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Is x^2 - y^2 an even integer?

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chetan2u
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Is x^2 - y^2 an even integer? [#permalink] New post   05 Jun 2016, 01:07
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Is \(x^2 - y^2\) an even integer?
(1) x-y is an even integer.
(2) x is an odd integer.


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Re: Is x^2 - y^2 an even integer? [#permalink] New post   05 Jun 2016, 04:51
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fantaisie wrote:
\(x^2\) - \(y^2\) = (x-y)*(x+y)

1) We know that (x - y) is even, but don't know if (x+y) is even or odd
If (x-y) & (x+y) is odd: (x+y)(x-y) = O*O = O
If (x-y) is odd & (x+y) is even: O*E = E
NOT SUFFICIENT


Pl recheck the highlighted portion..
If x-y is already given as EVEN, how can you take that as ODD

It is a bit tricky as Divyadisha has mentioned below....
The following point is very important from point of view of GMAT

Solution -


We are looking at whether x^2-y^2 is EVEN integer..

lets see the statements-

(1) x-y is an even integer.
Most of us will correctly factorize \(x^2-y^2\)as (x-y)*(x+y), but will say further since x-y is an EVEN integer... (x-y)(x+y) should also be even integer...
But NO.... are we told that x and y are integers...

a) \(x = 5, y = 1\)........ YES \(x^2-y^2\)is an even integer.....
b) \(x=\frac{25}{3} , y=\frac{1}{3}\).........\(.x-y =\frac{25}{3}-\frac{1}{3}= \frac{24}{3} = 8............\).But \(x+y = \frac{25}{3}+\frac{1}{3} =\frac{26}{3.}.............\) so\(x^2-y^2 = 8*\frac{26}{3}..........\)NO,\(x^2-y^2\) is not an even integer..

Different answers....
Insuff

(2) x is an odd integer.
Nothing about y..
Insuff...

Combined..
We know x-y is an EVEN integer and x is an integer, so y will also be an integer ..
and \(x^2-y^2\) will be EVEN integer
Suff

C
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Re: Is x^2 - y^2 an even integer? [#permalink] New post   05 Jun 2016, 01:25
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\(x^2\) - \(y^2\) = (x-y)*(x+y)

1) We know that (x - y) is even, but don't know if (x+y) is even or odd
If (x-y) & (x+y) is odd: (x+y)(x-y) = O*O = O
If (x-y) is odd & (x+y) is even: O*E = E
NOT SUFFICIENT

2) x is an odd integer
If x is odd, & y is odd, then:
(x-y) = O - O = E or 0
(x+y) = O + O = E or 0
(x+y)(x-y) = E * E = E

If x is odd, & y is even, then:
(x-y) = O - E = O
(x+y) = O + E = O
(x+y)(x-y) = O * O = O

NOT SUFFICIENT

1 & 2)
We know that (x-y) is even & that x is odd.
Given those circumstances, we can conclude that y is also odd:
(x-y) = O - O = E (In the formulas above we can see, that if y is even, then (x-y) would be odd)

If x is odd and y is odd, then:
(x-y)*(x+y) = E * E = E

Answer: C
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Re: Is x^2 - y^2 an even integer? [#permalink] New post   05 Jun 2016, 08:49
fantaisie wrote:
\(x^2\) - \(y^2\) = (x-y)*(x+y)

1) We know that (x - y) is even, but don't know if (x+y) is even or odd
If (x-y) & (x+y) is odd: (x+y)(x-y) = O*O = O
If (x-y) is odd & (x+y) is even: O*E = E
NOT SUFFICIENT

2) x is an odd integer
If x is odd, & y is odd, then:
(x-y) = O - O = E or 0
(x+y) = O + O = E or 0
(x+y)(x-y) = E * E = E

If x is odd, & y is even, then:
(x-y) = O - E = O
(x+y) = O + E = O
(x+y)(x-y) = O * O = O

NOT SUFFICIENT

1 & 2)
We know that (x-y) is even & that x is odd.
Given those circumstances, we can conclude that y is also odd:
(x-y) = O - O = E (In the formulas above we can see, that if y is even, then (x-y) would be odd)

If x is odd and y is odd, then:
(x-y)*(x+y) = E * E = E

Answer: C


Thanks for pointing out such a careless mistake! My new take on the question:

