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If x and y are integers and y^2(1+x)=1, what is the value of y?

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If x and y are integers and y^2(1+x)=1, what is the value of y?  [#permalink]

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New post Updated on: 13 Oct 2019, 08:35
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If x and y are integers and \(y^2\)(1+x)=1, what is the value of y?

I) x=0

II) \(y^{999}≠1\)

Source: TargetTestPrep

Originally posted by Xin Cho on 13 Oct 2019, 08:21.
Last edited by Xin Cho on 13 Oct 2019, 08:35, edited 2 times in total.
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Re: If x and y are integers and y^2(1+x)=1, what is the value of y?  [#permalink]

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New post 13 Oct 2019, 08:26
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Re: If x and y are integers and y^2(1+x)=1, what is the value of y?  [#permalink]

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New post 13 Oct 2019, 08:30
If x and y are integers and \(y^2\)(1+x)=1, what is the value of y?

I) x=0
\(y^2\)(1+x)=1......\(y^2*1=1...y^2=1\), so y = 1 or -1
Insuff

II) \(y^{999}=1\)
y=1
Suff

B
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Re: If x and y are integers and y^2(1+x)=1, what is the value of y?  [#permalink]

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New post 13 Oct 2019, 09:53
Xin Cho wrote:
If x and y are integers and \(y^2\)(1+x)=1, what is the value of y?

I) x=0

II) \(y^{999}≠1\)

Source: TargetTestPrep


Asked: If x and y are integers and \(y^2\)(1+x)=1, what is the value of y?

y^2 = 1/(1+x)
For y to be an integer, x=0 & y^2 = 1; y=1 or -1

I) x=0
y=1 or -1
NOT SUFFICIENT

II) \(y^{999}≠1\)
y ≠1
y = -1
SUFFICIENT

IMO B
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Re: If x and y are integers and y^2(1+x)=1, what is the value of y?  [#permalink]

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New post 13 Oct 2019, 12:25
Kinshook wrote:
Xin Cho wrote:
If x and y are integers and \(y^2\)(1+x)=1, what is the value of y?

I) x=0

II) \(y^{999}≠1\)

Source: TargetTestPrep


Asked: If x and y are integers and \(y^2\)(1+x)=1, what is the value of y?

y^2 = 1/(1+x)
For y to be an integer, x=0 & y^2 = 1; y=1 or -1

I) x=0
y=1 or -1
NOT SUFFICIENT

II) \(y^{999}≠1\)
y ≠1
y = -1
SUFFICIENT

IMO B


Hi, I don't understand the reasoning behind the second statement could you please explain why did you consider y= -1 ?
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Re: If x and y are integers and y^2(1+x)=1, what is the value of y?  [#permalink]

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New post 13 Oct 2019, 13:14
Fatineel

From the prompt we know that \(y^2\)(1+x)=1. If x is negative then the whole left side is negative \(y^2\)(1+x), which will not equal to 1.
If x is positive then we get \(y^2\)(positive) which will be inevitably greater than 1. Notice that the prompt states that both x and y are integers.
What does this mean? x has to equal to zero.

If x=0 we can simplify to \(y^2\)(1+0)=1 ==> \(y^2\)=1
Now, y can be either 1 or -1. As \(1^2\)=1 or \((-1)^2\)=1.

Makes sense?
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Re: If x and y are integers and y^2(1+x)=1, what is the value of y?  [#permalink]

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New post 13 Oct 2019, 13:55
From the prompt we know that \(y^2\)(1+x)=1. If x is negative then the whole left side is negative \(y^2\)(1+x), which will not equal to 1.
If x is positive then we get \(y^2\)(positive) which will be inevitably greater than 1. Notice that the prompt states that both x and y are integers.
What does this mean? x has to equal to zero.

If x=0 we can simplify to \(y^2\)(1+0)=1 ==> \(y^2\)=1
Now, y can be either 1 or -1. As \(1^2\)=1 or \((-1)^2\)=1.

Makes sense?[/quote]

Great stuff!
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Re: If x and y are integers and y^2(1+x)=1, what is the value of y?   [#permalink] 13 Oct 2019, 13:55
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