Is x^2 - y^2 divisible by 8? 1. x and y are even integers. : GMAT Data Sufficiency (DS)
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# Is x^2 - y^2 divisible by 8? 1. x and y are even integers.

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Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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07 Dec 2009, 02:15
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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.

I got B as x^2 - y^2 = (x+b) (x-b)
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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07 Dec 2009, 02:27
study wrote:
Is $$x^2 - y^2$$ divisible by 8?

1. $$x$$ and $$y$$ are even integers.
2. $$x + y$$ is divisible by 8.

I got B as x^2 - y^2 = (x+b) (x-b)

1. let x = 12 and y =10 then x^2 - y^2 = 144 -100 =44 not divisible by 8
let x = 6 and y =2 then x^2 - y^2 = 36 - 4 = 32 divisible by 8
hence not suf

2. now x^2 - y^2 = (x+y) (x-y) since (x+y) is divisble by 8 hence Suff

so B
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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07 Dec 2009, 02:29
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study wrote:
Is $$x^2 - y^2$$ divisible by 8?

1. $$x$$ and $$y$$ are even integers.
2. $$x + y$$ is divisible by 8.

I got B as x^2 - y^2 = (x+b) (x-b)

Think B is not correct. Note that we are not told that $$x$$ and $$y$$ are integers.

(1) Clearly insufficient;
(2) $$x + y$$ is divisible by $$8$$ --> $$x^2 - y^2=(x+y)(x-y)$$, if one of the multiples is divisible product also divisible: 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: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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07 Dec 2009, 04:46
Bunuel wrote:
study wrote:
Is $$x^2 - y^2$$ divisible by 8?

1. $$x$$ and $$y$$ are even integers.
2. $$x + y$$ is divisible by 8.

I got B as x^2 - y^2 = (x+b) (x-b)

Think B is not correct. Note that we are not told that $$x$$ and $$y$$ are integers.

(1) Clearly insufficient;
(2) $$x + y$$ is divisible by $$8$$ --> $$x^2 - y^2=(x+y)(x-y)$$, if one of the multiples is divisible product also divisible: 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 may be wrong here but even when we consider x and y as non integers. lets say we consider values for x and y same as assumed in example above x=4.8 and y=3.2. Then x^2 - y^2 will give a value of 12.8 which is divisible by 8 (1.6 times). Question - is the quotient expected to be 0 or +ve integer ONLY?
if we consider x = 17/2 y= -1/2 or 15/2 and 1/2 still x^2 - y^2 is divisible by 8
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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07 Dec 2009, 04:56
hi... divisible means it should give u an int when divided and not decimal(1.6)..... if that is the case all nos would be divisible by 2,4,5,8,10.......
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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07 Dec 2009, 04:59
kp 1811 - thanks...that is exactly my question as well.

the rule x^2 - y^2 = (x+y) (x - y) cannot be true to just one side of the equation. If it is true to (x + y) (x - y), it has to be valid for x^2 - y^2, coz it is nothing than simplifying the equation. The answer should be the same irrespective of which equation one is following.
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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07 Dec 2009, 05:03
A decimal numerator would most likely give a decimal answer when divided by an integer. However, an integer when divided by another integer should give you an integer: isn't that the rule of divisibility?

For eg:

3.6/3 = 1.2 - Hence 3.6 is divisible by 3

36/3 = 12 - Here an integer is divided by an integer
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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07 Dec 2009, 05:22
kp1811 wrote:
I may be wrong here but even when we consider x and y as non integers. lets say we consider values for x and y same as assumed in example above x=4.8 and y=3.2. Then x^2 - y^2 will give a value of 12.8 which is divisible by 8 (1.6 times). Question - is the quotient expected to be 0 or +ve integer ONLY?
if we consider x = 17/2 y= -1/2 or 15/2 and 1/2 still x^2 - y^2 is divisible by 8

If an integer x divided by another number y yields an integer, then the x is said to be divisible by y.

When we talk about the divisibility remainder should be zero.
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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07 Dec 2009, 05:34
Bunuel wrote:
kp1811 wrote:
I may be wrong here but even when we consider x and y as non integers. lets say we consider values for x and y same as assumed in example above x=4.8 and y=3.2. Then x^2 - y^2 will give a value of 12.8 which is divisible by 8 (1.6 times). Question - is the quotient expected to be 0 or +ve integer ONLY?
if we consider x = 17/2 y= -1/2 or 15/2 and 1/2 still x^2 - y^2 is divisible by 8

If an integer x divided by another number y yields an integer, then the x is said to be divisible by y.

When we talk about the divisibility remainder should be zero.

agreed on divisibility rule but we have assumed x and y as decimals to make statement 2 as insufficient.
if a decimal is divided by integer then we cannot get an integer as quotient similarly integer divided by a decimal [incase divisible] will give integer (e.g. 5 divided by 2.5).

Last edited by kp1811 on 07 Dec 2009, 05:44, edited 1 time in total.
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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07 Dec 2009, 05:39
btw study what is the OA?
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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07 Dec 2009, 05:46
kp1811 wrote:
agreed on divisibility rule but we have assumed x and y as decimals to make statement 2 as insufficient.
if a decimal is divided by integer then we cannot get an integer as quotient similarly integer divided by a decimal will give integer (e.g. 5 divided by 2.5).

OK back to our original question: we are asked whether x^2-y^2 is divisible by8. The answer YES will be ONLY if x^2-y^2 is evenly divisible by 8, which means remainder must be 0.

