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If z is an integer, is y an integer?

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Joined: 02 Aug 2009
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If z is an integer, is y an integer? [#permalink]

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New post Updated on: 29 Nov 2017, 20:40
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If z is an integer, is y an integer?

(1) \(y^2-z=0\)
(2) \(y\neq{1}\)

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Originally posted by chetan2u on 29 Nov 2017, 19:01.
Last edited by Bunuel on 29 Nov 2017, 20:40, edited 1 time in total.
Renamed the topic, edited the question
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Re: If z is an integer, is y an integer? [#permalink]

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New post 29 Nov 2017, 19:12
chetan2u wrote:
If z is an integer, is y an integer?
(1) \(y^2-z=0\)
(2) \(y\neq{1}\)


(1)
Suppose z = 3;
Then y = ±√3 = Not integer

again, suppose z= 4;
Then y = √4 = ±2 = Integer
Hence, Insuff.

(2) y≠1; Consider y = ±√3 or y = ±2
conditions in (1) still hold true. Insuff.

(1) & (2) together,
conditions in (1) still hold true. Insuff.
Ans: E.
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Re: If z is an integer, is y an integer? [#permalink]

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New post 29 Nov 2017, 19:21
chetan2u wrote:
If z is an integer, is y an integer?
(1) \(y^2-z=0\)
(2) \(y\neq{1}\)


Hi chetan2u,

From Stmnt 1: \(z\)= \(y^2\), y can be \(\sqrt{3}\), \(\sqrt{5}\)
From Stmnt 2: y # 1.. Insufficient

Combined.. Not Sufficient.. So E.
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Re: If z is an integer, is y an integer? [#permalink]

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New post 29 Nov 2017, 22:29
(1) y^2 - z = 0 or y^2 = z
This tells us that z is a non negative integer, but we cannot say anything about y.
Here y will be an integer if z is a perfect square, else not. Insufficient.

(2) y is not equal to 1. This doesnt tell us whether y is an integer or not. Insufficient.

Combining the two statements, y=√z and y is not equal to 1. Still there are multiple possibilities. We can have z=9, then y can take two values: 3 or -3 (integer values). But if z =8, then y = √8 or -√8 (non integer values). So insufficient.

Hence E answer
Re: If z is an integer, is y an integer?   [#permalink] 29 Nov 2017, 22:29
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