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m08-q26

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Re: m08-q26 [#permalink]

30 Aug 2010, 10:14

IMO B, tried plugging in values.....

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Re: m08-q26 [#permalink]

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, 07:27

Francois wrote:

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

XY divisible by 4:
X= √2, Y= 0 => XY = 0 (divisible by 4)...satisfied
X= √2, Y= 8√2 => XY = 32 (divisible by 4)...satisfied
X= √2, Y= -8√2 => XY = -32 (divisible by 4)...satisfied

In the first case, Y=0 - not a positive integer...satisfied
In the 2nd case, Y=8√2 -not a positive integer...satisfied
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Re: m08-q26 [#permalink]

30 Aug 2011, 02:45

how about C? can someone explain why it's not feasible?

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Re: m08-q26 [#permalink]

30 Aug 2011, 12:02

i agree , we can get answer by elimination method....but how come B is true.

If y is negative integer, then it will be any number...
0 is one value in that not positive number set which will have {0,-ve numbers....}

questions asked about which of the following is must be true...not could be true....even then B is not correct..actually all options are wrong.....

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Re: m08-q26q [#permalink]

31 Aug 2011, 03:38

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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!

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Re: m08-q26 [#permalink]

03 Sep 2011, 09:13

Wow. Really created a flutter this one! And it was so simple all the while! Great explanations everybody.
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Re: m08-q26 [#permalink]

30 Aug 2012, 05:23

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

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.
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Re: m08-q26 [#permalink] 30 Aug 2012, 05:23

m08-q26

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