Statement 2 is insufficient as we can get integer value for $$\frac{x^2-y^2}{8}$$, in this case answer is YES OR we can get non-integer value for $$\frac{x^2-y^2}{8}$$, in this case answer is NO. Two different answers, hence insufficient.
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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07 Dec 2009, 19:15
Bunuel wrote:
kp1811 wrote:
agreed on divisibility rule but we have assumed x and y as decimals to make statement 2 as insufficient.
if a decimal is divided by integer then we cannot get an integer as quotient similarly integer divided by a decimal will give integer (e.g. 5 divided by 2.5).

OK back to our original question: we are asked whether x^2-y^2 is divisible by8. The answer YES will be ONLY if x^2-y^2 is evenly divisible by 8, which means remainder must be 0.

Statement 2 is insufficient as we can get integer value for $$\frac{x^2-y^2}{8}$$, in this case answer is YES OR we can get non-integer value for $$\frac{x^2-y^2}{8}$$, in this case answer is NO. Two different answers, hence insufficient.

....rotfl....just deviated enough from the obvious...thanks man...
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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07 Dec 2009, 23:49
Im confused here

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

1. X and Y are even integers.----clearly not sufficient

Take stmt 2
2. X+Y is divisible by 8.

Pull in some sample numbers

If both x and y divisible by 8 then numbers should be 8,16,24,32,40....etc

Take a number "8"
It can be written as

8--4+4,5+3,6+2,7+1
If u take 4+4 (x+y) the answer will be zero...(4^2-4^2) wont be divisible by 8
If u take 7+1 (x+y) the answer will be 6...(7^2-1^2) will be divisible by 8

Not sufficient

Tke both stmt---X and Y should be even and sum shld be divisible by 8
If u take 4+4 (x+y) the answer will be zero...(4^2-4^2) wont be divisible by 8
If u take 6+2 (x+y) the answer will be 4...(4^2-4^2) will be divisible by 8

why cant this be E..PLZ HELP
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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08 Dec 2009, 00:41
raghavs wrote:
Im confused here

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

1. X and Y are even integers.----clearly not sufficient

Take stmt 2
2. X+Y is divisible by 8.

Pull in some sample numbers

If both x and y divisible by 8 then numbers should be 8,16,24,32,40....etc

Take a number "8"
It can be written as

8--4+4,5+3,6+2,7+1
If u take 4+4 (x+y) the answer will be zero...(4^2-4^2) wont be divisible by 8
If u take 7+1 (x+y) the answer will be 6...(7^2-1^2) will be divisible by 8

Not sufficient

Tke both stmt---X and Y should be even and sum shld be divisible by 8
If u take 4+4 (x+y) the answer will be zero...(4^2-4^2) wont be divisible by 8
If u take 6+2 (x+y) the answer will be 4...(4^2-4^2) will be divisible by 8

why cant this be E..PLZ HELP

Two things :

1) 0 is evenly divisible by any number since it leaves a remainder of 0 every time.

2) St. (2) is insufficient because 'x' and 'y' can hold decimal values since it not specified that they are integers. Thus while 'x = 7.75' and 'y = 0.25' satisfies St.(2), $$x^2 - y^2 = (x+y)(x-y) = 8*7.5 = 60$$ is not divisible by 8.

Since St.(2) gives conflicting results, it is insufficient.

Combining St.(1) and St.(2) together, we know that 'x' and 'y' must be even integers whose sum is 8. Since we know that 'x' and 'y' are integers, 'x - y' must also be an integer. Thus, $$x^2 - y^2$$ will always be evenly divisible by 8 since it will always leave a remainder of 0. Hence when taken together, the statements are sufficient.
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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02 Aug 2010, 23:18
Is x^2-y^2 divisible by 8?
1.x and y are even integers.
2.x+y is divisible by 8.
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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03 Aug 2010, 01:18

Statement 1:
(x,y)=(8,0) --> $$x^2-y^2=64$$ divisible by 8
(x,y)=(8,2) --> $$x^2-y^2=60$$ not divisible by 8

Statement 2:
(x,y)=(8,0) --> [ divisible by 8
(x,y)=(7.9,0.1) --> $$x^2-y^2=62.4$$ not divisible by 8

Both statement 1&2:
$$x^2-y^2=(x+y)(x-y)$$
x+y divisible by 8
x-y integer
so $$x^2-y^2$$ divisible by 8
Sufficient
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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03 Aug 2010, 01:57
Isn't there any other way besides plugging? I usually get confused with these kind of qs.. i guess its worthwhile to plug in decimals and check which i didn't do.
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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10 Aug 2016, 07:01
Regards
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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers. [#permalink]

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01 Jan 2017, 13:35
This is an Excellent Question Testing out knowledge on the Integer Properties

Here we need to check whether x^2-y^2 is divisible by 8 or not
Remember x,y can be any numbers on the number line (may/may nor be integers)

Statement 1
x and y are even
using test cases=>
x=2
y=2
x^2-y^2=> 0 which is divisible by 8
Now, let x=4 y=2
x^-y^2=> 6*2=> 12
Hence not divisible by 8
Hence This statement is clearly not sufficient.
Lets look at statement 2 then
Here x+y=> divisible by 8
With our pulses running high we could say that x^-y^2=(x+y)(x-y) => must be divisible by 8
This would have been true if x and y were integers
But they may be or may not be integers.
e.g=>
x=7.2
y=0.8
x+y=8
but x-y=6.4
hence x^2-y^2 wont be an integer
hence this statement is insufficient too.

Combing the two statement we can say that x and y are integers (from statement 1) and thus (x+y)(x-y) must be divisible by 8
Hence C

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Re: Is x^2 - y^2 divisible by 8? 1. x and y are even integers.   [#permalink] 01 Jan 2017, 13:35
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