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

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

by Max@Math Revolution » Thu Aug 25, 2016 5:29 pm
If (n+2)!/n!=132, n=?
A. 2/131
B. 9
C. 10
D. 11
E. 12

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by Brent@GMATPrepNow » Fri Aug 26, 2016 4:48 am
Max@Math Revolution wrote:If (n+2)!/n! = 132, n=?
A. 2/131
B. 9
C. 10
D. 11
E. 12
(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: C

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: C

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Brent
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by Max@Math Revolution » Tue Aug 30, 2016 7:03 pm
(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.