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Is x^2  y^2 divisible by 8? [#permalink]
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09 May 2009, 11:25
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Is x^2  y^2 divisible by 8? (1) x and y are even integers (2) x + y is divisible by 8 M1523
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Re: divisible by 8? [#permalink]
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09 May 2009, 12:51
ST1  x and y are even....even integers are from 0,2,4,6....etc ST2  x+y divisible by 8, suppose x+y=8 then x will be anything from 1 to 8 and same with y.
ST1 and 2 are also not possible, as per the above approach x may be 0,2,4,6 and y may be 0,2,4,6.
IMO E, what is OA.



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Re: divisible by 8? [#permalink]
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09 May 2009, 15:51
OA is not E.
(xy) (x+y) divisible by 8?
Why is it not B?
Given that (x+y) = 8L
(xy) X 8 L /8 = (xy) L
Answer is not B either.



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Re: divisible by 8? [#permalink]
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09 May 2009, 16:29
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icandy wrote: Is x^2  y^2 divisible by 8?
1. x and y are even integers 2. x + y is divisible by 8
* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient * Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient * BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient * EACH statement ALONE is sufficient * Statements (1) and (2) TOGETHER are NOT sufficient 1) x^2  y^2 xy *x+y x=4 y=2 > 2*6 > not divisible by 8 x=6 y=2 > 4*8 > divisble by 8 not sufficient 2) x + y is divisible by 8 here we are not sure whether x and y are integers or not. say x=7.6and y=0.4 x+y=8 but xy =7.2 8*7.2 is not disible by 8 not sufficient combined suffcient because x and y are integers,and x+y is divisible by 8 C
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Re: divisible by 8? [#permalink]
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09 May 2009, 16:34
x2suresh wrote: 2)
x + y is divisible by 8
here we are not sure whether x and y are integers or not.
say x=7.6and y=0.4 x+y=8 but xy =7.2 8*7.2 is not disible by 8
not sufficient
Wait! Isnt (8 X 7.2/8 = 7.2)?? and 8 divides 8 X 7.2 right? What I am missing?



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Re: divisible by 8? [#permalink]
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09 May 2009, 18:36
icandy wrote: x2suresh wrote: 2)
x + y is divisible by 8
here we are not sure whether x and y are integers or not.
say x=7.6and y=0.4 x+y=8 but xy =7.2 8*7.2 is not disible by 8
not sufficient
Wait! Isnt (8 X 7.2/8 = 7.2)?? and 8 divides 8 X 7.2 right? What I am missing? take this example x=7.75 y=0.25 8*7.5 = 60 is 60 divisible by 8 ?? No.
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Re: divisible by 8? [#permalink]
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09 May 2009, 23:33
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X^2Y^2 can be wriitten as (xy)(x+y)
statement 1. Insuffcient...x=4, y=2; X=6, Y=2
Statement 2. x+y is divisible by 8 that mean X+Y is factor of 8....so X+Y can be 8,16,24 etc...
so (Xy)(X+Y) is divisible by 8 and hence remainder is Zero
B



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Re: divisible by 8? [#permalink]
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10 May 2009, 09:06
lamhe and typhoidx,
I guess I am not alone. But I tell you what we are not alone. This is a bit crooked and twisted. I realized why x2suresh is correct. Thanks X2
say x=7.6and y=0.4 x+y=8 but xy =7.2 8*7.2 is not divisible by 8
When x2suresh is saying this, what he is alluding to is that yeah you will see 7.2 but what we are not realizing is that 7.2 itself is a fraction. 7 .2 is 72/10 and not just 7



