Magic square from $2010$












0












$begingroup$


A multiplicative magic square (MMS) is a square array of positive integers in which
the product of each row, column, and long diagonal is the same. The $16$ positive
factors of $2010$ can be formed into a $4times 4$ MMS. What is the common product of
every row, column, and diagonal? Write your answer in the corresponding blank on
the answer sheet



The answer:
That product raised to the 4th power should equal the product of the sixteen
factors of $2010$. Note that
$2010= (2)(3)(5)(67)$. Among the $16$ numbers, half of
them have a factor of $2$ with multiplicity $1$ while the rest don’t have the factor $2$.
Likewise, half of the 16 numbers have a factor of $3$ with multiplicity $1$ while the
rest don’t have the factor $3$. The same holds for $5$ and for $67$. I don't get why the product is to the $4$th power ??? And why it is equal to the $16$ factors ?? I DO get the part about the factors .. $8$ of them will have factor of $2$ etc. ..






magic-square







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  • $begingroup$
    Another way to find the product of all $16$ is to note you can form $8$ pairs, each with product $2010$. The product of all of them is then 2018^8$
    $endgroup$
    – Ross Millikan
    Jan 4 at 4:00
















0












$begingroup$


A multiplicative magic square (MMS) is a square array of positive integers in which
the product of each row, column, and long diagonal is the same. The $16$ positive
factors of $2010$ can be formed into a $4times 4$ MMS. What is the common product of
every row, column, and diagonal? Write your answer in the corresponding blank on
the answer sheet



The answer:
That product raised to the 4th power should equal the product of the sixteen
factors of $2010$. Note that
$2010= (2)(3)(5)(67)$. Among the $16$ numbers, half of
them have a factor of $2$ with multiplicity $1$ while the rest don’t have the factor $2$.
Likewise, half of the 16 numbers have a factor of $3$ with multiplicity $1$ while the
rest don’t have the factor $3$. The same holds for $5$ and for $67$. I don't get why the product is to the $4$th power ??? And why it is equal to the $16$ factors ?? I DO get the part about the factors .. $8$ of them will have factor of $2$ etc. ..






magic-square







share|cite|improve this question











$endgroup$












  • $begingroup$
    Another way to find the product of all $16$ is to note you can form $8$ pairs, each with product $2010$. The product of all of them is then 2018^8$
    $endgroup$
    – Ross Millikan
    Jan 4 at 4:00














0












0








0





$begingroup$


A multiplicative magic square (MMS) is a square array of positive integers in which
the product of each row, column, and long diagonal is the same. The $16$ positive
factors of $2010$ can be formed into a $4times 4$ MMS. What is the common product of
every row, column, and diagonal? Write your answer in the corresponding blank on
the answer sheet



The answer:
That product raised to the 4th power should equal the product of the sixteen
factors of $2010$. Note that
$2010= (2)(3)(5)(67)$. Among the $16$ numbers, half of
them have a factor of $2$ with multiplicity $1$ while the rest don’t have the factor $2$.
Likewise, half of the 16 numbers have a factor of $3$ with multiplicity $1$ while the
rest don’t have the factor $3$. The same holds for $5$ and for $67$. I don't get why the product is to the $4$th power ??? And why it is equal to the $16$ factors ?? I DO get the part about the factors .. $8$ of them will have factor of $2$ etc. ..






magic-square







share|cite|improve this question











$endgroup$




A multiplicative magic square (MMS) is a square array of positive integers in which
the product of each row, column, and long diagonal is the same. The $16$ positive
factors of $2010$ can be formed into a $4times 4$ MMS. What is the common product of
every row, column, and diagonal? Write your answer in the corresponding blank on
the answer sheet



The answer:
That product raised to the 4th power should equal the product of the sixteen
factors of $2010$. Note that
$2010= (2)(3)(5)(67)$. Among the $16$ numbers, half of
them have a factor of $2$ with multiplicity $1$ while the rest don’t have the factor $2$.
Likewise, half of the 16 numbers have a factor of $3$ with multiplicity $1$ while the
rest don’t have the factor $3$. The same holds for $5$ and for $67$. I don't get why the product is to the $4$th power ??? And why it is equal to the $16$ factors ?? I DO get the part about the factors .. $8$ of them will have factor of $2$ etc. ..






