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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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Question Stats:

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.)

Re: Are x and y consecutive perfect squares [#permalink]
16 Nov 2009, 00:27

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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

Re: If x, y, and z are positive integers, where x > y and [#permalink]
26 Sep 2013, 03:00

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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

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

gmatclubot

Re: If x, y, and z are positive integers, where x > y and
[#permalink]
30 Apr 2015, 18:18

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