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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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jlgdr
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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, 17:15
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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
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B
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Bunuel
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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, 17:22
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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 \(2^3*3^2*5*7\).

Answer: B.
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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, 04:43
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, 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
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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, 05:51
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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] New post   04 Jul 2015, 01:04
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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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Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink] New post   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...

Hence answer will be (B)
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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   29 Jul 2023, 06:36
7 x 6 x 5 x 4 x 3 = (2^3)(3^2)(5)(7)

and 240=(2^4)(3)(5)

It can be inferred that the product is not divisible by 240

Hence B
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Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]
29 Jul 2023, 06:36
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