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by goncaljj » Mon Nov 17, 2008 1:25 pm
If x is positive, is x prime?

a) x^3 has exactly 4 distinct positive integer factors
b) x^2-x-6=0

Just to make this brief, Kaplan's reason to reject statement a) is that, as a counterexample, "6" has 4 distinct integer factors and it is not prime, even less a number to the cubic power... that is my problem there...

...if it is not a number to the cubic power, then why using it as a counterexample? the statement is indicating that unknown number has to be such that you raise it to the power of 3, and then the result happens to have 4 positive factors.

Who wrote this explanation? :?
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Source: — Data Sufficiency |

by mals24 » Mon Nov 17, 2008 2:01 pm
I think the catch in this question is that x might not be an integer. It could be a fraction or a cube root of a number. Unless not specified in the question, we cannot assume that x is an integer.

For instance x can be cube root of 6

(cube root of 6)^3 = 6

6 has 4 distinct integers 1,2,3,6
But 6 is not a prime number.

Hope the explanation is clear now.
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by cramya » Mon Nov 17, 2008 3:35 pm
Goncalj,
Whats the official answer given? Please use spoiler to post this.

Mals already provided the answer to your question but if you could please post the OA.

Regards,
Cramya
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