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# m15 question 23

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m15 question 23 [#permalink]  11 Apr 2011, 10:36
In this question, wouldn't statements 1 and 2 together be insufficient?

Question:
Is x² - y² divisible by 8?

1. x and y are even integers

2. x + y is divisible by 8

According to the answer key, the correct answer is C, but aren't the statements together insufficient? Consider:

The answer is yes if, for example x=6 and y=2 because (6+2)(6-2) is divisible by 8

The answer is no if x=4 and y=4 because (4+4)(4-4) is not divisible by 8
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Re: m15 question 23 [#permalink]  11 Apr 2011, 10:59
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Fk5000 wrote:
In this question, wouldn't statements 1 and 2 together be insufficient?

Question:
Is x² - y² divisible by 8?

1. x and y are even integers

2. x + y is divisible by 8

According to the answer key, the correct answer is C, but aren't the statements together insufficient? Consider:

The answer is yes if, for example x=6 and y=2 because (6+2)(6-2) is divisible by 8

The answer is no if x=4 and y=4 because (4+4)(4-4) is not divisible by 8

Infamous "0" is a multiple of everything "Or" 0 is divisible by everything but itself "Or" every known number is a factor of "0".

Thus,
0 is divisible by 8. 0 when divided by 8 leaves a remainder 0.
"0/8=0"
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Re: m15 question 23 [#permalink]  12 Apr 2011, 08:46
Fk5000 wrote:
In this question, wouldn't statements 1 and 2 together be insufficient?

Question:
Is x² - y² divisible by 8?

1. x and y are even integers

2. x + y is divisible by 8

According to the answer key, the correct answer is C, but aren't the statements together insufficient? Consider:

The answer is yes if, for example x=6 and y=2 because (6+2)(6-2) is divisible by 8

The answer is no if x=4 and y=4 because (4+4)(4-4) is not divisible by 8

Hi Fluke,

If it was mentioned in the question itself that x and y are integers (not necessarily even integers), Option B would have sufficed right?
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Re: m15 question 23 [#permalink]  12 Apr 2011, 11:45
fluke wrote:
Fk5000 wrote:
In this question, wouldn't statements 1 and 2 together be insufficient?

Question:
Is x² - y² divisible by 8?

1. x and y are even integers

2. x + y is divisible by 8

According to the answer key, the correct answer is C, but aren't the statements together insufficient? Consider:

The answer is yes if, for example x=6 and y=2 because (6+2)(6-2) is divisible by 8

The answer is no if x=4 and y=4 because (4+4)(4-4) is not divisible by 8

Infamous "0" is a multiple of everything "Or" 0 is divisible by everything but itself "Or" every known number is a factor of "0".

Thus,
0 is divisible by 8. 0 when divided by 8 leaves a remainder 0.
"0/8=0"

Still Didn't get you fluke
isn't statement 2 sufficient
as x² - y²=(x+y)(x-y)
now if (x+y) is divisible by 8 then whole term x² - y² will also be divisible by 8

did not understand how st 1 is useful
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Re: m15 question 23 [#permalink]  12 Apr 2011, 11:58
@ Warlock007,

We are not told that x and y are integers. If x=3/4 and y= 3/2, statement B alone will not suffice. We need to be sure that x and y are integers. Thats what statement 1 tells us and thats how it is useful
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Re: m15 question 23 [#permalink]  12 Apr 2011, 12:23
Warlock007 wrote:
Still Didn't get you fluke
isn't statement 2 sufficient
as x² - y²=(x+y)(x-y)
now if (x+y) is divisible by 8 then whole term x² - y² will also be divisible by 8

did not understand how st 1 is useful

I agree with gmatpapa.

x=6.8
y=1.2

x+y=6.8+1.2=8
x-y=6.8-1.2=5.6

x^2-y^2=(x+y)(x-y)=8*5.6=44.8

44.8 is not divisible by 8

44.8/8 = 448/80=28/5 will leave a remainder of 3.

If both x and y were even integers, then x+y=even integer, x-y=even integer
And (x+y) is divisible by 8, then (x+y)/8= integer

$$\frac{x^2-y^2}{8}=\frac{(x+y)(x-y)}{8}$$
$$\frac{x+y}{8}*(x-y)=integer*integer=integer$$
Thus 8 will divide $$x^2-y^2$$ leaving an integer as quotient and no remainder.
Ans: "C"
*************

Raise concern, if any.
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Re: m15 question 23 [#permalink]  12 Apr 2011, 13:05
I think Is x² - y² divisible by 8?

can be rephrased as Is (x-y)(x+y) divisible by 8 where x and y are integers. Is this correct?
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Re: m15 question 23 [#permalink]  12 Apr 2011, 13:09
gmat1220 wrote:
I think Is x² - y² divisible by 8?

can be rephrased as Is (x-y)(x+y) divisible by 8 where x and y are integers. Is this correct?

Why did you add the extra bit "where x and y are integers". I am not able to gather your thought process. I would just rephrase it as, "Is (x-y)(x+y) divisible by 8?".
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Re: m15 question 23 [#permalink]  12 Apr 2011, 20:51
Hello fluke

Is x² - y² divisible by 8?
I think the necessary condition (x+y)or(x-y) or their product divisible by 8 is not sufficient to answer the question. We also need the condition - Is x and y integers? to conclusively answer.

yes to both questions - the answer is yes
yes and no to questions - the answer is no
no and no to questions - the answer is no

I meant we need answers to two questions to answer this question - Is x² - y² divisible by 8? Isnt it?
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Re: m15 question 23 [#permalink]  12 Apr 2011, 21:18
gmat1220 wrote:
Hello fluke

Is x² - y² divisible by 8?
I think the necessary condition (x+y)or(x-y) or their product divisible by 8 is not sufficient to answer the question. We also need the condition - Is x and y integers? to conclusively answer.

yes to both questions - the answer is yes
yes and no to questions - the answer is no
no and no to questions - the answer is no

I meant we need answers to two questions to answer this question - Is x² - y² divisible by 8? Isnt it?