magic-square




magic-square






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edited Jan 4 at 4:23









max_zorn

3,36361329




3,36361329










asked Jan 4 at 3:23









RandinRandin

347116




347116












  • $begingroup$
    Another way to find the product of all $16$ is to note you can form $8$ pairs, each with product $2010$. The product of all of them is then 2018^8$
    $endgroup$
    – Ross Millikan
    Jan 4 at 4:00


















  • $begingroup$
    Another way to find the product of all $16$ is to note you can form $8$ pairs, each with product $2010$. The product of all of them is then 2018^8$
    $endgroup$
    – Ross Millikan
    Jan 4 at 4:00
















$begingroup$
Another way to find the product of all $16$ is to note you can form $8$ pairs, each with product $2010$. The product of all of them is then 2018^8$
$endgroup$
– Ross Millikan
Jan 4 at 4:00




$begingroup$
Another way to find the product of all $16$ is to note you can form $8$ pairs, each with product $2010$. The product of all of them is then 2018^8$
$endgroup$
– Ross Millikan
Jan 4 at 4:00










2 Answers
2






active

oldest

votes


















1












$begingroup$

If the product of the values in each of the $4$ rows is $k$, it follows that the product of the whole square is $k^4$. That's quite plain to see, and is what the 4th power bit is trying to say.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Dang I saw this about 5 minutes ago when I got off work and looked back this problem!! But i didnt see the 4 rows , I saw if we consider the center of the MMS and look at the 2 diagonals and the row and the collumn I, -, /, these all will have the same product right? so multiplying them all together will give the big product of ALL 16 factors ..right?
    $endgroup$
    – Randin
    Jan 4 at 9:30



















0












$begingroup$

As all $16$ factors are in the array, the product of all $4$ row products, for example, would be the same as the product of the $16$ factors. In particular, if $A_{ij}$ is the value of the $i$th row and $j$th column, then the $i$th row product would be



$$prod_{j = 1}^{4} A_{ij}$$



Multiplying each of these row products gives



$$prod_{i = 1}^{4}left(prod_{j = 1}^{4} A_{ij}right)$$



This is the same as the product of all values in the array as the double product uses each element once.



Note that as each row product is the same, this means that row product to the $4$th power is the same as the product of all $16$ factors.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    If the product of the values in each of the $4$ rows is $k$, it follows that the product of the whole square is $k^4$. That's quite plain to see, and is what the 4th power bit is trying to say.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Dang I saw this about 5 minutes ago when I got off work and looked back this problem!! But i didnt see the 4 rows , I saw if we consider the center of the MMS and look at the 2 diagonals and the row and the collumn I, -, /, these all will have the same product right? so multiplying them all together will give the big product of ALL 16 factors ..right?
      $endgroup$
      – Randin
      Jan 4 at 9:30
















    1












    $begingroup$

    If the product of the values in each of the $4$ rows is $k$, it follows that the product of the whole square is $k^4$. That's quite plain to see, and is what the 4th power bit is trying to say.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Dang I saw this about 5 minutes ago when I got off work and looked back this problem!! But i didnt see the 4 rows , I saw if we consider the center of the MMS and look at the 2 diagonals and the row and the collumn I, -, /, these all will have the same product right? so multiplying them all together will give the big product of ALL 16 factors ..right?
      $endgroup$
      – Randin
      Jan 4 at 9:30














    1












    1








    1





    $begingroup$

    If the product of the values in each of the $4$ rows is $k$, it follows that the product of the whole square is $k^4$. That's quite plain to see, and is what the 4th power bit is trying to say.






    share|cite|improve this answer









    $endgroup$



    If the product of the values in each of the $4$ rows is $k$, it follows that the product of the whole square is $k^4$. That's quite plain to see, and is what the 4th power bit is trying to say.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 4 at 3:31









