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Official Solution:


(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 \(8\). Now, \(x^2 - y^2=(x+y)(x-y)\), 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 \((x+y)(x-y)\) is divisible by \(8\). Sufficient.


Answer: C
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Re: M15-23 [#permalink]

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New post 07 Jan 2015, 09:51
Bunuel wrote:
Official Solution:


(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 \(8\). Now, \(x^2 - y^2=(x+y)(x-y)\), 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 \((x+y)(x-y)\) is divisible by \(8\). Sufficient.


Answer: C


what if X=Y
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New post 07 Jan 2015, 10:07
him1985 wrote:
Bunuel wrote:
Official Solution:


(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 \(8\). Now, \(x^2 - y^2=(x+y)(x-y)\), 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 \((x+y)(x-y)\) is divisible by \(8\). Sufficient.


Answer: C


what if X=Y


0 is divisible by every integer (except 0 itself).

I think you should go through the basics and brush up fundamentals once more. For a start check here: divisibility-multiples-factors-tips-and-hints-174998.html

Hope it helps.
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Re: M15-23 [#permalink]

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New post 07 Jan 2015, 11:00
Bunuel wrote:
Official Solution:


(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 \(8\). Now, \(x^2 - y^2=(x+y)(x-y)\), 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 \((x+y)(x-y)\) is divisible by \(8\). Sufficient.


Answer: C


Hi Bunuel
If \(x=4.8\) and \(y=3.2\) then also we can write \(4.8^2 - 3.2^2=(4.8+3.2)(4.8-3.2)\)
and 8.0(1.6) can be divisible by 8 as well (if not than why?)
I know you have mentioned that if one of the multiples is divisible by 8 then so is the product: true for integers,
but it seems that with \(x=4.8\) and \(y=3.2\) , its working in this case.
puzzled :|
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New post 07 Jan 2015, 11:12
Ankur9 wrote:
Bunuel wrote:
Official Solution:


(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 \(8\). Now, \(x^2 - y^2=(x+y)(x-y)\), 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 \((x+y)(x-y)\) is divisible by \(8\). Sufficient.


Answer: C


Hi Bunuel
If \(x=4.8\) and \(y=3.2\) then also we can write \(4.8^2 - 3.2^2=(4.8+3.2)(4.8-3.2)\)
and 8.0(1.6) can be divisible by 8 as well (if not than why?)
I know you have mentioned that if one of the multiples is divisible by 8 then so is the product: true for integers,
but it seems that with \(x=4.8\) and \(y=3.2\) , its working in this case.
puzzled :|


4.8^2 - 3.2^2 = 12.8 is not divisible by 8 because 12.8 is not an integer.
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Re: M15-23 [#permalink]

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New post 11 Apr 2015, 19:32
Bunuel wrote:
Ankur9 wrote:
Bunuel wrote:
Official Solution:


(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 \(8\). Now, \(x^2 - y^2=(x+y)(x-y)\), 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 \((x+y)(x-y)\) is divisible by \(8\). Sufficient.


Answer: C


Hi Bunuel
If \(x=4.8\) and \(y=3.2\) then also we can write \(4.8^2 - 3.2^2=(4.8+3.2)(4.8-3.2)\)
and 8.0(1.6) can be divisible by 8 as well (if not than why?)
I know you have mentioned that if one of the multiples is divisible by 8 then so is the product: true for integers,
but it seems that with \(x=4.8\) and \(y=3.2\) , its working in this case.
puzzled :|


4.8^2 - 3.2^2 = 12.8 is not divisible by 8 because 12.8 is not an integer.



we have x = 4.8 and y = 3.2, in which case (x+y)(x-y) = (4.8+3.2)(4.8-3.2) = (8)(1.6) = 12.8
Now We know that a no. is divisible by another no. only if the remainder is an integer. Here, if we accept your expalantion, then remainder is 1.6, which is not an integer. Therefore not divisible by 8.

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New post 12 Apr 2015, 03:39
abhi1785 wrote:
Bunuel wrote:
Ankur9 wrote:

Hi Bunuel
If \(x=4.8\) and \(y=3.2\) then also we can write \(4.8^2 - 3.2^2=(4.8+3.2)(4.8-3.2)\)
and 8.0(1.6) can be divisible by 8 as well (if not than why?)
I know you have mentioned that if one of the multiples is divisible by 8 then so is the product: true for integers,
but it seems that with \(x=4.8\) and \(y=3.2\) , its working in this case.
puzzled :|


4.8^2 - 3.2^2 = 12.8 is not divisible by 8 because 12.8 is not an integer.



we have x = 4.8 and y = 3.2, in which case (x+y)(x-y) = (4.8+3.2)(4.8-3.2) = (8)(1.6) = 12.8
Now We know that a no. is divisible by another no. only if the remainder is an integer. Here, if we accept your expalantion, then remainder is 1.6, which is not an integer. Therefore not divisible by 8.