1) We know that (x - y) is even, but don't know if (x+y) is even or odd
If (x-y) even & (x+y) is odd: (x+y)(x-y) = E*O = E
If (x-y) even & (x+y) is even: E*E = E
SUFFICIENT

2) x is an odd integer

If x is odd, & y is odd, then:
(x-y) = O - O = E or 0
(x+y) = O + O = E or 0
(x+y)(x-y) = E * E = E

If x is odd, & y is even, then:
(x-y) = O - E = O
(x+y) = O + E = O
(x+y)(x-y) = O * O = O
NOT SUFFICIENT

Answer: A
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Re: Is x^2 - y^2 an even integer? [#permalink] New post   05 Jun 2016, 15:14
chetan2u wrote:
Is \(x^2 - y^2\) an even integer?
(1) x-y is an even integer.
(2) x is an odd integer.


Self Made


\(x^2 - y^2\) an even integer?
(x-y) (x+y) is even integer?

If (x-y) or (x+y) or both are odd or any one is '0' then the x^2 - y^2 will be even.

(1) x-y is an even integer.
x-y is even integer. It can be any +ve even integer/ -ve even integer/ 0

(x+y) can be even/odd/ 0

(x-y) (x+y) can either be even or 0 (even)

Sufficient.

(2) x is an odd integer

y can be odd, even or 0 and x^2 - y^2 can be odd or even.

Not sufficient.

A is the answer

Though the question is made by chetan2u, I expect it to have some hidden trick :P
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Re: Is x^2 - y^2 an even integer? [#permalink] New post   07 Jun 2016, 00:43
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Divyadisha wrote:
chetan2u wrote:
Is \(x^2 - y^2\) an even integer?
(1) x-y is an even integer.
(2) x is an odd integer.


Self Made


\(x^2 - y^2\) an even integer?
(x-y) (x+y) is even integer?

If (x-y) or (x+y) or both are odd or any one is '0' then the x^2 - y^2 will be even.

(1) x-y is an even integer.
x-y is even integer. It can be any +ve even integer/ -ve even integer/ 0

(x+y) can be even/odd/ 0

(x-y) (x+y) can either be even or 0 (even)

Sufficient.

(2) x is an odd integer

y can be odd, even or 0 and x^2 - y^2 can be odd or even.

Not sufficient.

A is the answer

Though the question is made by chetan2u, I expect it to have some hidden trick :P


You are correct Divya.... the Q is a bit tricky....
No one has caught the POINT, so I am posting the answer above..
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Re: Is x^2 - y^2 an even integer? [#permalink] New post   07 Jun 2016, 01:04
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chetan2u wrote:
Is \(x^2 - y^2\) an even integer?
(1) x-y is an even integer.
(2) x is an odd integer.


Self Made


Question: is \(x^2 - y^2\) an even integer.

That is: (x-y)*(x+y) an even integer?

Which means that we need to show that either (x-y) is even, or (x+y) is even.

Statement (1)

x-y is an even integer.

Sufficient.

Statement (1)
x is an odd integer.
\(x^2\) will be odd.

Then for \(x^2 - y^2\) to be even.

\(y^2\) will have to odd [ Odd - Odd is even]

which mans that y will have to be odd --> This is not given

Not sufficient.

(A) is the answer.
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Re: Is x^2 - y^2 an even integer? [#permalink] New post   07 Jun 2016, 06:03
adiagr wrote:
chetan2u wrote:
Is \(x^2 - y^2\) an even integer?
(1) x-y is an even integer.
(2) x is an odd integer.


Self Made


Question: is \(x^2 - y^2\) an even integer.

That is: (x-y)*(x+y) an even integer?

Which means that we need to show that either (x-y) is even, or (x+y) is even.

Statement (1)

x-y is an even integer.

Sufficient.

Statement (1)
x is an odd integer.
\(x^2\) will be odd.

Then for \(x^2 - y^2\) to be even.

\(y^2\) will have to odd [ Odd - Odd is even]

which mans that y will have to be odd --> This is not given

Not sufficient.

(A) is the answer.


Hi,

Answer cannot be A.

It is not stated in the question that x and y are integers.

St1: (x - y) is even --> If (x + y) is a fraction there is a possibility that (x - y)(x + y) will be odd

We need St2 to ensure that both x and y are integers.
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Re: Is x^2 - y^2 an even integer? [#permalink] New post   07 Jun 2016, 09:16
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chetan2u wrote:
Divyadisha wrote:
chetan2u wrote:
Is \(x^2 - y^2\) an even integer?
(1) x-y is an even integer.
(2) x is an odd integer.


