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The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol

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The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink] New post 20 Sep 2013, 16:15
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57% (02:04) correct 42% (01:22) wrong based on 47 sessions
The product 7 x 6 x 5 x 4 x 3 is divisible by all of the following EXCEPT:

(A) 120
(B) 240
(C) 360
(D) 840
(E) 1,260
[Reveal] Spoiler: OA
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Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink] New post 20 Sep 2013, 16:22
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Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink] New post 08 Jan 2014, 03:43
Hi there!

I didn't get the answer right, because neither 120, nor 240 include the factor 7. Why do we just focus on the exponential term of 2 and not on the 7?
Thank you very much for your response.
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Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink] New post 08 Jan 2014, 04:42
jlgdr wrote:
The product 7 x 6 x 5 x 4 x 3 is divisible by all of the following EXCEPT:

(A) 120
(B) 240
(C) 360
(D) 840
(E) 1,260


5*4*3= 120
240=5*4*3*2
hence 7*6*5*4*3 is not divisible by 240
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Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink] New post 08 Jan 2014, 04:51
Expert's post
Emanuel2410 wrote:
Hi there!

I didn't get the answer right, because neither 120, nor 240 include the factor 7. Why do we just focus on the exponential term of 2 and not on the 7?
Thank you very much for your response.


The question is: 3*4*5*6*7=2^3*3^2*5*7 is divisible by all of the following EXCEPT... So, 2^3*3^2*5*7 is dividend and the options are divisors, not vise-versa.

2^3*3^2*5*7 is NOT divisible only by 240=2^4*3*5, because the power of 2 in the divisor is higher than the power of 2 in dividend.

Hope it's clear.
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Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol   [#permalink] 08 Jan 2014, 04:51
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The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol

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