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# If (n+2)!/n!=156, n=?

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
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14 Dec 2016, 01:31
1
1
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Difficulty:

5% (low)

Question Stats:

84% (01:37) correct 16% (01:27) wrong based on 118 sessions

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If (n+2)!/n!=156, n=?

A. 2/131
B. 9
C. 10
D. 11
E. 12

_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Intern Joined: 02 Apr 2016 Posts: 1 Re: If (n+2)!/n!=156, n=? [#permalink] ### Show Tags 14 Dec 2016, 01:51 1 it should be D. (n+2)! = (n+2)*(n+1)*(n)! Therefore (n+2)!/n! = 156 can be written as (n+2)(n+1) = 156-> n = -14,11 From options n=11 is correct Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4834 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: If (n+2)!/n!=156, n=? [#permalink] ### Show Tags 14 Dec 2016, 11:04 1 1 MathRevolution wrote: If (n+2)!/n!=156, n=? A. 2/131 B. 9 C. 10 D. 11 E. 12 $$156$$ = $$2^2$$ x $$3^1$$ x $$13^1$$ Or, $$156$$ = $$12$$ x $$13$$ Thus, $$\frac{(n+2)!}{n!} = \frac{( 11 + 2 )!}{11!}$$ = $$\frac{13!}{11!}$$ = $$13*12$$ Hence, answer will be (D) 11 _________________ 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 ) Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8235 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If (n+2)!/n!=156, n=? [#permalink] ### Show Tags 16 Dec 2016, 00:36 ==> From (n+2)!/n!=156 and (n+2)(n+1)n!/n!=(n+2)(n+1)=156=13*12, you get n+2=13, n=11. Therefore, the answer is D. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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16 Dec 2016, 10:32
MathRevolution wrote:
If (n+2)!/n!=156, n=?

A. 2/131
B. 9
C. 10
D. 11
E. 12

(n+2)(n+1)n!/n! = 156
(n+2)(n+1)=156
solving n=11
Manager
Joined: 05 Dec 2016
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20 Dec 2016, 01:47
Using the factorial main rule (n+2)!/n!=156 ; n!(n+2)(n+1)/n!=156
divide both sides by n! we get (n+2)(n+1) = 156
Then using substitution approach it is clear that Answer D is correct.
Director
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11 May 2017, 02:39
MathRevolution wrote:
If (n+2)!/n!=156, n=?

A. 2/131
B. 9
C. 10
D. 11
E. 12

(n+2)!/n!=156
-> (n+2)(n+1) = 156 = 13x12
-> n+1 = 12
-> n =11
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2809

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16 May 2017, 18:15
MathRevolution wrote:
If (n+2)!/n!=156, n=?

A. 2/131
B. 9
C. 10
D. 11
E. 12

(n + 2)!/n! = (n + 2)(n + 1)n!/n! = (n + 2)(n + 1) = 156

n^2 + 3n + 2 = 156

n^2 + 3n - 154 = 0

(n + 14)(n - 11) = 0

n = -14 or n = 11

We see that only n = 11 is among the answer choices.

Alternate Solution:

(n + 2)!/n! = (n + 2)(n + 1)n!/n! = (n + 2)(n + 1) = 156

We observe that 156 is the product of n + 2 and n + 1, which are two consecutive integers. Scanning the answer choices, we see that all but one are positive integers. Let’s prime factorize 156 to see if it can be expressed as the product of two consecutive integers:

156 = 2^2 x 3 x 13

Since 2^2 x 3 = 12, we observe that 156 = 12 x 13. Thus, n + 1 = 12 and n + 2 = 13. This yields n = 11.

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Manager
Joined: 18 Jun 2017
Posts: 58

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16 Jul 2017, 23:18
Since n! in numerator and denominator would cancel out to leave the product of (n+1).(n+2) which is 156 here.
The trick is to get the factors of 156 to identify n. Option D.
Re: If (n+2)!/n!=156, n=?   [#permalink] 16 Jul 2017, 23:18
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