    Rhys HughesRhys Hughes

    5,8031529




    5,8031529












    • $begingroup$
      Dang I saw this about 5 minutes ago when I got off work and looked back this problem!! But i didnt see the 4 rows , I saw if we consider the center of the MMS and look at the 2 diagonals and the row and the collumn I, -, /, these all will have the same product right? so multiplying them all together will give the big product of ALL 16 factors ..right?
      $endgroup$
      – Randin
      Jan 4 at 9:30


















    • $begingroup$
      Dang I saw this about 5 minutes ago when I got off work and looked back this problem!! But i didnt see the 4 rows , I saw if we consider the center of the MMS and look at the 2 diagonals and the row and the collumn I, -, /, these all will have the same product right? so multiplying them all together will give the big product of ALL 16 factors ..right?
      $endgroup$
      – Randin
      Jan 4 at 9:30
















    $begingroup$
    Dang I saw this about 5 minutes ago when I got off work and looked back this problem!! But i didnt see the 4 rows , I saw if we consider the center of the MMS and look at the 2 diagonals and the row and the collumn I, -, /, these all will have the same product right? so multiplying them all together will give the big product of ALL 16 factors ..right?
    $endgroup$
    – Randin
    Jan 4 at 9:30




    $begingroup$
    Dang I saw this about 5 minutes ago when I got off work and looked back this problem!! But i didnt see the 4 rows , I saw if we consider the center of the MMS and look at the 2 diagonals and the row and the collumn I, -, /, these all will have the same product right? so multiplying them all together will give the big product of ALL 16 factors ..right?
    $endgroup$
    – Randin
    Jan 4 at 9:30











    0












    $begingroup$

    As all $16$ factors are in the array, the product of all $4$ row products, for example, would be the same as the product of the $16$ factors. In particular, if $A_{ij}$ is the value of the $i$th row and $j$th column, then the $i$th row product would be



    $$prod_{j = 1}^{4} A_{ij}$$



    Multiplying each of these row products gives



    $$prod_{i = 1}^{4}left(prod_{j = 1}^{4} A_{ij}right)$$



    This is the same as the product of all values in the array as the double product uses each element once.



    Note that as each row product is the same, this means that row product to the $4$th power is the same as the product of all $16$ factors.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      As all $16$ factors are in the array, the product of all $4$ row products, for example, would be the same as the product of the $16$ factors. In particular, if $A_{ij}$ is the value of the $i$th row and $j$th column, then the $i$th row product would be



      $$prod_{j = 1}^{4} A_{ij}$$



      Multiplying each of these row products gives



      $$prod_{i = 1}^{4}left(prod_{j = 1}^{4} A_{ij}right)$$



      This is the same as the product of all values in the array as the double product uses each element once.



      Note that as each row product is the same, this means that row product to the $4$th power is the same as the product of all $16$ factors.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        As all $16$ factors are in the array, the product of all $4$ row products, for example, would be the same as the product of the $16$ factors. In particular, if $A_{ij}$ is the value of the $i$th row and $j$th column, then the $i$th row product would be



        $$prod_{j = 1}^{4} A_{ij}$$



        Multiplying each of these row products gives



        $$prod_{i = 1}^{4}left(prod_{j = 1}^{4} A_{ij}right)$$



        This is the same as the product of all values in the array as the double product uses each element once.



        Note that as each row product is the same, this means that row product to the $4$th power is the same as the product of all $16$ factors.






        share|cite|improve this answer











        $endgroup$



        As all $16$ factors are in the array, the product of all $4$ row products, for example, would be the same as the product of the $16$ factors. In particular, if $A_{ij}$ is the value of the $i$th row and $j$th column, then the $i$th row product would be



        $$prod_{j = 1}^{4} A_{ij}$$



        Multiplying each of these row products gives



        $$prod_{i = 1}^{4}left(prod_{j = 1}^{4} A_{ij}right)$$



        This is the same as the product of all values in the array as the double product uses each element once.



        Note that as each row product is the same, this means that row product to the $4$th power is the same as the product of all $16$ factors.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 3:36

























        answered Jan 4 at 3:29









        John OmielanJohn Omielan

        1,961210




        1,961210






























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