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Are the positive integers x and y consecutive?

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Are the positive integers x and y consecutive? [#permalink]

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Are the positive integers x and y consecutive?

(1) x^2 - y^2 = 2y + 1
(2) x^2 - xy - x = 0
[Reveal] Spoiler: OA

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Re: Are the positive integers x and y consecutive? [#permalink]

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New post 06 Jan 2013, 23:24
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You can simplify the first statement in following way:
(1) \(x^2-Y^2=2y+1\)
i.e. \(x^2 = Y^2 + 2y+1\)
i.e. \(x^2 = (y+1)^2\)
i.e. \(x=|y+1| = y + 1\)(taking positive root as both x & y are positive integers)
Hence (1) is SUFFICIENT as x & y are consecutive

(2) \(x^2-xy-x=0\)
i.e. \(x(x-y-1)=0\)
i.e. \(x=0\) or \(x=y+1\)
As x is positive integer x<>0, thus \(x=y+1\)
Hence (2) is SUFFICIENT.

Choice (D) is the answer.
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Re: Are the positive integers x and y consecutive? [#permalink]

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New post 06 Jan 2013, 11:00
daviesj wrote:
Are the positive integers x and y consecutive?
(1)\(x^2 - y^2 = 2y + 1\)
(2) \(x^2 - xy - x = 0\)


The question is basically asking whether \(x=y+1\)
Statement 1)
\(x^2 - y^2=2y+1\) can be written as \((x+y)(x-y)=2y+1\).----equation 1
If we put \(x=y+1\), then LHS must be equal to RHS.

Equation 1 can be written, after substituting x=y+1, as \((2y+1)(1)=2y+1\). They are equal. Hence x and y are consecutive.

Statement 2)
\(x(x-y-1)=0\)
The above equation can be equal to 0 only when \(x-y=1\) because x is given to be positive.

Hope I am correct.
+1D
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Re: Are the positive integers x and y consecutive? [#permalink]

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Re: Are the positive integers x and y consecutive? [#permalink]

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Re: Are the positive integers x and y consecutive?   [#permalink] 01 Oct 2014, 00:21
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