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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # If x and y are integers, is x^2+x+y an even number?  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6495 GMAT 1: 760 Q51 V42 GPA: 3.82 If x and y are integers, is x^2+x+y an even number? [#permalink] ### Show Tags 05 Dec 2017, 01:00 00:00 Difficulty: 25% (medium) Question Stats: 84% (01:00) correct 16% (00:59) wrong based on 52 sessions ### HideShow timer Statistics [GMAT math practice question] If $$x$$ and $$y$$ are integers, is $$x^2+x+y$$ an even number? 1) $$x=5$$ 2) $$y=5$$ _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Concentration: Strategy, Finance
GMAT 1: 620 Q46 V29
Re: If x and y are integers, is x^2+x+y an even number?  [#permalink]

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05 Dec 2017, 01:34
First, we can easily get rid of powers, because it doesn't affect the process of reasomning

(1) x=5=Odd, so we have
Odd+Odd+ Even/Odd =Even+ Even/Odd=Even/Odd
Insufficient
(2) y=5=Odd
1 case: x=Even
Even+Even+Odd=Odd
2 case: x=Odd
Odd+Odd+Odd=Odd
Sufficient

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6495
GMAT 1: 760 Q51 V42
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Re: If x and y are integers, is x^2+x+y an even number?  [#permalink]

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06 Dec 2017, 23:58
=>

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, and then recheck the question.

Since $$x^2+x=x(x+1)$$ is always even, the question is really asking, “is y even?”. Condition 1) tells us nothing about the value of y, so it is not sufficient. Since condition 2) tells us that y = 5, it is sufficient, and the answer is B.
Condition 2)
Setting $$y = 5 yields$$
$$x^2+x+y = x^2+x+5 = x(x+1) + 5,$$
which is odd since x(x+1) is the product of two consecutive integers, one of which must be even.
This is sufficient by CMT(Common Mistake Type) 1.

Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E).

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Re: If x and y are integers, is x^2+x+y an even number? &nbs [#permalink] 06 Dec 2017, 23:58
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