Self Made


\(x^2 - y^2\) an even integer?
(x-y) (x+y) is even integer?

If (x-y) or (x+y) or both are odd or any one is '0' then the x^2 - y^2 will be even.

(1) x-y is an even integer.
x-y is even integer. It can be any +ve even integer/ -ve even integer/ 0

(x+y) can be even/odd/ 0

(x-y) (x+y) can either be even or 0 (even)

Sufficient.

(2) x is an odd integer

y can be odd, even or 0 and x^2 - y^2 can be odd or even.

Not sufficient.

A is the answer

Though the question is made by chetan2u, I expect it to have some hidden trick :P


You are correct Divya.... the Q is a bit tricky....
No one has caught the POINT, so I am posting the answer above..


The best questions to blow the minds!! Thanks chetan2u

I have to remind myself again and again not to forget fractions, -ve and 0 :beat
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Re: Is x^2 - y^2 an even integer? [#permalink] New post   20 Nov 2016, 08:42
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One of the Best Questions on Evens-odds

Here we are not told that x and y are both integers
We need to to check if x^2-y^2 is even or not

Lets look at statements

Statement 1
x-y=even
Using test cases
x=4
y=2
x-y=2
x+y=6
so x^2-y^2 will be even
let x=4.3
y=0.3
x-y=4
and x+y=4.3
hence x^2-y^2 won't be an integer=> Neither even nor odd
Hence insufficient

Statement 2
x is odd
no clue on y
hence not sufficient
combining the two statements

x+y=even
and x is odd
so y=even-odd => odd
so x and y are both odd
Great
Thus x^2-y^2 will be Odd^2-Odd^2 => Odd-Odd=> Even
Hence the answer to the Question is YES
Hence sufficient

Hence C
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Re: Is x^2 - y^2 an even integer? [#permalink] New post   22 Aug 2019, 09:03
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chetan2u wrote:
Is \(x^2 - y^2\) an even integer?
(1) x-y is an even integer.
(2) x is an odd integer.


Self Made


Asked: Is \(x^2 - y^2\) an even integer?
Is (x+y)(x-y) an even integer?

(1) x-y is an even integer.
Since it is not given that x & y are integers.
NOT SUFFICIENT

(2) x is an odd integer.
No information is provided for y.
NOT SUFFICIENT

Combining (1) & (2)
(1) x-y is an even integer.
(2) x is an odd integer.
If x is odd and (x-y) is even => y is odd
(x-y) is even and (x+y) is even
(x-y)(x+y) is even
SUFFICIENT

IMO C
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Re: Is x^2 - y^2 an even integer? [#permalink] New post   26 Sep 2019, 23:22
Hi Chetnau,

In ur solution for the option 1 you took second example of fraction that cannot be possible where u shown x+y = 26/8 or something as its already mentioned its integer and integer cannot be in fraction hence I guess you need to give better answer for this
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Re: Is x^2 - y^2 an even integer? [#permalink] New post   29 Oct 2019, 06:50
try x= 4.6 y = 0.6 that means x-y is even but x=y is not even integer.
integer*not integer = not integer. A is insufficient
B is insufficient, y not commented upon.

Combining x is odd integer and x-y is even, hence y is an odd integer. Hence, x+y must be odd + odd = odd integer .
and x-y = even integer . Hence x^2-y^2 = even integer * odd integer = even integer = Sufficient.
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Is x^2 - y^2 an even integer? [#permalink] New post   17 Nov 2020, 16:22
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chetan2u wrote:
Is \(x^2 - y^2\) an even integer?
(1) x-y is an even integer.
(2) x is an odd integer.

Chetan's questions


(1) Both x and y can be odd; 5-3=2 (even); then \(x^2 - y^2\) will be even. YES

\(Consdieral factional numbers\)
Both x and y can be even: 3.5-1.5=2; \(x^2 - y^2\) will not be even. No
Insufficient.

(2) No Information about y. Insufficient.

Combining both information:
X is an odd integer from option 2 then Y will be also an odd integer. So, \(x^2 - y^2\) will be even.

Sufficient.
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Is x^2 - y^2 an even integer? [#permalink]
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