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If x and y are positive integers and (x + y)^2 is an odd integer, is x

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If x and y are positive integers and (x + y)^2 is an odd integer, is x [#permalink]

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11 Apr 2016, 03:01
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If x and y are positive integers and (x + y)^2 is an odd integer, is x an even integer?

(1) x^4 – y^3 is an odd integer
(2) 0 < y^3 – y < 24
[Reveal] Spoiler: OA

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If x and y are positive integers and (x + y)^2 is an odd integer, is x [#permalink]

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11 Apr 2016, 10:12
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(x + y)^2 = odd --> (x + y) = odd
Is x even?

St1: x^4 - y^3 = odd
even - odd = odd
odd - even = odd
Not Sufficient as x can be even or odd.

St2: 0 < (y - 1) * y * (y + 1) < 24 --> Product of 3 consecutive positive integers is between 0 and 24.
Only value y can take is 2.
y = 2 = even
x + y = odd
x = odd - even = odd. --> x is not even
Sufficient.

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Re: If x and y are positive integers and (x + y)^2 is an odd integer, is x [#permalink]

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13 Apr 2016, 00:54
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given :
(x+y)^2 = odd
(x+y)(x+y) = odd, ---> (x+y) must be odd ---> x is even, y is odd
or y is odd, x is even

1) x^4 - y^3 = odd
x.x.x.x - y.y.y = odd , we know that Even - Odd= Odd So x can be EVEN and y ODD or
Odd - Even = Odd x can be ODD and y EVEN
INSUFFICIENT

2) 0< y^3 - y <24
0< y(y^2 - 1) <24
0< y(y+1)(y-1)<24
--> bingo, we have 3 consecutive integers so the product is even.
--> y^3 - y = even -->y is even.

given --> x+y = odd
thus, x is odd
SUFFICIENT
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Re: If x and y are positive integers and (x + y)^2 is an odd integer, is x [#permalink]

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08 Aug 2017, 14:32
1. it can be either odd or even.
2. if y=1, then doesn't satisfy the condition. if y=2, it works. if y=3, then y^3 - y = 24. therefore, y must be 2.
sufficient.

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Re: If x and y are positive integers and (x + y)^2 is an odd integer, is x [#permalink]

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08 Aug 2017, 16:39
Abkhazian wrote:
given :
(x+y)^2 = odd
(x+y)(x+y) = odd, ---> (x+y) must be odd ---> x is even, y is odd
or y is odd, x is even

1) x^4 - y^3 = odd
x.x.x.x - y.y.y = odd , we know that Even - Odd= Odd So x can be EVEN and y ODD or
Odd - Even = Odd x can be ODD and y EVEN
INSUFFICIENT

2) 0< y^3 - y <24
0< y(y^2 - 1) <24
0< y(y+1)(y-1)<24
--> bingo, we have 3 consecutive integers so the product is even.
--> y^3 - y = even -->y is even.

given --> x+y = odd
thus, x is odd
SUFFICIENT
y can be odd as odd-odd is even i think you must solve to get y=2 so even

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If x and y are positive integers and (x+y)^2 is an odd integer, is x [#permalink]

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15 Oct 2017, 15:09
If x and y are positive integers and $$(x+y)^2$$ is an odd integer, is x an even integer?

1) $$x^4 - y^3$$ is an odd integer

2) $$0<y^3-y<24$$

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Re: If x and y are positive integers and (x+y)^2 is an odd integer, is x [#permalink]

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15 Oct 2017, 19:54
khalid228 wrote:
If x and y are positive integers and $$(x+y)^2$$ is an odd integer, is x an even integer?

1) $$x^4 - y^3$$ is an odd integer

2) $$0<y^3-y<24$$

Hi...
If $$(x+y)^2$$ is an odd integer and both x and y are integers, one of them will be odd and other even.

Let's see the statements..
1) $$x^4 - y^3$$ is an odd integer
Still not clear.
It gives the same info as the main statement, that is, one is odd and other is even.
Insufficient

2) $$0<y^3-y<24$$
Now what does $$y^3-y$$ mean..
$$y^3-y=y(y^2-1)=(y-1)*y*(y+1)$$
y is positive integer so least value of equation will be when y=1..
So (y-1)y(y+1)=0*1*2=0.... But it has to be >0... So y is not 1

When y=2, y(y-1)(y+1)=1*2*3=6...... possible
When y=3, #=2*3*4=24....but the # has to be <24... Not possible

So y =2, and x will be ODD
Suff

B
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: If x and y are positive integers and (x+y)^2 is an odd integer, is x [#permalink]

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15 Oct 2017, 19:55
Merging posts.
Pl search before posting.
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: If x and y are positive integers and (x+y)^2 is an odd integer, is x   [#permalink] 15 Oct 2017, 19:55
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