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M06-23

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Math Expert
Joined: 02 Sep 2009
Posts: 50058

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16 Sep 2014, 00:28
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Difficulty:

25% (medium)

Question Stats:

67% (00:49) correct 33% (00:48) wrong based on 67 sessions

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If $$x$$ is a positive integer, what is the least possible value of $$x$$?

(1) $$x^2 - 5x + 5 = 1 - x$$

(2) $$x^2 = 4$$

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Math Expert
Joined: 02 Sep 2009
Posts: 50058

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16 Sep 2014, 00:28
Official Solution:

Statement (1) by itself is sufficient. Solving the equation, we get $$x= 2$$.

Statement (2) is by itself sufficient. Solving the equation, we get $$x = \pm 2$$, but since $$x$$ is a positive integer, $$x = 2$$.

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Joined: 16 Dec 2016
Posts: 8
Location: India
GMAT 1: 720 Q49 V39
GPA: 3

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03 Apr 2017, 09:30
Won't statement 2 be sufficient even if we don't know that x is positive? As x can hold only two values, 2 and -2, it has a minimum already at -2
Re: M06-23 &nbs [#permalink] 03 Apr 2017, 09:30
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M06-23

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