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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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Current Student
Joined: 06 Sep 2013
Posts: 1997
Concentration: Finance
The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

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20 Sep 2013, 17:15
1
KUDOS
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Difficulty:

25% (medium)

Question Stats:

69% (02:06) correct 31% (01:23) wrong based on 214 sessions

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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
Math Expert
Joined: 02 Sep 2009
Posts: 39725
Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

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20 Sep 2013, 17:22
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

$$3*4*5*6*7=2^3*3^2*5*7$$.

(B) $$240=2^4*3*5$$ --> 2 has higher power here than in $$2^3*3^2*5*7$$.

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Intern
Joined: 11 Dec 2013
Posts: 1
Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

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08 Jan 2014, 04:43
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.
Intern
Joined: 17 Oct 2013
Posts: 41
Schools: HEC Dec"18
GMAT Date: 02-04-2014
Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

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08 Jan 2014, 05:42
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
Math Expert
Joined: 02 Sep 2009
Posts: 39725
Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

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08 Jan 2014, 05:51
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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Posts: 586
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Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

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04 Jul 2015, 01:04
Eliminate 0 in all options and 5*2 in multiple to get

7*3*2*2*3

12=3*2*2, Yes
24=2*2*2*3, No
36=2*3*3*2, Yes
84=2*7*3*2, Yes
126=2*7*3*3, Yes

B
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Joined: 09 Sep 2013
Posts: 16014
Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

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26 Oct 2016, 09:47
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Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

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26 Oct 2016, 11:13
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

7 x 6 x 5 x 4 x 3 = 2^3 x 3^2 x 5 x 7

Now, check the options -

(A) 120 = 2^3 x 3^1 x 5
(B) 240 = 2^4 x 3^1 x 5^1
(C) 360 = 2^3 x 3^2 x 5^1
(D) 840 = 2^3 x 3^1 x 5^1 x 7^1
(E) 1,260 = 2^2 x 3^2 x 5^1 x 7^1

Thus all except option (B) can completely divide the number...

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Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol   [#permalink] 26 Oct 2016, 11:13
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