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
_________________
5-star rated online GMAT quant
self study course
See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.