If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, what is the value of n?
Choose the correct choice from the following:  A) 4  B) 5  C) 6  D) 7  E) 3
Notice that:   1! = 1   2! = 2   3! = 6   4! = 24   5! = 120   So, the smallest factorial must be 5! for the GCD to be 120   The smallest of the three factorials is (n-1)!   So, we need (n-1)! to equal 5!   This means that n = 6   If n = 6, then (n+2)!, (n-1)!, and (n+4)! become 8!, 5! and 10!   Let's CONFIRM this conclusion   8! = (1)(2)(3)(4)(5)(6)(7)(8) = (120)(6)(7)(8)   5! = (1)(2)(3)(4)(5) = 120   10! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10) = (120)(6)(7)(8)(9)(10)   As you can see, 120 is a divisor of all three factorials.   In fact, 120 is the GREATEST COMMON divisor of all three factorials.   Answer: 6
<result>C</result>
If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, what is the value of n?