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M32-11

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M32-11  [#permalink]

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New post 17 Jul 2017, 05:56
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

62% (01:05) correct 38% (01:03) wrong based on 21 sessions

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Posts: 47880
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.
Re: M32-11 &nbs [#permalink] 25 Jan 2018, 14:04
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