Sorry, but your question is not clear. Please elaborate.

As for the red part: an integer is divisible by an integer if the remainder is 0.
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Re: M15-23 [#permalink]

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New post 03 May 2015, 02:11
the definition of being divisible has to be mentioned:
A is divisible by B if A/B = C is an integer. 12,8/8 = 1,6 which isn't an integer.

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Re: M15-23 [#permalink]

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New post 28 May 2015, 20:06
Bunuel wrote:
Official Solution:


(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 \(8\). Now, \(x^2 - y^2=(x+y)(x-y)\), 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 \((x+y)(x-y)\) is divisible by \(8\). Sufficient.


Answer: C




But even if we add 3.2 and 4.8 it equals 8. And if we expand x^2-y^2, it will be divisible by 8. So irrespective of x & y being integer or rational or decimal, if it gives x+y or x-y as 8 or a multiple of 8, i guess it should work. That's why I marked B. :( :(

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New post 29 May 2015, 04:02
usadude05 wrote:
Bunuel wrote:
Official Solution:


(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 \(8\). Now, \(x^2 - y^2=(x+y)(x-y)\), 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 \((x+y)(x-y)\) is divisible by \(8\). Sufficient.


Answer: C




But even if we add 3.2 and 4.8 it equals 8. And if we expand x^2-y^2, it will be divisible by 8. So irrespective of x & y being integer or rational or decimal, if it gives x+y or x-y as 8 or a multiple of 8, i guess it should work. That's why I marked B. :( :(


Please read the whole thread: m15-184055.html#p1466250
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New post 19 Jun 2015, 06:19
Great question Bunnel..

Totally Stumped :)

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New post 17 Mar 2016, 12:56
Cadaver wrote:
Great question Bunnel..

Totally Stumped :)


Totally agree with Cadaver on this one.

Hat off Bunuel.
Lovely question. Really awesome solution/explanation.

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New post 01 Jun 2016, 19:41
desiwolverine wrote:
Cadaver wrote:
Great question Bunnel..

Totally Stumped :)


Totally agree with Cadaver on this one.

Hat off Bunuel.
Lovely question. Really awesome solution/explanation.


Yep, agreed! I was trapped as well - what a good question! :-D
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New post 24 Jul 2016, 18:10
hi!
when (x+y)(x-y) = xsq- ysq...
then one of the multiple that is x+y is divisible by 8.. why do we care about second one.
does it matter we write xsq - y sq or (x+y)(x-y)?
thanks

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New post 24 Jul 2016, 19:31
Celestial09 wrote:
hi!
when (x+y)(x-y) = xsq- ysq...
then one of the multiple that is x+y is divisible by 8.. why do we care about second one.
does it matter we write xsq - y sq or (x+y)(x-y)?
thanks

Celestial09 : the key here is whether x & y are Integer or not, because divisibility is defined by remainder = 0.

So in the case that, x=3.8, y=4.2 --> \(x+y=8 (divisible by 8), x-y=0.4 -> (x^2-y^2) = 4.2*0.4=1.68\) --> this number is not divisible by 8 (since remainder is #0)
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New post 20 Aug 2016, 03:40
I think this is a poor-quality question and I don't agree with the explanation. if x+y is a multiple of 8 as stated by statement 2, then x^2 - y^2 is divisible by 8 as well since the 8 divides the (x+y) in (x+y)(x-y) form of the expression. Why is this logic wrong , this is not addressed in this thread.

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New post 21 Aug 2016, 02:19
Senthil7 wrote:
I think this is a poor-quality question and I don't agree with the explanation. if x+y is a multiple of 8 as stated by statement 2, then x^2 - y^2 is divisible by 8 as well since the 8 divides the (x+y) in (x+y)(x-y) form of the expression. Why is this logic wrong , this is not addressed in this thread.


Even if x+y is a multiple of 8, (x+y)(x-y) might not be an integer. Divisibility is applied to integers on the GMAT.
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New post 26 Oct 2016, 06:47
hi
12.8/8=4*3.2/8=3.2/2=32/10*2=16/10=1.6

8 is completely divided in 12.8. Why should we care whether (x-y) is integer or not?

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Re: M15-23   [#permalink] 26 Oct 2016, 06:51

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