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Re: M15-23 [#permalink]
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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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Re: M15-23 [#permalink]
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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Re: M15-23 [#permalink]
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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Re: M15-23 [#permalink]
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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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Re: M15-23 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M15-23 [#permalink]
I think this is a high-quality question and I agree with explanation. Wow !! Eye opener..
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Re: M15-23 [#permalink]
The question did not mention if x>y, may I ask if it would be considered divisible if x^2-y^2 is negative. Thanks.
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Re: M15-23 [#permalink]
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jessiemjx wrote:
The question did not mention if x>y, may I ask if it would be considered divisible if x^2-y^2 is negative. Thanks.


Yes. For example, -8, -16, -24, ... are divisible by 8.
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Re: M15-23 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M15-23 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M15-23 [#permalink]
One of my favourite GMAT questions. This will get even the most prudent test takers. Love it !

Posted from my mobile device
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Re: M15-23 [#permalink]
I fail to understand why the answer is not E because given both equations say x=4 and y=4 then (x+y)(x-y) = (8)(0) and would not be divisible by 8.
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Re: M15-23 [#permalink]
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GmatSchro wrote:
I fail to understand why the answer is not E because given both equations say x=4 and y=4 then (x+y)(x-y) = (8)(0) and would not be divisible by 8.


Some facts about 0.

ZERO:

1. Zero is an INTEGER.

2. Zero is an EVEN integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. Zero is neither positive nor negative (the only one of this kind).

4. Zero is divisible by EVERY integer except 0 itself (\(\frac{x}{0} = 0\), so 0 is a divisible by every number, x).

5. Zero is a multiple of EVERY integer (\(x*0 = 0\), so 0 is a multiple of any number, x).

6. Zero is NOT a prime number (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed: anything/0 is undefined.

8. Any non-zero number to the power of 0 equals 1 (\(x^0 = 1\))

9. \(0^0\) case is NOT tested on the GMAT.

10. If the exponent n is positive (n > 0), \(0^n = 0\).

11. If the exponent n is negative (n < 0), \(0^n\) is undefined, because \(0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0}\), which is undefined. You CANNOT take 0 to the negative power.

12. \(0! = 1! = 1\).


Hope it helps.
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M15-23 [#permalink]
Bunuel wrote:
Official Solution:


Is \(x^2 - y^2\) an integer divisible by 8?

(1) \(x\) and \(y\) are even integers.

This information is clearly insufficient. For instance, if \(x=y=0\), the answer is YES, whereas if \(x=2\) and \(y=0\), the answer is NO.

(2) \(x + y\) is an integer divisible by \(8\).

Since \(x^2 - y^2=(x+y)(x-y)\), if one of the factors is divisible by 8, then the product is as well—this is true for integers. However, we aren't given that \(x\) and \(y\) are integers. For example, if \(x=4.8\) and \(y=3.2\), while \(x+y\) is divisible by 8, \(x^2 - y^2\) is not. Not sufficient.

(1)+(2) Given that \(x\) and \(y\) are integers, and that \(x+y\) is divisible by 8, it follows that \((x+y)(x-y)\) is equal to a multiple of 8 multiplied by an integer, and is therefore divisible by 8. Sufficient.


Answer: C

Bunuel

Can't we say for option B X+Y= 8K so when we divide X+Y*X-Y it will be divisible irrespective of whether they are integer or not.What I'm missing in understanding
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Re: M15-23 [#permalink]
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saby1410 wrote:
Bunuel wrote:
Official Solution:


Is \(x^2 - y^2\) an integer divisible by 8?

(1) \(x\) and \(y\) are even integers.

This information is clearly insufficient. For instance, if \(x=y=0\), the answer is YES, whereas if \(x=2\) and \(y=0\), the answer is NO.

(2) \(x + y\) is an integer divisible by \(8\).

Since \(x^2 - y^2=(x+y)(x-y)\), if one of the factors is divisible by 8, then the product is as well—this is true for integers. However, we aren't given that \(x\) and \(y\) are integers. For example, if \(x=4.8\) and \(y=3.2\), while \(x+y\) is divisible by 8, \(x^2 - y^2\) is not. Not sufficient.

(1)+(2) Given that \(x\) and \(y\) are integers, and that \(x+y\) is divisible by 8, it follows that \((x+y)(x-y)\) is equal to a multiple of 8 multiplied by an integer, and is therefore divisible by 8. Sufficient.


Answer: C

Bunuel

Can't we say for option B X+Y= 8K so when we divide X+Y*X-Y it will be divisible irrespective of whether they are integer or not.What I'm missing in understanding


The question asks whether \(x^2 - y^2\) is an integer divisible by 8. If \(x = 4.8\) and \(y = 3.2\), then \(x^2 - y^2\) is not an integer, hence we cannot talk about its divisibility at all.
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