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Is n^2 an even integer?

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Is n^2 an even integer? [#permalink] New post 01 Nov 2013, 23:56
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Is n^2 an even integer?

(1) n/2 is an odd integer.

(2) (n^2)/2 is not an even integer.

GH-06.12.13 | OE to follow
[Reveal] Spoiler: OA

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Re: Is n^2 an even integer? [#permalink] New post 02 Nov 2013, 03:56
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Is n^2 an even integer?

Notice that we are NOT told that n is an integer.

(1) n/2 is an odd integer --> \(\frac{n}{2}=odd\) --> \(n=2*odd=even\) --> n^2=even^2=even. Sufficient.

(2) (n^2)/2 is not an even integer --> if n=1, then the answer is NO but if \(n=\sqrt{2}\), then the answer is YES. Not sufficient.

Answer: A.

Hope it's clear.
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Director
Director
User avatar
Status: 1,750 Q's attempted and counting
Affiliations: University of Florida
Joined: 09 Jul 2013
Posts: 514
Location: United States (FL)
Schools: UFL (A)
GMAT 1: 600 Q45 V29
GMAT 2: 590 Q35 V35
GMAT 3: 570 Q42 V28
GMAT 4: 610 Q44 V30
GPA: 3.45
WE: Accounting (Accounting)
Followers: 24

Kudos [?]: 437 [0], given: 630

Re: Is n^2 an even integer? [#permalink] New post 05 Nov 2013, 21:42
Official Explanation

Answer: A
- Statement (1) is sufficient. If n/2 is an odd integer, we can find n by multiplying both sides by 2. n is 2 times an odd integer, which is always an even. If n is an even, then n^2 is an even -- an even times an even is always an even.

Statement (2) is not sufficient. Note that "not an even integer" does not mean "an odd integer." It could also refer to any non-integer, such as 4.5. If n = 3, then n^2 = 9 and n^2/2 = 4.5, which is not an even integer. In that case, the answer is "no." However, if n^2 = 2, then n^2/2 = 1, which is not an even integer. But in this case, n^2 is even, so the answer is "yes." Choice (A) is correct.
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Re: Is n^2 an even integer?   [#permalink] 05 Nov 2013, 21:42
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