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M15-23

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Re: M15-23  [#permalink]

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New post 27 Jan 2017, 06:54
Good catch ... thanks for posting this question.
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Re: M15-23  [#permalink]

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New post 15 Dec 2017, 08:41
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 is Negative? We know they're even integers but don't know if X>Y or not.
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Re: M15-23  [#permalink]

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New post 15 Dec 2017, 08:43
sambha 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 is Negative? We know they're even integers but don't know if X>Y or not.


Can't a negative integer be divisible by 8? For example, -8 IS divisible by 8.
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Re: M15-23  [#permalink]

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New post 21 May 2018, 13:57
what if Y>X?
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Re: M15-23  [#permalink]

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New post 07 Jun 2018, 07:40
Bunuel wrote:
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.

Hi bunuel,
By this you mean to say that "only an integer value is divisible by an other integer value" even though the non integer value when divided with an integer gives a remainder which equals zero ?
Please give an assertion to this
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Re: M15-23  [#permalink]

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New post 08 Jun 2018, 02:19
PreethiVishwanathan wrote:
Bunuel wrote:
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.

Hi bunuel,
By this you mean to say that "only an integer value is divisible by an other integer value" even though the non integer value when divided with an integer gives a remainder which equals zero ?
Please give an assertion to this


Yes. x is divisible by y means that x is integer, y is an integer and x/y is an integer.
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Re M15-23  [#permalink]

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New post 16 Jun 2018, 20:37
I think this the explanation isn't clear enough, please elaborate. Even if x and y are not integers x^2-y^2 could still be written as (x+y)(x-y) and since x+y is divisible by 8 (x+y)(x-y) should be divisible by 8.
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Re: M15-23  [#permalink]

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New post 16 Jun 2018, 22:09
Anishajha wrote:
I think this the explanation isn't clear enough, please elaborate. Even if x and y are not integers x^2-y^2 could still be written as (x+y)(x-y) and since x+y is divisible by 8 (x+y)(x-y) should be divisible by 8.


a is divisible by b means that a is integer, b is an integer and a/b is an integer. In the case you mention we cannot be sure that (x+y)(x-y) and (x+y)(x-y)/8 are integers.
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Re M15-23  [#permalink]

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New post 23 Oct 2018, 04:11
I think this the explanation isn't clear enough, please elaborate.
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Re M15-23  [#permalink]

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New post 26 Oct 2018, 06:10
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.
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Re M15-23 &nbs [#permalink] 26 Oct 2018, 06:10

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