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Re: divisible by 8? [#permalink]
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14 May 2009, 14:26
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icandy wrote: Is x^2  y^2 divisible by 8?
1. x and y are even integers 2. x + y is divisible by 8
* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient * Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient * BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient * EACH statement ALONE is sufficient * Statements (1) and (2) TOGETHER are NOT sufficient Question:(\((x^2y^2)/8\) an integer)? Question:(\(((xy)(x+y))/8\) an integer)? (1) x & y are even Insufficient If we look at it from an even/odd perspective, eveneven=even & even+even=even, and even*even=even, which ensures we have \(2^2\), but we need 3 2's in the prime factorizations to be sure. Here is a case that discards it. x=2,y=0, (xy)(x+y) = 2*2 which is NOT divisible by 8, so we have a NO answer. and of course 4 & 0, which gives us a YES answer. YES & NO ==> Insufficient. (2) \(x+y\) divisible by 8 Let's rewrite this to: \(x+y = 8(\lambda)\), where \(\lambda\) is any integer. \(y = 8(\lambda)  x\) Substitute above equation into question, Question:(\(((x(8(\lambda)  x))(x+(8(\lambda)  x)))/8\) an integer)? Question:(\(((2x8(\lambda))(8(\lambda)))/8\) an integer)? Question:(\(((8(\lambda))(2x8(\lambda)))/8\) an integer)? Question:(\((\lambda)(2x8(\lambda))\) an integer)? Question:(\(2x8(\lambda)\) an integer)? Question:(\(2x\) an integer)? Question:(\(x\) an integer)? We don't know. Insufficient. (1&2) Sufficient, as if X & Y are even this implies they are both integers, sufficient. Final Answer, C.
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Re: divisible by 8? [#permalink]
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21 May 2009, 09:53
But is it necessary that x and y are different. Consider this case: x = 8 and y = 8 In this case both conditions are satisfies but \(x^2  y^2\)is not divisible by 8
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Re: divisible by 8? [#permalink]
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21 May 2009, 11:24
0 is divisible by 8... 0 is divisible by any number.
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Re: divisible by 8? [#permalink]
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17 Aug 2009, 09:06
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x2suresh wrote: icandy wrote: Is x^2  y^2 divisible by 8?
1. x and y are even integers 2. x + y is divisible by 8
* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient * Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient * BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient * EACH statement ALONE is sufficient * Statements (1) and (2) TOGETHER are NOT sufficient 1) x^2  y^2 xy *x+y x=4 y=2 > 2*6 > not divisible by 8 x=6 y=2 > 4*8 > divisble by 8 not sufficient 2) x + y is divisible by 8 here we are not sure whether x and y are integers or not. say x=7.6and y=0.4 x+y=8 but xy =7.2 8*7.2 is not disible by 8 not sufficient combined suffcient because x and y are integers,and x+y is divisible by 8 C Hi Agree with C, works for even negative integer choices



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Re: divisible by 8? [#permalink]
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05 Sep 2009, 20:00
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Re: divisible by 8? [#permalink]
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05 Sep 2009, 21:23
Is x^2  y^2 divisible by 8? 1. x and y are even integers 2. x + y is divisible by 8 hi... 1) if x is 8...y is 2 ans is 'no' and if x is 8 and y is 4 ans is yes .. so insufficient 2) x^2y^2=(x+y)(xy) .. and if (x+y) is divisible by 8 .... and it is not given they r integers , it is not sufficient... for eg... nos are 11.6 and 4.4 ...so (11.6+4.4)(11.64.4)=16(7.2)... now 16(7.2) is divisible by 8 but in decimals.. combining two sufficient..... C pl post original ans
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Re: divisible by 8? [#permalink]
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05 Sep 2009, 22:49
If and only x and y are integers..
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Re: divisible by 8? [#permalink]
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12 Nov 2009, 19:31
icandy wrote: Is x^2  y^2 divisible by 8?
1. x and y are even integers 2. x + y is divisible by 8
* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient * Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient * BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient * EACH statement ALONE is sufficient * Statements (1) and (2) TOGETHER are NOT sufficient This question is tricky  most people will pick B because most of GMAT DS questions state that X, Y are integers. However, the correct answer should be C. XY might not be an integer and 1) will ensure that XY is an integer. I would have totally picked B on the test given the time constraint on each problem. It also looks like such an easy question too!
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Re: divisible by 8? [#permalink]
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02 Sep 2010, 16:22
I x, y are even then (x+y) is even (xy) is even and their product is even . not sufficient
II. x+y div by 8 means that I can write x=8a y =8b so x+y =8a+8b=8(a+b) that is div by 8. (xy)=8a8b=8(ab) that is div by 8. (x+y)(xy)=8(a+b)8(ab) this is div by 8 so the answer is b



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Re: divisible by 8? [#permalink]
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08 Oct 2011, 07:52
nice question and nice explanation tooo... considering the time constraints, i might have picked B.. thnx
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Re: Is x^2  y^2 divisible by 8? 1. x and y are even integers 2. [#permalink]
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Re: divisible by 8? [#permalink]
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15 Jun 2014, 11:31
Bunuel wrote: Is \(x^2  y^2\) divisible by 8?
(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer.
(2) \(x + y\) is divisible by \(8\) > \(x^2  y^2=(x+y)(xy)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2  y^2\) is not. Not sufficient.
(1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(xy)\) is divisible by \(8\). Sufficient.
Answer: C. I don't know, I think I'm missing something here. If x=4.8 and y=3.2, Still the product would be divisible by 8 IMO. (x+y)(xy)/8 (4.8+3.2)(4.83.2)/8 8(1.6)/8 =1.6 What am I doing wrong here?




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