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If (n+2)!/n!=132, n=?

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

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New post Updated on: 25 Aug 2016, 20:14
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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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New post 25 Aug 2016, 20:11
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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.

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

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New post 26 Aug 2016, 05:51
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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
Answer:

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
Answer:

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

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New post 30 Aug 2016, 20:03
(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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Re: If (n+2)!/n!=132, n=?  [#permalink]

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New post 11 May 2017, 02:06
Already answered above, but I'd like to add :

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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New post 11 May 2017, 02:35
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
10 .. Answer C

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

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New post 17 Jul 2018, 16:42
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.

Answer: C
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