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M23-26

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M23-26 [#permalink]

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New post 16 Sep 2014, 01:19
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New post 16 Sep 2014, 01:19
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Re: M23-26 [#permalink]

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New post 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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Re: M23-26 [#permalink]

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New post 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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Re: M23-26 [#permalink]

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New post 06 Oct 2017, 12:33
1 is also divisible by 6 isn't it?
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New post 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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Re: M23-26 [#permalink]

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New post 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.
Re: M23-26   [#permalink] 03 Feb 2018, 00:39
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