The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 19 Jan 2017, 10:29

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 62

Kudos [?]: 594 [1] , given: 355

The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

### Show Tags

20 Sep 2013, 16:15
1
KUDOS
00:00

Difficulty:

25% (medium)

Question Stats:

71% (02:06) correct 29% (01:22) wrong based on 208 sessions

### HideShow timer Statistics

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: 36566
Followers: 7080

Kudos [?]: 93185 [0], given: 10553

Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

### Show Tags

20 Sep 2013, 16:22
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 $$2^3*3^2*5*7$$.

_________________
Intern
Joined: 11 Dec 2013
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

### Show Tags

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.
Intern
Joined: 17 Oct 2013
Posts: 40
Schools: HEC '14
GMAT Date: 02-04-2014
Followers: 0

Kudos [?]: 13 [0], given: 21

Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

### Show Tags

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
Math Expert
Joined: 02 Sep 2009
Posts: 36566
Followers: 7080

Kudos [?]: 93185 [0], given: 10553

Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

### Show Tags

08 Jan 2014, 04:51
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.
_________________
Director
Joined: 23 Jan 2013
Posts: 579
Schools: Cambridge'16
Followers: 1

Kudos [?]: 42 [0], given: 40

Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

### Show Tags

04 Jul 2015, 00:04
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
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13456
Followers: 575

Kudos [?]: 163 [0], given: 0

Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

### Show Tags

26 Oct 2016, 08:47
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Math Forum Moderator
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 2302
Location: India
GPA: 3.5
Followers: 92

Kudos [?]: 633 [0], given: 317

Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol [#permalink]

### Show Tags

26 Oct 2016, 10:13
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...

_________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )

Re: The product 7 x 6 x 5 x 4 x 3 is divisible by all of the fol   [#permalink] 26 Oct 2016, 10:13
Similar topics Replies Last post
Similar
Topics:
11 If x^2−5x+6=2−|x−1|, what is the product of all possible values of x? 7 29 Apr 2016, 01:34
8 What is the product of all the solutions of x^2+4x+7=|x+2|+3? 8 21 Dec 2014, 20:17
6 If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3) 6 13 Oct 2013, 19:13
23 If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit 17 25 Feb 2013, 13:13
66 What is the product of all the solutions of x^2 - 4x + 6=3 31 13 Feb 2013, 04:22
Display posts from previous: Sort by