GMATGuruNY wrote:

lheiannie07 wrote:If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, what is the value of n?

A. 4
B. 5
C. 6
D. 7
E. 3

We can PLUG IN THE ANSWERS, which represent the value of n.
120 = 2*3*4*5 = 5!.
Since 5! is the GCF of the three given values, (n-1)! must be at least 5!.
Eliminate A, B and E, which will render too small a value for (n-1)!.

D: n=7
Here, (n+2)! = 9!, (n-1)! = 6!, and (n+4)! = 11!, with the result that all three values are divisible by 6!.
Since the GCF must be less than 6!, eliminate D.

The correct answer is C.

[email protected] wrote:Hi lheiannie07,

We're told that the greatest common divisor of (N+2)!, (N-1)!, and (N+4)! is 120. We're asked for the value of N. This question can be approached in a number of different ways. Here's how you can use a bit of logic and Prime Factorization to get the answer.

To start, it's worth noting that a larger factorial will be divisible by ALL of the factors that appear in a smaller factorial. For example: 4! = 24 and 3! = 6. ALL of the factors of 3! (re: 1, 2, 3 and 6) are also factors of 4! (re: 1, 2, 3, 4, 6, 8, 12, and 24). Thus, (N+2)! and (N+4)! are both bigger than (N-1)! so they will both be divisible by the same factors as (N-1)! This is meant to say that to get the answer, we just have to make sure that we make the N as small as possible AND that (N-1)! is divisible by 120.

If we 'prime factor' 120, we get....
120 =
(10)(12) =
(2)(5)(2)(2)(3)

For (N-1)! to be divisible by 120, it has to be divisible by 5... which means that '5' has to be a part of the factorial calculation. The smallest value for N in which that will happen is when N=6:
(6-1)! = 5! = (5)(4)(3)(2)(1)

Final Answer: C

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