I see. Yes, correct. I somehow thought that you were trying to rephrase the question stem. Gotcha.
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Re: m15 question 23 [#permalink]  20 Feb 2013, 22:14
Why is B not sufficient...

x^2-y^2 =(x+y)(x-y)

If x+y is divisible by 8 .. so is (x+y)(x-y).

Kindly help..
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Re: m15 question 23 [#permalink]  21 Feb 2013, 03:32
Expert's post
Sachin9 wrote:
Why is B not sufficient...

x^2-y^2 =(x+y)(x-y)

If x+y is divisible by 8 .. so is (x+y)(x-y).

Kindly help..

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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Re: m15 question 23 [#permalink]  21 Feb 2013, 04:16
Bunuel wrote:
Sachin9 wrote:
Why is B not sufficient...

x^2-y^2 =(x+y)(x-y)

If x+y is divisible by 8 .. so is (x+y)(x-y).

Kindly help..

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.

x^2-y^2 is ultimately (x+y)(x-y) .. So irrespective of whether x and y are integers or not.. if (x+y) is divisible by 8 , why can't (x+y)* ( x-y) be divisible by 8?

What am I missing?
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Re: m15 question 23 [#permalink]  21 Feb 2013, 04:20
Expert's post
Sachin9 wrote:
Bunuel wrote:
Sachin9 wrote:
Why is B not sufficient...

x^2-y^2 =(x+y)(x-y)

If x+y is divisible by 8 .. so is (x+y)(x-y).

Kindly help..

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.

x^2-y^2 is ultimately (x+y)(x-y) .. So irrespective of whether x and y are integers or not.. if (x+y) is divisible by 8 , why can't (x+y)* ( x-y) be divisible by 8?

What am I missing?

In my post above there is an example for which x^2 - y^2 is NOT divisible by 8 (x=4.8 and y=3.2) because it's not an integer at all.
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Re: m15 question 23 [#permalink]  21 Feb 2013, 07:44
Bunuel wrote:

In my post above there is an example for which x^2 - y^2 is NOT divisible by 8 (x=4.8 and y=3.2) because it's not an integer at all.

So do you mean to say that x^2 - y^2 = (x+y)*(x-y) only when x and y are integers?
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Re: m15 question 23 [#permalink]  22 Feb 2013, 01:15
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Expert's post
Sachin9 wrote:
Bunuel wrote:

In my post above there is an example for which x^2 - y^2 is NOT divisible by 8 (x=4.8 and y=3.2) because it's not an integer at all.

So do you mean to say that x^2 - y^2 = (x+y)*(x-y), only when x and y are integers?

No, I didn't say that!

Consider this: if x=4.8 and y=3.2, then x^2-y^2=12.8. Is 12.8 divisible by 8?
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Re: m15 question 23 [#permalink]  22 Feb 2013, 01:17
Bunuel wrote:
Sachin9 wrote:
Bunuel wrote:

In my post above there is an example for which x^2 - y^2 is NOT divisible by 8 (x=4.8 and y=3.2) because it's not an integer at all.

So do you mean to say that x^2 - y^2 = (x+y)*(x-y), only when x and y are integers?

No, I didn't say that!

Consider this: if x=4.8 and y=3.2, then x^2-y^2=12.8. Is 12.8 divisible by 8?

I am getting you..
So , x^2 - y^2 will be divisible by 8 only if x and y are integers..
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Re: m15 question 23 [#permalink]  22 Feb 2013, 01:27
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No. x^2 - y^2 can be divisible by 8 even if x and y are not integers. For example, x=7.5 and y=0.5.
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Re: m15 question 23 [#permalink]  22 Feb 2013, 01:36
Bunuel wrote:
No. x^2 - y^2 can be divisible by 8 even if x and y are not integers. For example, x=7.5 and y=0.5.

ok. so no such rule for divisibility as such.. we gotto plug in numbers and check .. numbers to be plugged in would depend on the scope of the variables..
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Re: m15 question 23 [#permalink]  22 Feb 2013, 02:09
Is x² - y² divisible by 8?

1. x and y are even integers

2. x + y is divisible by 8

It is a Yes No Data Sufficiency question wherein you should try to disprove the statements rather than prove them. To disprove a statement there should be at least two values which will satisfy and not satisfy the condition in the question or simply you need a yes and a no from the statement.

Lets see:

Statement (1): Lets substitute for x and y
Let us say x = 4 and y = 2 (Satisfies the statement)
Putting it in x² - y² we get, 12 which is not divisible by 8, hence we get a 'NO'.

Now lets try for a yes, let us say x = 16 and y = 8(Satisfies the statement)
x² - y² = 192, which is divisible by 8, so we have a 'YES'

Since we have a yes and a no so this statement is not sufficient.

Statement (2): Let us do the same for statement 2
X = 6 and Y = 2 (Satisfies the statement)
x² - y² = 32, so we get a YES

X = 7.6 and Y = 0.4
x² - y² = 57.6, hence it is not divisible by 8

If we combine both we get the answer as it tells us that x and y are integers.
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Re: m15 question 23   [#permalink] 22 Feb 2013, 02:09
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# m15 question 23

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