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# If x, y, and z are positive integers, where x > y and

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Manager
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If x, y, and z are positive integers, where x > y and [#permalink]  15 Nov 2009, 21:47
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42% (03:21) correct 58% (02:36) wrong based on 131 sessions
If x, y, and z are positive integers, where x > y and z=x^(1/2), are x and y consecutive perfect squares? (A perfect square is defined as the square of an integer. For example, 36 is a perfect square since it equals 6 squared, while 38 is not a perfect square since it is not equal to the square of any integer.)

(1) x + y = 8z +1
(2) x – y = 2z – 1
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Jun 2013, 04:51, edited 1 time in total.
Senior Manager
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Re: Are x and y consecutive perfect squares [#permalink]  16 Nov 2009, 00:27
1
KUDOS
ctrlaltdel wrote:
If x, y, and z are positive integers, where x > y and z=x^(1/2), are x and y consecutive perfect squares? (A perfect square is defined as the square of an integer. For example, 36 is a perfect square since it equals 6 squared, while 38 is not a perfect square since it is not equal to the square of any integer.)

(1) x + y = 8z +1
(2) x – y = 2z – 1

Happy Solving

B

1. x+ y = 8z+1
the values of x and y satisfying this equation is 25 and 16 which are consecutive perfect squares
the values of x and y satisfying this equation is 64 and 1 which are NOT consecutive perfect squares
Hence insuff

2. x -y = 2z-1
the values x and y satisfying this equation will be all consecutive perfect squares (be it [4,1] [9,4][2500,2401]
non consecutive perfect squares will not satisfy the equation
hence suff
Manager
Joined: 24 Jul 2009
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Kudos [?]: 59 [0], given: 124

Re: Are x and y consecutive perfect squares [#permalink]  16 Nov 2009, 00:42
[Reveal] Spoiler: OA
B
Kudos to kp1811
Director
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Kudos [?]: 343 [0], given: 18

Re: Are x and y consecutive perfect squares [#permalink]  16 Nov 2009, 06:55
What is the source?? This is a heck of a problem. Plugging numbers is painful. Do we have any algebraic approach?
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Re: Are x and y consecutive perfect squares [#permalink]  16 Nov 2009, 07:40
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Economist wrote:
What is the source?? This is a heck of a problem. Plugging numbers is painful. Do we have any algebraic approach?

Here you go.
If $$x$$ and $$y$$ are consecutive perfect squares, then $$\sqrt{y}$$ and $$\sqrt{x}$$ must be consecutive integers.

So, the question: is $$\sqrt{y}=\sqrt{x}-1$$?

Square both sides: $$y=x-2\sqrt{x}+1$$?

(1) $$x+y=8z+1$$
$$x+y=8\sqrt{x}+1$$
$$y=8\sqrt{x}+1-x$$

So basically question transforms to is $$8\sqrt{x}+1-x=x-2\sqrt{x}+1$$?
$$10\sqrt{x}=2x$$?
$$5\sqrt{x}=x$$?
$$\sqrt{x}=5$$?
$$x=25$$?

If x=25 then yes, if not then no. But we don't know the value of x, hence insufficient.

(2) $$x-y=2\sqrt{x}-1$$
$$y=x-2\sqrt{x}+1$$, which is exactly what we are asked in the question. Hence sufficient.

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Re: Are x and y consecutive perfect squares [#permalink]  16 Nov 2009, 09:08
Thanks Bunuel ! Great job..+1K
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Re: Are x and y consecutive perfect squares [#permalink]  17 Nov 2009, 19:07
bunuel - sometimes i wonder how easily you solve these questions - you are awesome
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Re: If x, y, and z are positive integers, where x > y and [#permalink]  26 Sep 2013, 03:00
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If x, y, and z are positive integers, where x > y and [#permalink]  11 Mar 2014, 17:31
IM BACK!

The difference of perfect squares can be expressed as an odd number. Or viceversa, an odd number can be expressed as the difference of two consecutive perfect squares

Namely,

(k+1)^2 - (k)^2 = 2k + 1

Hope this clarifies
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Re: If x, y, and z are positive integers, where x > y and [#permalink]  30 Apr 2015, 18:18
If x, y, and z are positive integers, where x > y and z=x^(1/2), are x and y consecutive perfect squares? (A perfect square is defined as the square of an integer. For example, 36 is a perfect square since it equals 6 squared, while 38 is not a perfect square since it is not equal to the square of any integer.)

(1) x + y = 8z +1
(2) x – y = 2z – 1

Alt way

2. z^2-2z+1 =y
(z-1)^2 = y, z-1 = y^1/2, z=y^1/2+1, since z=X^1/2 , x^1/2 and y^1/2 are consecutive
1. z^2-8z-1 =-y NO solution
Hence B
Re: If x, y, and z are positive integers, where x > y and   [#permalink] 30 Apr 2015, 18:18
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