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

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If (n+2)!/n!=156, n=?  [#permalink]

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New post 14 Dec 2016, 01:31
1
1
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A
B
C
D
E

Difficulty:

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Question Stats:

84% (01:35) correct 16% (01:27) wrong based on 115 sessions

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

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

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Re: If (n+2)!/n!=156, n=?  [#permalink]

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New post 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
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Re: If (n+2)!/n!=156, n=?  [#permalink]

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New post 14 Dec 2016, 11:04
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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

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Re: If (n+2)!/n!=156, n=?  [#permalink]

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New post 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.
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Re: If (n+2)!/n!=156, n=?  [#permalink]

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New post 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
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Re: If (n+2)!/n!=156, n=?  [#permalink]

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New post 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.
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Re: If (n+2)!/n!=156, n=?  [#permalink]

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New post 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
11... Answer D

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Re: If (n+2)!/n!=156, n=?  [#permalink]

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New post 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.

Answer: D
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Re: If (n+2)!/n!=156, n=?  [#permalink]

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New post 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.
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Re: If (n+2)!/n!=156, n=?   [#permalink] 16 Jul 2017, 23:18
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