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How many positive integer divisors of 23,328 are perfect cubes?

 Dec 24, 2018
 #1
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How many positive integer divisors of 23,328 are perfect cubes?

 

23328^(1/3) = 28.573218935427587   Well less than 29 anyway.

 

1 is ok

factor(23328) = 2^5*3^6

 

 

\(2*2*2*2*2\quad * \quad 3*3*3*3*3*3\\~\\ 1^3,2^3,3^3,6^3,9^3,18^3 \;\;\text{that seems to be it.} \)

 

I just got those from visual examination of the factors.

So that is  6 of them.

 Dec 24, 2018
 #2
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23,328 = 2^5 * 3^6

Will divide the exponents by 3 and add 1 and then multiply them:

5 /3 + 1 =2

6/3  + 1 =3

So, 3 x 2  = 6 perfect cubes.

 Dec 24, 2018

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