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# If x^(1/3) is a positive integer, does x^(1/3) have more

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If x^(1/3) is a positive integer, does x^(1/3) have more [#permalink]

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08 Apr 2012, 19:36
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If x^(1/3) is a positive integer, does x^(1/3) have more than two distinct integer factors?

(1) 64 < x < 216
(2) x is divisible by 5.

Cannot Find an OA: My Solution

[Reveal] Spoiler:
1) 64^(1/3)=4 , 125^(1/3)=5, and 216^(1/3)=6. Since x^(1/3) is an integer, x must equal 5. 5 is a prime number and only has two factors 1 and 5. So the answer is NO. Hence sufficient.

2) X=5k. Well 1000=5*200. x^(1/3)=1000^(1/3)=10 , which has 1,2,5,10 as its factors. So the answer the question is YES.
But we can also pick x=125=5*25. x^(1/3)=125^(1/3)=5, which has only two factors: 1 and 5. Hence the answer is NO. Hence INSUFF.
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Re: Divisibility , Cubic Power [#permalink]

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08 Apr 2012, 22:51
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b. x can be 125 or 125^3
thus x ^ (1/3) = 5 or 125, factors 1,5 or 1,5,25,125 so on.

hence Insufficient.

A it is.
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Re: If x^(1/3) is a positive integer, does x^(1/3) have more [#permalink]

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09 Apr 2012, 00:58
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If $$\sqrt[3]{x}$$ is a positive integer, does $$\sqrt[3]{x}$$ have more than two distinct integer factors?

(1) 64 < x < 216 --> take the cube root: $$4<\sqrt[3]{x}<6$$ --> since given that $$\sqrt[3]{x}=integer$$ then: $$\sqrt[3]{x}=5$$. 5 has only 2 distinct positive factors: 1 and 5. Sufficient.

(2) x is divisible by 5 --> if $$\sqrt[3]{x}=5$$ the answer is NO but if $$\sqrt[3]{x}=10$$ the answer is YES. Not sufficient.

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Re: If x^(1/3) is a positive integer, does x^(1/3) have more [#permalink]

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13 Sep 2012, 04:04
alphabeta1234 wrote:
If x^(1/3) is a positive integer, does x^(1/3) have more than two distinct integer factors?

(1) 64 < x < 216
(2) x is divisible by 5.

Cannot Find an OA: My Solution

[Reveal] Spoiler:
1) 64^(1/3)=4 , 125^(1/3)=5, and 216^(1/3)=6. Since x^(1/3) is an integer, x must equal 5. 5 is a prime number and only has two factors 1 and 5. So the answer is NO. Hence sufficient.

2) X=5k. Well 1000=5*200. x^(1/3)=1000^(1/3)=10 , which has 1,2,5,10 as its factors. So the answer the question is YES.
But we can also pick x=125=5*25. x^(1/3)=125^(1/3)=5, which has only two factors: 1 and 5. Hence the answer is NO. Hence INSUFF.

Another way to look at the question.
Basically question says Is x^(1/3) a prime number, if x^(1/3) is a positive integer?
(1) x=5, which is a prime. -----> Sufficient
(2) x=5k, where is a positive integer -----> x= 5,10....----->Insufficient

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Re: If x^(1/3) is a positive integer, does x^(1/3) have more [#permalink]

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12 Jun 2016, 04:30
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Re: If x^(1/3) is a positive integer, does x^(1/3) have more [#permalink]

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12 Jun 2016, 08:10
A

1) x=125 .so x^1/3 =5 only 1&5 as factors.Suff

2)x^1/3 can be 25 or 5 which has more than 2 and only 2 factors.Not Suff
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Re: If x^(1/3) is a positive integer, does x^(1/3) have more [#permalink]

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13 Jun 2016, 09:26
alphabeta1234 wrote:
If x^(1/3) is a positive integer, does x^(1/3) have more than two distinct integer factors?

(1) 64 < x < 216
(2) x is divisible by 5.

x^1/3 is prime or not?

(1) 64 < x < 216
64^1/3 < x^1/3 < 216 ^1/3
4<< x^1/3 <6

Since x^1/3 is an integer it will be 5 (prime). sufficient

(2) x is divisible by 5. Well it can be 5 or any multiple of 5. not sufficient.

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Re: If x^(1/3) is a positive integer, does x^(1/3) have more   [#permalink] 13 Jun 2016, 09:26
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