looking at the above table. I don't know if I was lucky, but apart from the observations above, I guessed all the rest right. Just for the sake of it, here is another one: $$D24E1879A5B6C3F$$








                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 8 hours ago

























                  answered 8 hours ago









                  Arnaud MortierArnaud Mortier

                  5,76913 silver badges49 bronze badges




                  5,76913 silver badges49 bronze badges


























                      2














                      $begingroup$

                      As an addendum to the answer from @Arnaud:




                      The smallest such number is 375B6E19C4D2A8F.
                      According to the brute-force program I made, there are just $787$ solutions.







                      share|improve this answer











                      $endgroup$



















                        2














                        $begingroup$

                        As an addendum to the answer from @Arnaud:




                        The smallest such number is 375B6E19C4D2A8F.
                        According to the brute-force program I made, there are just $787$ solutions.







                        share|improve this answer











                        $endgroup$

















                          2














                          2










                          2







                          $begingroup$

                          As an addendum to the answer from @Arnaud:




                          The smallest such number is 375B6E19C4D2A8F.
                          According to the brute-force program I made, there are just $787$ solutions.







                          share|improve this answer











                          $endgroup$



                          As an addendum to the answer from @Arnaud:




                          The smallest such number is 375B6E19C4D2A8F.
                          According to the brute-force program I made, there are just $787$ solutions.








                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited 8 hours ago

























                          answered 8 hours ago









                          JensJens

                          3966 bronze badges




                          3966 bronze badges































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