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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 $4\times 4$ 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. ..

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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
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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.

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  • $\begingroup$ 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
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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.

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