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# Is x^2 - y^2 divisible by 8?

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VP
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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

M15-23
[Reveal] Spoiler: OA
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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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09 May 2009, 15:51
OA is not E.

(x-y) (x+y) divisible by 8?

Why is it not B?

Given that (x+y) = 8L

(x-y) X 8 L /8 = (x-y) L

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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

x-y *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 x-y =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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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 x-y =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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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 x-y =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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09 May 2009, 23:33
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X^2-Y^2 can be wriitten as (x-y)(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 (X-y)(X+Y) is divisible by 8 and hence remainder is Zero

B
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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 x-y =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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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:($$(x^2-y^2)/8$$ an integer)?
Question:($$((x-y)(x+y))/8$$ an integer)?

(1) x & y are even
Insufficient

If we look at it from an even/odd perspective, even-even=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, (x-y)(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:($$((2x-8(\lambda))(8(\lambda)))/8$$ an integer)?
Question:($$((8(\lambda))(2x-8(\lambda)))/8$$ an integer)?
Question:($$(\lambda)(2x-8(\lambda))$$ an integer)?
Question:($$2x-8(\lambda)$$ 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.

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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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21 May 2009, 11:24
0 is divisible by 8... 0 is divisible by any number.
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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

x-y *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 x-y =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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05 Sep 2009, 20:00
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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 $$8$$ --> $$x^2 - y^2=(x+y)(x-y)$$, 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.

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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^2-y^2=(x+y)(x-y) .. 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)(11.6-4.4)=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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05 Sep 2009, 22:49
If and only x and y are integers..
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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. X-Y might not be an integer and 1) will ensure that X-Y 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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02 Sep 2010, 16:22
I x, y are even then (x+y) is even (x-y) 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.
(x-y)=8a-8b=8(a-b) that is div by 8.
(x+y)(x-y)=8(a+b)8(a-b) this is div by 8 so the answer is b
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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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09 Jun 2014, 01:06
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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 $$8$$ --> $$x^2 - y^2=(x+y)(x-y)$$, 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.

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)(x-y)/8
(4.8+3.2)(4.8-3.2)/8
8(1.6)/8
=1.6
What am I doing wrong here?
Re: divisible by 8?   [#permalink] 15 Jun 2014, 11:31

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