Is P odd? (1) The cube root of P is odd (2) P^3 is odd

3*√3 is not even, √3 is not an integer.

P^3 if we equal P = 3 will be 3^3 = 27, odd.

vijk

I mean like this: 3(small on the root) √3

You can't see questions like this on the GMAT, where they ask if something is odd, and the "trap" is that the thing might not be an integer at all. If, from Statement 2, p^3 is an odd integer, then p is either an odd integer, or it is not an integer at all (it might be something like \( \sqrt[3]{7} \) ). But if p is a non-integer, then the question "is p odd?" is meaningless. It's like asking "is p purple?" or "what is the value of p/0?" The concept of evenness and oddness is not defined for non-integers, so a properly-designed GMAT question can't even allow as a possibility that you might need to consider whether some non-integer is odd. I've seen dozens of prep company questions with this 'trap' built into them, and there is not a single real GMAT question with the same 'trap'. A similar question on the GMAT would always need to ask "Is p an odd integer?", so you know you're trying to answer two things: is p an integer at all, and if so, is it an odd integer. But with the correct wording, the 'trap' becomes much more obvious.

So it's not a good question. Their justification for the OA (A) will be that S1 guarantees p is the cube of an odd integer, so is itself odd, while for S2, p can be an odd integer, or might not be an integer at all.

Is P odd?

(1) The cube root of P is odd --> suff: \(P^{\frac{1}{3}}=odd => p = {odd}^3 = odd\)
(2) P^3 is odd --> insuff: \(P^3=odd\) => P can be an integer or not. if \(P^3=27\), then P=3 i.e. the answer is YES, but if \(P^3=9\), then \(P=9^{\frac{1}{3}}\) i.e. the answer is NO
Answer: A

Is p Odd?

1) The cube root of p is odd
2) p^3 is odd


Answer: A


Why is statement 2 not sufficient? Wouldn't p need to be an odd integer?

Asked: Is p Odd?

1) The cube root of p is odd
pˆ(1/3) is odd
p is oddˆ3 = odd
SUFFICIENT

2) p^3 is odd
If pˆ3 = 3 ; p=3ˆ(1/3) is NOT odd
But if pˆ3 = 27; p = 3 is odd
NOT SUFFICIENT

IMO A

But we have to treat statement 2 as a fact and p^3 is odd and odd numbers are integers than we can’t use 1/3 as a test case right? Our only option would be to use a odd integer.

Posted from my mobile device

Because nothing is mentioned of "P" being an interger. we are free to use any values to test the statement.
So, in statement B, \(P^3\) could be 81 or it could be 19. We can see P results in odd in the former case and a decimal value in the later.
Had the question stem mentioned that "P" be an integer, we would have been bound to take such \(P^3\) values that might result a integer value of P.

Hope it helps.