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Re: divisible by 8? [#permalink]
09 May 2009, 11:51
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.
Re: divisible by 8? [#permalink]
09 May 2009, 15:29
1
This post received KUDOS
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 _________________
Your attitude determines your altitude Smiling wins more friends than frowning
Re: divisible by 8? [#permalink]
10 May 2009, 08:06
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
Re: divisible by 8? [#permalink]
14 May 2009, 13:26
1
This post received KUDOS
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.
Re: divisible by 8? [#permalink]
17 Aug 2009, 08:06
1
This post received KUDOS
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
Re: divisible by 8? [#permalink]
05 Sep 2009, 19:00
4
This post received KUDOS
Expert's post
7
This post was BOOKMARKED
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.
Re: divisible by 8? [#permalink]
05 Sep 2009, 20:23
Expert's post
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 _________________
Re: divisible by 8? [#permalink]
12 Nov 2009, 18:31
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! _________________
Re: divisible by 8? [#permalink]
02 Sep 2010, 15:22
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
Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers 2. [#permalink]
09 Jun 2014, 00:06
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Re: divisible by 8? [#permalink]
15 Jun 2014, 10:31
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.
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)(x-y)/8 (4.8+3.2)(4.8-3.2)/8 8(1.6)/8 =1.6 What am I doing wrong here?
gmatclubot
Re: divisible by 8?
[#permalink]
15 Jun 2014, 10:31
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