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# m08-q26

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Manager
Joined: 05 Jan 2009
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30 May 2009, 19:09
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If $$XY$$ is divisible by 4, which of the following must be true?

(A) If $$X$$ is even then $$Y$$ is odd.
(B) If $$X = \sqrt{2}$$ then $$Y$$ is not a positive integer.
(C) If $$X$$ is 0 then $$X + Y$$ is not 0.
(D) $$X^Y$$ is even.
(E) $$\frac{X}{Y}$$ is not an integer.

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REVISED VERSION OF THIS QUESTION WITH SOLUTION IS HERE: m08-q26-78982-20.html#p1117292
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01 Jun 2009, 15:05
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Because it is divisible by 4
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01 Jun 2009, 15:12
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XY must be an integer because when a number is divisible by an integer, this means that the quotient MUST be an integer. Sure you can divide 8.12 by 4, you get 2.03, but that does not mean that 8.12 is divisible by 4. divisibility implies no decimal remainders.

SO XY must be an integer. What I'm not sure of though with this question is why must Y NOT be a positive integer?

pmal04 wrote:
The OE is not making sense to me. Why XY has to be an integer?
why can't it be 8*root2?

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Manager Joined: 05 Jan 2009 Posts: 81 Followers: 1 Kudos [?]: 116 [1] , given: 2 Re: m08-q26 [#permalink] ### Show Tags 07 Jun 2009, 12:29 1 This post received KUDOS Thanks jallenmorris for your reply. +1 for you. Also, I am not sure why must Y NOT be a positive integer. Can founder please explain it? SVP Joined: 30 Apr 2008 Posts: 1874 Location: Oklahoma City Schools: Hard Knocks Followers: 42 Kudos [?]: 590 [1] , given: 32 Re: m08-q26 [#permalink] ### Show Tags 07 Jun 2009, 14:49 1 This post received KUDOS I agree with you regarding "positive integer". This doesn't make any sense. Maybe there is something that I do not understand, but I see no reason that "positive" matters. pmal04 wrote: Thanks jallenmorris for your reply. +1 for you. Also, I am not sure why must Y NOT be a positive integer. Can founder please explain it? _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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21 Aug 2009, 06:03
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As already told by bipolarbear, if $$X=\sqrt{2}$$ and $$Y=0$$, then $$XY=0$$ and it is divisible by 4. If option B doesn't have "positive integer", then B is violated as 0 is an integer (neither positive nor negative).

