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# M23-26

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Math Expert
Joined: 02 Sep 2009
Posts: 60678

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16 Sep 2014, 01:19
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Difficulty:

35% (medium)

Question Stats:

65% (00:54) correct 35% (01:02) wrong based on 113 sessions

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If $$|x| \lt 4$$, what is the value of $$x$$ ?

(1) $$x$$ is an integer divisible by 3

(2) $$x$$ is an integer divisible by 2

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:19
1
2
Official Solution:

Statement (1) by itself is insufficient. Consider $$x = 3$$ and $$x = -3$$.

Statement (1) by itself is insufficient. Consider $$x = 2$$ and $$x = -2$$.

Statements (1) and (2) combined are sufficient. $$x = 0$$.

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Joined: 05 Dec 2015
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07 Aug 2016, 15:14
how do statements 1 and 2 lead us to zero? How about 6 and -6 as they are both divisible by 2 and 3. so i thought both statements combined are not sufficient.
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08 Aug 2016, 01:05
ehiszele wrote:
how do statements 1 and 2 lead us to zero? How about 6 and -6 as they are both divisible by 2 and 3. so i thought both statements combined are not sufficient.

It's given in the stem that |x| < 4, which means that -4 < x < 4.
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Joined: 07 Feb 2016
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17 Apr 2017, 02:02
NaijaBoy234 wrote:
how do statements 1 and 2 lead us to zero? How about 6 and -6 as they are both divisible by 2 and 3. so i thought both statements combined are not sufficient.

$$\frac{0}{3}=0$$

$$\frac{0}{2}=0$$

and $$0$$ is an integer as well.
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Joined: 17 May 2017
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06 Oct 2017, 12:33
1 is also divisible by 6 isn't it?
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07 Oct 2017, 02:00
haardiksharma wrote:
1 is also divisible by 6 isn't it?

Integer x is divisible by integer y means that x/y = integer, so x when divided by y leaves no remainder. 1/6 is not an integer, 1 divided by 6 gives the remaimder of 1, so NO 1 is NOT divisible by 6.
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Joined: 06 Jan 2018
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03 Feb 2018, 00:39
From (1) we know that x could be -3, 0 or 3 (all numbers between -4 and 4 that are divisible by 3). Not sufficient.

From (2) we know that x could be -2, 0 or 2. Not sufficient.

From (1) and (2) combined, the only number which appears in both sets is 0, so this is sufficient.

0 is divisible by everything. The trap would be forgetting to include 0 in the original lists.
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Joined: 26 Sep 2018
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28 Aug 2019, 13:34
This is a very nicely crafted and GMAT-like math. Kudos to the maker.
Re: M23-26   [#permalink] 28 Aug 2019, 13:34
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# M23-26

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