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

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ctrlaltdel
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If x, y, and z are positive integers, where x > y and [#permalink] New post   Updated on: 13 Jun 2013, 05:51
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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
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B

Originally posted by ctrlaltdel on 15 Nov 2009, 22:47.
Last edited by Bunuel on 13 Jun 2013, 05:51, edited 1 time in total.
OA added.
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Bunuel
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Re: Are x and y consecutive perfect squares [#permalink] New post   16 Nov 2009, 08:40
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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.

Answer: B.
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Re: Are x and y consecutive perfect squares [#permalink] New post   16 Nov 2009, 01:27
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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
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Re: Are x and y consecutive perfect squares [#permalink] New post   16 Nov 2009, 01:42
Show SpoilerOA
B
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Re: Are x and y consecutive perfect squares [#permalink] New post   16 Nov 2009, 07:55
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] New post   16 Nov 2009, 10:08
Thanks Bunuel :) ! Great job..+1K
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Re: Are x and y consecutive perfect squares [#permalink] New post   17 Nov 2009, 20:07
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] New post   11 Mar 2014, 18:31
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] New post   30 Apr 2015, 19:18
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
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Re: If x, y, and z are positive integers, where x > y and [#permalink] New post   30 May 2017, 05:05
thanks that was a good one... pretty weak in ds got a get the basics first i guess
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If x, y, and z are positive integers, where x > y and [#permalink] New post   22 May 2020, 21:40
Bunuel wrote:
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.

Answer: B.



So, the question: is \(\sqrt{y}=\sqrt{x}-1\)?
Bunnel can you help me on above line. how do you get that from the provided question
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If x, y, and z are positive integers, where x > y and [#permalink] New post   23 May 2020, 02:56
Expert Reply
chandra73 wrote:
Bunuel wrote:
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.

Answer: B.



So, the question: is \(\sqrt{y}=\sqrt{x}-1\)?
Bunnel can you help me on above line. how do you get that from the provided question


The questions asks: are x and y consecutive perfect squares (given x > y)?

If \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{y}\) and \(\sqrt{x}\) must be consecutive integers.

So, the question: is \(\sqrt{y}=\sqrt{x}-1\)?
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If x, y, and z are positive integers, where x > y and [#permalink]
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