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If n is an integer, is (n+1)^2 an even integer?

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If n is an integer, is (n+1)^2 an even integer?  [#permalink]

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New post 13 Sep 2018, 00:36
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A
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D
E

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Question Stats:

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[Math Revolution GMAT math practice question]

If \(n\) is an integer, is \((n+1)^2\) an even integer?

1) \(n-1\) is an even integer
2) \((n-1)^2\) is an even integer

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Re: If n is an integer, is (n+1)^2 an even integer?  [#permalink]

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New post 13 Sep 2018, 00:46
For \((n+1)^2\) to be even, \(n+1\) should be even.
A) \(n-1\) is even implies that \(n+1\) will be even and hence sufficient.
B) \((n-1)^2\) is even implies that \(n-1\) is even and thus \(n+1\) will be even and hence sufficient.
So answer is D
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Re: If n is an integer, is (n+1)^2 an even integer?  [#permalink]

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New post 13 Sep 2018, 01:12
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If \(n\) is an integer, is \((n+1)^2\) an even integer?

1) \(n-1\) is an even integer
2) \((n-1)^2\) is an even integer



we are looking for whether \((n + 1 )^2\) is even or not.

Statement 1: n - 1 = even. we know odd - odd = even. So, n is odd. So n +1 = even. Sufficient.

Statement 2 : \((n - 1 )^2\)= even. We know odd - odd = even and even times even is even. So, n is odd and n + 1 = even. Sufficient .

Note: There is no effect of Exponents in odd even question .

The best answer is D.
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Re: If n is an integer, is (n+1)^2 an even integer?  [#permalink]

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New post 13 Sep 2018, 07:25
Top Contributor
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If \(n\) is an integer, is \((n+1)^2\) an even integer?

1) \(n-1\) is an even integer
2) \((n-1)^2\) is an even integer


Target question: Is (n+1)² an even integer?
This is a good candidate for rephrasing the target question.

Aside: At the bottom of this point, you'll find a video with tips on rephrasing the target question

(n+1)² = (n+1)(n+1). So, in order for (n+1)² to be even, it must be the case that n+1 is EVEN.
Why is this?
Well, if n+1 were ODD, then (n+1)² = (ODD)² = (ODD)(ODD) = ODD, but we want (n+1)² to be EVEN
However, if n+1 were EVEN, then (n+1)² = (EVEN)² = (EVEN)(EVEN) = EVEN. Perfect.
From here, we can see that if n+1 is EVEN, then it must be the case that n is ODD
So, asking Is (n+1)² an even integer? is the same as asking Is n odd?

REPHRASED target question: Is n odd?

Statement 1: n-1 is an even integer
If n-1 is an even integer, then we can be certain that n is odd
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: (n-1)² is an even integer
If (n-1)² is an even integer, then we know that (n-1) is EVEN
If (n-1) is EVEN, then we can be certain that n is odd
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

Answer: D

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Re: If n is an integer, is (n+1)^2 an even integer?  [#permalink]

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New post 13 Sep 2018, 08:56
I would like to add some comments, related to dangerous situations usually explored in traps for the uncautious students...

\({x^2}\,\,{\text{even}}\,\,\,\,{\text{does}}\,\,{\text{NOT}}\,\,{\text{imply}}\,\,\,\,\,x\,\,{\text{even}}\,\,\,\,\,\,\,{\text{ }}\left( {x = \sqrt 2 \,\,\,{\text{for}}\,\,{\text{example}}} \right)\)

\({y^2}\,\,{\text{odd}}\,\,\,\,{\text{does}}\,\,{\text{NOT}}\,\,{\text{imply}}\,\,\,\,\,y\,\,{\text{odd}}\,\,\,\,\,\,\,{\text{ }}\left( {y = \sqrt 3 \,\,\,{\text{for}}\,\,{\text{example}}} \right)\)

On the other hand, as it is the case in the problem proposed,

\(x\,\,\,\operatorname{int} \,,\,\,\,{x^2}\,\,{\text{even}}\,\,\,\,{\text{imply}}\,\,\,\,\,\,x\,\,{\text{even}}\,\,\,\,\,\,\,{\text{ }}\left( {x\,\,{\text{is}}\,\,{\text{odd}}\,\,{\text{or}}\,\,{\text{even,}}\,\,{\text{but}}\,\,\,{\text{it}}\,\,{\text{is}}\,\,{\text{not}}\,\,{\text{odd}}...} \right)\)

\(y\,\,\,\operatorname{int} \,,\,\,\,{y^2}\,\,{\text{odd}}\,\,\,\,{\text{imply}}\,\,\,\,\,\,y\,\,{\text{odd}}\,\,\,\,\,\,\,{\text{ }}\left( {y\,\,{\text{is}}\,\,{\text{odd}}\,\,{\text{or}}\,\,{\text{even,}}\,\,{\text{but}}\,\,\,{\text{it}}\,\,{\text{is}}\,\,{\text{not}}\,\,{\text{even}}...} \right)\)


This kind of observation follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: If n is an integer, is (n+1)^2 an even integer?  [#permalink]

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New post 16 Sep 2018, 17:28
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

\((n+1)^2\) is an even integer
=> \(n+1\) is an even integer
=> \(n\) is an odd integer

Condition 1)
Since “\(n-1\) is an even integer” is equivalent to “\(n\) is an odd integer”, condition 1) is sufficient.

Condition 2)
“\((n-1)^2\) is an even integer” is equivalent to “\(n-1\) is an even integer”, which is condition 1).
Condition 2) is also sufficient.

Therefore, D is the answer.

Answer: D

Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2).
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Re: If n is an integer, is (n+1)^2 an even integer? &nbs [#permalink] 16 Sep 2018, 17:28
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