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

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by Max@Math Revolution » Thu Sep 13, 2018 12:30 am

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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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Source: — Data Sufficiency |

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by Brent@GMATPrepNow » Thu Sep 13, 2018 7:23 am
Max@Math Revolution 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
Some important rules:
1. ODD +/- ODD = EVEN
2. ODD +/- EVEN = ODD
3. EVEN +/- EVEN = EVEN

4. (ODD)(ODD) = ODD
5. (ODD)(EVEN) = EVEN
6. (EVEN)(EVEN) = EVEN

Target question: Is (n+1)² an even integer?
This is a good candidate for rephrasing the target question.
Aside: Here's a video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100

(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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by fskilnik@GMATH » Thu Sep 13, 2018 8:51 am
Brent´s solution is perfect and full of important details.

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.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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Portuguese-speakers :: https://www.gmath.com.br
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by Max@Math Revolution » Sun Sep 16, 2018 5:24 pm
=>

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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by Ferguson259 » Mon Sep 17, 2018 1:12 am
Theorem: If n is an odd integer, then n2 is an odd integer. Proof: Since n is an odd integer, there exists an integer k such that n=2k+1. Therefore, n2 = (2k+1)2 = 4k2+4k+1 = 2(2k2+2k)+1

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