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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.

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.

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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