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Re: If n is a positive integer and n^3 is odd, which of the following [#permalink]

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26 Oct 2017, 10:07

chetan2u wrote:

genxer123 wrote:

Bunuel wrote:

If n^3 is odd, which of the following statements are true?

I. n is odd. II. n^2 is odd. III. n^2 is even.

(A) I only (B) II only (C) III only (D) I and II only (E) I and III only

\(n^3\) is odd. From that we can deduce that \(n\) must be odd.

There is only one way a number, \(n\), multiplied by itself 3 times or 100 times, can be odd: \(n\) is odd.

Odd * odd is odd. Odd * even is even.

In fact, the only way to get an odd product is to have only odd factors. If there just one even term (or two or 50), the product is even.

There is an odd product? Then there are ONLY odd factors.

Which statements are true?

I: n is odd. See above. For \(n^3\) to be odd, n must be odd. TRUE

II. \(n^2\) is odd. Odd * odd = ODD. TRUE

III. \(n^2\) is even. If n is odd, as it is here, \(n^2\) can never be even. Odd * odd = odd. FALSE

Answer D

hi..

what if n= \(\sqrt[3]{x}\) then neither n nor n^2 is ODD..

Had the Q been Could be true... YES Although this is what it means here

chetan2u , as is often the case, you are provocative, which I appreciate and respect.

Perhaps I do not understand what the x is in your n= \(\sqrt[3]{x}\)

If x = 27, then \(\sqrt[3]{27}\) = 3 If x = 8, then \(\sqrt[3]{8}\) = 2

What cube root of an even number is odd? More important, what even number, cubed, is odd? I am confused.

And we are told that n^3 is ODD. From that fact we are asked to infer number properties.

I considered n= \(\sqrt[3]{8}\) (similar to what you are suggesting, where x = 8, and n = 2).

I decided that such an approach was to reason the wrong way: That is, to reason as such is to reason from a possibility for n to the given \(n^3\), rather than from the given \(n^3\) (= ODD) to the possible properties of \(n\).

Neither does the question's imperative -- TRUE (not "could be possible" or "could be true") -- seem to allow for such reasoning.

So I decided to answer the question the way I thought it was intended, a perceived intention upon which you agree.

Also, not wanting to sound too full of myself, or too esoteric, I decided not to mention the alternative I had considered, which seemed . . . inapposite. (I'm not an expert. You are.)

It is an interesting point. Is it apposite?

Were I to see a formulation such as that which you postulate, of course my answer would be different. Thanks, and kudos for noting subtleties (even though I am slightly confused about whether or not your suggestion applies to this question and its answer).

(P.S. This group can be a tough crowd. I still would not include your possibility in my answer.)

My point was that if this Q had a choice "none of the above", may be that would have been correct.

Say n=\(\sqrt[3]{3}\) then n^3=3 So here n^3 is an odd integer but n is not even an integer. It is nowhere mentioned that n is an integer and the Q has not asked "could be true".

But since these two things are not mentioned but the choices too does not contain "none of these", I would take n as an integer and answer the Q as you have answered.

The point was basically for all to know that there can be cases where n need not be an integer, if n^3 is odd integer.
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My point was that if this Q had a choice "none of the above", may be that would have been correct.

Say n=\(\sqrt[3]{3}\) then n^3=3 So here n^3 is an odd integer but n is not even an integer. It is nowhere mentioned that n is an integer and the Q has not asked "could be true".

But since these two things are not mentioned but the choices too does not contain "none of these", I would take n as an integer and answer the Q as you have answered.

The point was basically for all to know that there can be cases where n need not be an integer, if n^3 is odd integer.

Now, with your reference to integers, I understand.

Re: If n is a positive integer and n^3 is odd, which of the following [#permalink]

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26 Oct 2017, 21:14

This is a simple question that deals basically with fundamentals and concepts. For rational numbers If n^3 is odd then n must be odd. (odd*odd*odd is odd) now if n is odd then n^2 must also be odd since odd*odd is odd hence I and II is correct but we are not given any indication that its a rational number. lets assume its an irrational number It can be the case that n^3 is 2 then still n cannot be odd or even hence out of context. hence according to me answer is D

If n is a positive integer and n^3 is odd, which of the following statements are true?

I. n is odd. II. n^2 is odd. III. n^2 is even.

(A) I only (B) II only (C) III only (D) I and II only (E) I and III only

We should recall that when an even number is raised to an exponent, the result is always even, and when an odd number is raised to an exponent, the result is always odd.

Thus, if n^3 is odd, both n and n^2 are also odd.

Answer: D
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