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# M24-28

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Math Expert
Joined: 02 Sep 2009
Posts: 49231

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16 Sep 2014, 01:22
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Difficulty:

45% (medium)

Question Stats:

71% (01:05) correct 29% (01:00) wrong based on 48 sessions

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Is $$\frac{xyz}{12}$$ an integer?

(1) $$x$$, $$y$$, $$z$$ are consecutive integers

(2) $$x$$ and $$z$$ are prime numbers

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Joined: 02 Sep 2009
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16 Sep 2014, 01:22
Official Solution:

S1 + S2 is not sufficient. Consider $$x = 3$$, $$y = 4$$, $$z = 5$$ (the answer is YES) and $$x = 5$$, $$y = 6$$, $$z = 7$$ (the answer is NO).

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Joined: 23 Feb 2015
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03 Nov 2015, 07:01
Bunuel wrote:
Official Solution:

S1 + S2 is not sufficient. Consider $$x = 3$$, $$y = 4$$, $$z = 5$$ (the answer is YES) and $$x = 5$$, $$y = 6$$, $$z = 7$$ (the answer is NO).

Hi Bunuel,

345 is not divisible by 12 right?

I think C is OK.
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03 Nov 2015, 09:24
Mickjagger wrote:
Bunuel wrote:
Official Solution:

S1 + S2 is not sufficient. Consider $$x = 3$$, $$y = 4$$, $$z = 5$$ (the answer is YES) and $$x = 5$$, $$y = 6$$, $$z = 7$$ (the answer is NO).

Hi Bunuel,

345 is not divisible by 12 right?

I think C is OK.

xyz means x*y*z.
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Joined: 26 Mar 2012
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Concentration: Marketing
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04 Nov 2015, 04:43
Bunuel

I have a general question that popped up in my mind because of this question.

In this case, is it safe to assume that 'x' will always be the smallest integer when we say that x,y,z are consecutive integers (Statement 1) ?
Or, can we assume 'z' or 'y' the smallest too ?

Thinking in this way for statement 2, can we take, y=1, x=2, and z=3 ? (x and z are prime numbers)
Using statement 2, we get the value of xyz/12 NOT EQUAL to an integer and makes option C as the final answer to the question.

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Puneet.

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04 Nov 2015, 05:20
Bunuel

I have a general question that popped up in my mind because of this question.

In this case, is it safe to assume that 'x' will always be the smallest integer when we say that x,y,z are consecutive integers (Statement 1) ?
Or, can we assume 'z' or 'y' the smallest too ?

Thinking in this way for statement 2, can we take, y=1, x=2, and z=3 ? (x and z are prime numbers)
Using statement 2, we get the value of xyz/12 NOT EQUAL to an integer and makes option C as the final answer to the question.

No, you cannot assume that x<y<z if you are not explicitly told that. But the answer is not C because no matter whether x<y<z or not, you can get a NO as well as an YES answer to the question asked (check examples in the solution).
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04 Nov 2015, 06:15
Ok, Bunuel. Yes, I understand that C cannot be the answer.
Thanks a lot for the prompt reply.
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Puneet.

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25 Aug 2018, 06:44
Hi Bunuel,
can we rewrite the stem as is xyz=12* int=int?
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26 Aug 2018, 04:08
Satya91 wrote:
Hi Bunuel,
can we rewrite the stem as is xyz=12* int=int?

You can re-phrase as follows: is xyz an integer which is a multiple of 12?
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26 Aug 2018, 04:32
Hi bunuel,
Sorry that i wqs not specific in asking my doubt. As 12 is an integer, whats wrong in rephrasing the question as follows: Is xyz=12*int=int i.e is xyz an integer?

By doing so Answer would be A.Could u explain why the question should not be rephrased this way?

Posted from my mobile device
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26 Aug 2018, 04:35
1
Satya91 wrote:
Hi bunuel,
Sorry that i wqs not specific in asking my doubt. As 12 is an integer, whats wrong in rephrasing the question as follows: Is xyz=12*int=int i.e is xyz an integer?

By doing so Answer would be A.Could u explain why the question should not be rephrased this way?

Posted from my mobile device

xyz could be an integer and not a multiple of 12. So, the question should be rephrased as written in my previous post.
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26 Aug 2018, 04:58
Yeah got it. Thank you bunuel.

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M24-28 &nbs [#permalink] 26 Aug 2018, 04:58
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# M24-28

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