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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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If (n+2)!/n!=132, n=?  [#permalink]

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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

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Originally posted by MathRevolution on 25 Aug 2016, 18:29.
Last edited by Bunuel on 10 May 2017, 23:59, edited 1 time in total.
Edited the question and the tags.
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GRE 1: Q169 V154 Re: If (n+2)!/n!=132, n=?  [#permalink]

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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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STONE COLD
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Originally posted by stonecold on 25 Aug 2016, 20:10.
Last edited by stonecold on 25 Aug 2016, 20:14, edited 1 time in total.
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Re: If (n+2)!/n!=132, n=?  [#permalink]

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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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Posts: 4003
Re: If (n+2)!/n!=132, n=?  [#permalink]

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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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Re: If (n+2)!/n!=132, n=?  [#permalink]

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(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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GMAT 1: 760 Q51 V41 Re: If (n+2)!/n!=132, n=?  [#permalink]

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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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If (n+2)!/n!=132, n=?  [#permalink]

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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
10 .. Answer C

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Re: If (n+2)!/n!=132, n=?  [#permalink]

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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=?  [#permalink]

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