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How many factors of \(2^5\cdot3^6\) are perfect squares?

 Sep 27, 2018
 #1
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Any product made up of pairs of like prime factors will be a perfect square

 

there are 2 unique pairs, (2,2), (3,3)

 

there are 3 unique tetrads (2,2,2,2), (2,2,3,3), (3,3,3,3)

 

there are 3 unique hexads (2,2,2,2,3,3), (2,2,3,3,3,3), (3,3,3,3,3,3)

 

there are 2 unique octads (2,2,2,2,3,3,3,3), (2,2,3,3,3,3,3,3)

 

there is 1 unique decad (2,2,2,2,3,3,3,3,3,3)

 

So 11 perfect square factors 1, 4, 9, 16, 36, 81, 144, 324, 729, 1296, 2916, 11664

 Sep 27, 2018
 #2
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+1

Number of perfect squares in: 2^5 x 3^6 =
2^0, 2^2, 2^4=3
3^0, 3^2, 3^4, 3^6=4. Therefore the number of perfect squares that are factors of: 2^5 x 3^6= 3 x 4  =12 

Note: Rom made a small typo by not adding 1 to the total of 11, even though he listed it.

 Sep 27, 2018
edited by Guest  Sep 27, 2018

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