How to fill $4320$ multiplicative semi-magic square?












0












$begingroup$


How can we fill $4*4$ Matrix with distinct positive integers such that product of the entries in each row & column is equal to $4320$ ?



My Strategy:
$$4320=2^5 times 3^3 times 5$$
First I filled the Matrix in such a way that each column and row gets one $5$

One such way of doing so is as follows: 



5 1 1 1

1 5 1 1

1 1 5 1

1 1 1 5


Now,There are twelve 1's & four 5's .



Then,I started to fill the three 3's in such a way that

i) It reduces maximum possible number of repetition,&

ii) product of the entries in each row & column is $27$

One Such way of doing so is as follows: 



1 9 3 1

1 1 9 3

3 1 1 9

9 3 1 1


Multiplying by this grid the previous one, we have:



$$ 5 quad 9 quad 3 quad 1 $$
$$ 1 quad 5 quad 9 quad 3 $$
$$ 3 quad 1 quad 5 quad 9 $$
$$ 9 quad 3 quad 1 quad 5 $$
Now,There are four 1's , four 5's , four 9's & four 3's.



Now the repetition pattern was like this:
$color{green}{text{Green}} text{ represents } 5$
$color{red}{text{Red}} text{ represents } 1$
$color{yellow}{text{Yellow}} text{ represents } 3$
$color{blue}{text{Blue}} text{ represents } 9$
enter image description here

Here, same color can't have same powers of 2

and also upper limit of 2's power in any box is 3 as product of the entries in each row & column should be equal to $2^5$
$implies$ we have to fill the same color of boxes using $0,1,2 & 3$ ; where $0$ represents $2^0$ , and so on...

I started filling the powers of 2 in the same color but could not find any such combination that will satisfy our problem.So any help please...






magic-square construction






share









$endgroup$

















    0












    $begingroup$


    How can we fill $4*4$ Matrix with distinct positive integers such that product of the entries in each row & column is equal to $4320$ ?



    My Strategy:
    $$4320=2^5 times 3^3 times 5$$
    First I filled the Matrix in such a way that each column and row gets one $5$

    One such way of doing so is as follows: 



    5 1 1 1
    
1 5 1 1
    
1 1 5 1
    
1 1 1 5


    Now,There are twelve 1's & four 5's .



    Then,I started to fill the three 3's in such a way that

    i) It reduces maximum possible number of repetition,&

    ii) product of the entries in each row & column is $27$

    One Such way of doing so is as follows: 



    1 9 3 1
    
1 1 9 3
    
3 1 1 9
    
9 3 1 1


    Multiplying by this grid the previous one, we have:



    $$ 5 quad 9 quad 3 quad 1 $$
    $$ 1 quad 5 quad 9 quad 3 $$
    $$ 3 quad 1 quad 5 quad 9 $$
    $$ 9 quad 3 quad 1 quad 5 $$
    Now,There are four 1's , four 5's , four 9's & four 3's.



    Now the repetition pattern was like this:
    $color{green}{text{Green}} text{ represents } 5$
    $color{red}{text{Red}} text{ represents } 1$
    $color{yellow}{text{Yellow}} text{ represents } 3$
    $color{blue}{text{Blue}} text{ represents } 9$
    enter image description here

    Here, same color can't have same powers of 2

    and also upper limit of 2's power in any box is 3 as product of the entries in each row & column should be equal to $2^5$
    $implies$ we have to fill the same color of boxes using $0,1,2 & 3$ ; where $0$ represents $2^0$ , and so on...

    I started filling the powers of 2 in the same color but could not find any such combination that will satisfy our problem.So any help please...






    magic-square construction






    share









    $endgroup$















      0












      0








      0





      $begingroup$


      How can we fill $4*4$ Matrix with distinct positive integers such that product of the entries in each row & column is equal to $4320$ ?



      My Strategy:
      $$4320=2^5 times 3^3 times 5$$
      First I filled the Matrix in such a way that each column and row gets one $5$

      One such way of doing so is as follows: 



      5 1 1 1
      
1 5 1 1
      
1 1 5 1
      
1 1 1 5


      Now,There are twelve 1's & four 5's .



      Then,I started to fill the three 3's in such a way that

      i) It reduces maximum possible number of repetition,&

      ii) product of the entries in each row & column is $27$

      One Such way of doing so is as follows: 



      1 9 3 1
      
1 1 9 3
      
3 1 1 9
      
9 3 1 1


      Multiplying by this grid the previous one, we have:



      $$ 5 quad 9 quad 3 quad 1 $$
      $$ 1 quad 5 quad 9 quad 3 $$
      $$ 3 quad 1 quad 5 quad 9 $$
      $$ 9 quad 3 quad 1 quad 5 $$
      Now,There are four 1's , four 5's , four 9's & four 3's.



      Now the repetition pattern was like this:
      $color{green}{text{Green}} text{ represents } 5$
      $color{red}{text{Red}} text{ represents } 1$
      $color{yellow}{text{Yellow}} text{ represents } 3$
      $color{blue}{text{Blue}} text{ represents } 9$
      enter image description here

      Here, same color can't have same powers of 2

      and also upper limit of 2's power in any box is 3 as product of the entries in each row & column should be equal to $2^5$
      $implies$ we have to fill the same color of boxes using $0,1,2 & 3$ ; where $0$ represents $2^0$ , and so on...

      I started filling the powers of 2 in the same color but could not find any such combination that will satisfy our problem.So any help please...






      magic-square construction






      share









      $endgroup$




      How can we fill $4*4$ Matrix with distinct positive integers such that product of the entries in each row & column is equal to $4320$ ?



      My Strategy:
      $$4320=2^5 times 3^3 times 5$$
      First I filled the Matrix in such a way that each column and row gets one $5$

      One such way of doing so is as follows: 



      5 1 1 1
      
1 5 1 1
      
1 1 5 1
      
1 1 1 5


      Now,There are twelve 1's & four 5's .



      Then,I started to fill the three 3's in such a way that

      i) It reduces maximum possible number of repetition,&

      ii) product of the entries in each row & column is $27$

      One Such way of doing so is as follows: 



      1 9 3 1
      
1 1 9 3
      
3 1 1 9
      
9 3 1 1


      Multiplying by this grid the previous one, we have:



      $$ 5 quad 9 quad 3 quad 1 $$
      $$ 1 quad 5 quad 9 quad 3 $$
      $$ 3 quad 1 quad 5 quad 9 $$
      $$ 9 quad 3 quad 1 quad 5 $$
      Now,There are four 1's , four 5's , four 9's & four 3's.



      Now the repetition pattern was like this:
      $color{green}{text{Green}} text{ represents } 5$
      $color{red}{text{Red}} text{ represents } 1$
      $color{yellow}{text{Yellow}} text{ represents } 3$
      $color{blue}{text{Blue}} text{ represents } 9$
      enter image description here

      Here, same color can't have same powers of 2

      and also upper limit of 2's power in any box is 3 as product of the entries in each row & column should be equal to $2^5$
      $implies$ we have to fill the same color of boxes using $0,1,2 & 3$ ; where $0$ represents $2^0$ , and so on...

      I started filling the powers of 2 in the same color but could not find any such combination that will satisfy our problem.So any help please...






      magic-square construction




      magic-square construction





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      asked 8 mins ago









      SureshSuresh

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