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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # If (n+2)!/n!=132, n=?  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6497 GMAT 1: 760 Q51 V42 GPA: 3.82 If (n+2)!/n!=132, n=? [#permalink] ### Show Tags Updated on: 10 May 2017, 22:59 1 1 00:00 Difficulty: 15% (low) Question Stats: 78% (01:14) correct 22% (01:34) wrong based on 205 sessions ### HideShow timer Statistics If $$\frac{(n+2)!}{n!}= 132$$, n=? A. 2/131 B. 9 C. 10 D. 11 E. 12 *An answer will be posted in 2 days _________________ 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$99 for 3 month Online Course"
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Originally posted by MathRevolution on 25 Aug 2016, 17:29.
Last edited by Bunuel on 10 May 2017, 22:59, edited 1 time in total.
Edited the question and the tags.
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Updated on: 25 Aug 2016, 19:14
1
Hi MathRevolution
Firstly this Ain't a probability Question
Kindly change that
Here is my approach => (n+2)(n+1)*n!/n!=132=> (n+2)*(n+1)=132
Checking the values => n must be 10 => 12 *11=132

SMASH THAT C

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Originally posted by stonecold on 25 Aug 2016, 19:10.
Last edited by stonecold on 25 Aug 2016, 19:14, edited 1 time in total.
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25 Aug 2016, 19:11
1
1
MathRevolution wrote:
If (n+2)!/n!=132, n=?
A. 2/131
B. 9
C. 10
D. 11
E. 12

*An answer will be posted in 2 days

$$\frac{(n+2)!}{n!}$$ can be written as $$\frac{(n+2)(n+1)n!}{n!}$$

This gives $$(n+2)(n+1) = 132 = 11*12$$
Out of the choices given C) 10, would be the right value to solve.

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26 Aug 2016, 04:51
1
Top Contributor
1
MathRevolution wrote:
If (n+2)!/n!=132, n=?
A. 2/131
B. 9
C. 10
D. 11
E. 12

*An answer will be posted in 2 days

(n+2)!/n! = 132
Rewrite as: [(n+2)(n+1)(n)(n-1)(n-2)....(3)(2)(1)]/[(n)(n-1)(n-2)....(3)(2)(1)] = 132
Cancel out terms: (n+2)(n+1) = 132
From here, we might just TEST the answer choices.
Since (12)(11) = 132, we can see that n = 10

Alternatively, we can take (n+2)(n+1) = 132 and expand it to get: n² + 3n + 2 = 132
Subtract 132 from both sides to get: n² + 3n - 130 = 0
Factor to get: (n + 13)(n - 10) = 0
So, n = -13 or n = 10
Since n cannot be negative in a factorial, n must equal 10

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30 Aug 2016, 19:03
(n+2)!/n!=(n+2)(n+1)n!/n!=(n+2)(n+1)=132, n^2+3n-130=0, (n+12)(n-10)=0. Hence, the correct answer is C.
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11 May 2017, 01:06

Once you get to (n+2)*(n+1)=132 (or further n² + 3n + 2 = 132)

It's important to develop a "feel" for such equations. I'd just immediately look at the biggest term n² , we need an n that's at least close to 132 when squared, so 10^2 = 100 seems like a good starting point. If you just mentally plug that into either of those equations, you would realized you've hit jackpot. So definitely no factorization of quadratic or going through all the options required.
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11 May 2017, 01:35
MathRevolution wrote:
If $$\frac{(n+2)!}{n!}= 132$$, n=?

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

*An answer will be posted in 2 days

$$\frac{(n+2)!}{n!}= 132$$
-> (n+2)(n+1) = 132 = 12 x 11
-> n+1 = 11
-> n = 10

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17 Jul 2018, 15:42
MathRevolution wrote:
If $$\frac{(n+2)!}{n!}= 132$$, n=?

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

(n+2)!/n! = 132

(n+2)(n+1)n!/n! = 132

(n+2)(n+1) = 132

n^2 + 3n + 2 = 132

n^2 + 3n - 130 = 0

(n - 10)(n + 13) = 0

n = 10 or n = -13

Since the factorial of a negative number is undefined, n can’t be -13, so n = 10.

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Re: If (n+2)!/n!=132, n=? &nbs [#permalink] 17 Jul 2018, 15:42
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