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17 Jul 2017, 04:56
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Difficulty:

55% (hard)

Question Stats:

53% (01:02) correct 47% (01:02) wrong based on 34 sessions

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

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

(2) $$xy = 9$$

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17 Jul 2017, 04:56
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.

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25 Jan 2018, 13:04
Excellent question... I picked E because i got confused.
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03 Oct 2018, 21:59
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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03 Oct 2018, 22:04
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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22 Nov 2018, 21:53
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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23 Nov 2018, 01:23
abhinav2892 wrote:
if X*Y = 9 , why is X= Y ? , Is X = \sqrt{3} and Y=3\sqrt{3} not a right case ?

@bunnell ???

From (1) we got that x = y or x = -y. Your numbers do not satisfy this.
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Re: M32-11 &nbs [#permalink] 23 Nov 2018, 01:23
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