Please tell me if I'm wrong.
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Manager
Joined: 24 Aug 2010
Posts: 189
Location: Finland
Schools: Admitted: IESE($$),HEC, RSM,Esade WE 1: 3.5 years international Followers: 6 Kudos [?]: 94 [1] , given: 18 Re: m08-q26 [#permalink] ### Show Tags 25 Aug 2010, 06:46 1 This post received KUDOS dzyubam wrote: Thanks guys. +1 to both of you. We deleted the "positive" and now it reads "not an integer" in option B. Then the question becomes much easier to solve. Intern Joined: 07 Jul 2010 Posts: 13 Followers: 0 Kudos [?]: 8 [1] , given: 3 Re: m08-q26 [#permalink] ### Show Tags 26 Aug 2010, 02:15 1 This post received KUDOS If XY is divisible by 4, which of the following must be true? (A) If X is even then Y is odd --> X=4 Y =4 XY is disivible by 4, X is even and Y is not odd--> Eliminate B) If X=2^1/2 then Y is not a positive integer --> keep (C) If X is 0 then X + Y is not 0 --> X = 0 and Y =0 , then XY divisible by 4 --> eliminate (D) X^Y is even --> X=3, Y = 4 then XY is divisible by 4 and X^Y=81 even--> eliminate (E) X/Y is not an integer.--> X=4 and Y=4 then X/Y=1--> eliminate Hence can only be B by elimination. Prove that B is true: if X=2^1/2 there is no integer such as XY is divisible by 4 except 0. In fact Y needs to be divisible by 4 but XY divided by 4 will have a reminder as X = 2^1/2 Hence saying Y is not a positive integer is correct as 0 is the only solution. Please correct me if I am wrong Senior Manager Joined: 08 Jun 2010 Posts: 392 Location: United States Concentration: General Management, Finance GMAT 1: 680 Q50 V32 Followers: 3 Kudos [?]: 93 [1] , given: 13 Re: m08-q26q [#permalink] ### Show Tags 31 Aug 2011, 03:38 1 This post received KUDOS Brilliant question!! My initial thought was sqrt2 is a valid remainder. Haha. Surprising how assumptions can kill you!! Although I chose B (a good guess!) I plugged in x is sqrt2, then argued with myself that Y can be four and the statement will hold true. So, takeaway from this is REMAINDERS HAVE TO BE INTEGERS! Posted from my mobile device Math Expert Joined: 02 Sep 2009 Posts: 38841 Followers: 7718 Kudos [?]: 105918 [1] , given: 11593 Re: m08-q26 [#permalink] ### Show Tags 30 Aug 2012, 05:23 1 This post received KUDOS Expert's post 3 This post was BOOKMARKED pmal04 wrote: If $$XY$$ is divisible by 4, which of the following must be true? (A) If $$X$$ is even then $$Y$$ is odd. (B) If $$X = \sqrt{2}$$ then $$Y$$ is not a positive integer. (C) If $$X$$ is 0 then $$X + Y$$ is not 0. (D) $$X^Y$$ is even. (E) $$\frac{X}{Y}$$ is not an integer. [Reveal] Spoiler: OA B Source: GMAT Club Tests - hardest GMAT questions The OE is not making sense to me. Why XY has to be an integer? why can't it be 8*root2? This question is also discussed here: if-xy-is-divisible-by-4-which-of-the-following-must-be-true-73389.html Revised version of this question is below: If $$x$$ and $$y$$ are positive integer and $$xy$$ is divisible by 4, which of the following must be true? A. If $$x$$ is even then $$y$$ is odd. B. If $$x$$ is odd then $$y$$ is a multiple of 4. C. If $$x+y$$ is odd then $$\frac{y}{x}$$ is not an integer. D. If $$x+y$$ is even then $$\frac{x}{y}$$ is an integer. E. $$x^y$$ is even. Notice that the question asks which of the following MUST be true not COULD be true. A. If $$x$$ is even then $$y$$ is odd --> not necessarily true, consider: $$x=y=2=even$$; B. If $$x$$ is odd then $$y$$ is a multiple of 4 --> always true: if $$x=odd$$ then in order $$xy$$ to be a multiple of 4 y mst be a multiple of 4; C. If $$x+y$$ is odd then $$\frac{y}{x}$$ is not an integer --> not necessarily true, consider: $$x=1$$ and $$y=4$$; D. If $$x+y$$ is even then $$\frac{x}{y}$$ is an integer --> not necessarily true, consider: $$x=2$$ and $$y=4$$; E. $$x^y$$ is even --> not necessarily true, consider: $$x=1$$ and $$y=4$$; Answer: B. _________________ SVP Joined: 30 Apr 2008 Posts: 1874 Location: Oklahoma City Schools: Hard Knocks Followers: 42 Kudos [?]: 590 [0], given: 32 Re: m08-q26 [#permalink] ### Show Tags 21 Aug 2009, 05:40 Why is it back to "...not a positive integer" ? dzyubam wrote: :) it's back to "... not a positive integer" bipolarbear wrote: dzyubam wrote: Thanks guys. +1 to both of you. We deleted the "positive" and now it reads "not an integer" in option B. Dzyubam, I think you were right at the beginning. The reason you need "positive" in (B), if X is SQRT(2), Y is not a "POSITIVE" integer, is because if X is SQRT(2), Y can still be zero, which does not violate the condition. However, if the answer choice is changed to Y is not an integer, then X = SQRT (2), Y = 0, and XY = 0 and divisible by 4. Y IS an integer so the statement isn't true. The other statements don't work either, so I think you need to change it back to "Positive." _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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21 Aug 2009, 08:00
