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M32-11

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M32-11  [#permalink]

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New post 17 Jul 2017, 05:56
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

48% (01:07) correct 53% (01:17) wrong based on 40 sessions

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Re M32-11  [#permalink]

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New post 17 Jul 2017, 05:56
Official Solution:


Is \(\frac{x + y}{2}\) an integer?

First of all, notice that we are NOT told that \(x\) and \(y\) are integers.

(1) \(|x| = \sqrt {y^2}\)

\(|x| = |y|\). So, \(x=-y\) or \(x=y\). If \(x=-y\), then \(\frac{x + y}{2}=\frac{-y + y}{2}=0=integer\) but if \(x=y\), then \(\frac{x + y}{2}=\frac{x + x}{2}=x\), which will be an integer if \(x\) is an integer but won't be an integer if \(x\) is not an integer. Not sufficient.

(2) \(xy = 9\). If \(x=y=3\), then \(\frac{x + y}{2}=\frac{3 + 3}{2}=3=integer\) but if \(x=\frac{1}{3}\) and \(y=27\), then \(\frac{x + y}{2} \neq integer\)

(1)+(2) If from (1) \(x=-y\), then from (2) we'll have \(-x^2=9\), which is the same as \(x^2=-9\) but this is not possible (the square of a number cannot be negative). Thus, it must be true that \(x=y\), so \(x^2=9\), which gives \(x=y=3\) or \(x=y=-3\). In any case \(\frac{x + y}{2}=integer\). Sufficient.


Answer: C
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Re: M32-11  [#permalink]

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New post 25 Jan 2018, 14:04
Excellent question... I picked E because i got confused.
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M32-11  [#permalink]

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New post 03 Oct 2018, 22:59
Quote:
|x|=|y||x|=|y| . So, x=−yx=−y or x=yx=y. If x=−yx=−y, then x+y2=−y+y2=0=integer

May i ask, the way i thought was: x= -y , only with condition that both are 0. if we put 1 =(-1), it be wrong. or any other integer. Can you help to find where i am wrong? Thank you!
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Re: M32-11  [#permalink]

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New post 03 Oct 2018, 23:04
Yana2018 wrote:
Quote:
|x|=|y||x|=|y| . So, x=−yx=−y or x=yx=y. If x=−yx=−y, then x+y2=−y+y2=0=integer

May i ask, the way i thought was: x= -y , only with condition that both are 0. if we put 1 =(-1), it be wrong. or any other integer. Can you help to find where i am wrong? Thank you!


|x| = |y| means that the distance from x to 0 is the same as the distance of y from 0. For example, x = 2 and y = 2 or x = -3 and y = 3 or, as you correctly said, x = y = 0.
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Re: M32-11  [#permalink]

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New post 22 Nov 2018, 22:53
if X*Y = 9 , why is X= Y ? , Is X = \sqrt{3} and Y=3\sqrt{3} not a right case ?

@bunnell ???
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New post 23 Nov 2018, 02:23
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Re: M32-11   [#permalink] 23 Nov 2018, 02:23
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