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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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New post 11 Apr 2016, 04:01
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Question Stats:

62% (01:55) correct 38% (02:23) wrong based on 173 sessions

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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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New post 11 Apr 2016, 11:12
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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.

Answer: B
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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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New post 13 Apr 2016, 01:54
1
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
answer: B
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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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New post 08 Aug 2017, 15:32
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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New post 08 Aug 2017, 17:39
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
answer: B
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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New post 15 Oct 2017, 16:09
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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New post 15 Oct 2017, 20:54
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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Re: If x and y are positive integers and (x+y)^2 is an odd integer, is x [#permalink]

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New post 15 Oct 2017, 20:55
Merging posts.
Pl search before posting.
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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, 20:55
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