Yeah, that makes sense. I wasn't thinking about 0 being divisible by any number, including 4.
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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Manager Joined: 14 Aug 2009 Posts: 123 Followers: 2 Kudos [?]: 105 [0], given: 13 Re: m08-q26 [#permalink] ### Show Tags 21 Aug 2009, 12:06 Y=0 or $$\sqrt{2}*2*n$$, n is an integer. so Y is not a positive integer. _________________ Kudos me if my reply helps! SVP Joined: 29 Aug 2007 Posts: 2476 Followers: 70 Kudos [?]: 774 [0], given: 19 Re: m08-q26 [#permalink] ### Show Tags 21 Aug 2009, 14:00 pmal04 wrote: The OE is not making sense to me. Why XY has to be an integer? why can't it be 8*root2? However B implies that y can be a -ve integer, which is not true. If x = sqrt2, y is neither a +ve nor a -ve integer. Yes it could be 0 or n(sqrt2) where n is an even integer. Looks like answer choice "B" needs to be revised in a more absolute way. _________________ Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html GT Manager Joined: 11 Sep 2009 Posts: 101 Followers: 2 Kudos [?]: 28 [0], given: 0 Re: m08-q26 [#permalink] ### Show Tags 08 Nov 2009, 07:20 I think that B is correct as it is because at least we know that Y cannot be a positive integer. We do not know whether Y may or may not be a negative integer at this point. Manager Joined: 09 Sep 2009 Posts: 57 Followers: 2 Kudos [?]: 11 [0], given: 13 Re: m08-q26 [#permalink] ### Show Tags 25 Aug 2010, 05:46 Hmm... since x=sqrt(2), the value of y can well be y=4/sqrt(2) or y=8/sqrt(2) or so on...(or, as mentioned before y=0) In that case y is not even an integer (forget it being positive or negative)... Any opinions?? Intern Joined: 07 Jul 2010 Posts: 13 Followers: 0 Kudos [?]: 8 [0], given: 3 Re: m08-q26 [#permalink] ### Show Tags 25 Aug 2010, 09:32 Shouldn t it be D. In fact if X is odd. Then for XY to be divisible by four we need Y to be divisble by four hence Y is even hence x^ y is even if x is even then whatever y such as xy divisble by 4 x^ y will be even Posted from my mobile device SVP Joined: 30 Apr 2008 Posts: 1874 Location: Oklahoma City Schools: Hard Knocks Followers: 42 Kudos [?]: 590 [0], given: 32 Re: m08-q26 [#permalink] ### Show Tags 25 Aug 2010, 10:07 I'm not sure I completely understand your specific question, but I'll attempt to explain why the correct answer to this problem is not D. D. X^Y is Even. Well, the stem tesll us to assume that XY is divisible by 4, and then we have to determine if D must be true. We know that D is not necessarily true. You are correct in that if X is ODD, then Y could be divisible by 4 and therefore XY would be divisible by 4. However, this isn't exaclty what the question is getting at. X^Y could be ODD. Lets say you have a situation where XY = 104 (X=13 Y = 8) 13^8 = ODD number (an odd*odd will always = odd). Therefore it is false to say "If XY is dividisble by for, then X^Y must always be odd." Francois wrote: Shouldn t it be D. In fact if X is odd. Then for XY to be divisible by four we need Y to be divisble by four hence Y is even hence x^ y is even if x is even then whatever y such as xy divisble by 4 x^ y will be even Posted from my mobile device _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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25 Aug 2010, 11:23
It cannot be D.

If XY = 12(3*4), X^Y = 3*4 . then 3 ^ 4 = 81 (odd).
If X = 8 (2*4) , X^Y= 2^4 = 16(even)

i feel like we shud beware of MUST BE TRUE q's. these q's demand that the correct answer choice should be true in all cases. this is not same as CAN BE TRUE q's.

Also, in B option , it says Y cannot be a positive integer if X=sqrt(2). it is not possible for Y to be positive or negative integer. but as others have already mentioned Y can be 0. so i feel like it should be " Y cannot be positive or negative integer" .

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31 Aug 2010, 06:52
sampatkumar wrote:
I do not understand the logic here..

If XY has to be zero to be right answer, Why option C is not correct?

If X = 0 then XY= 0, So devisible by 4

(C): If X is 0 then X + Y is not 0.
(a) Consider X=0, Y=1; XY=0, hence divisible by 4 and X+Y = 1
(b) Consider X=0, Y=0; XY=0, hence divisible by 4 and X+Y = 0
You'll agree that option C contradicts (b), and is thus wrong.
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Re: m08-q26   [#permalink] 31 Aug 2010, 06:52

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