Given the following factorials (x+3)!, (x-2)!, and (x+4)! what is the value of x if the greatest common divisor is 720?	Pick.
A) 8 B) 9 C) 7 D) 6 E) 10
GCD of (x+3)!, (x-2)!, and (x+4)! is 720. Find x.   Best way to find n is substitute each of the answer and find out the correct one.   I started with A.   Try 8 and find GCD or 11!, 6! and 12!. 6! itself is 720 and 6! which is contained in the other two factorials. This satisfies the condition, hence answer is 7. Also, we can eliminate all other answers as they are less than 7 and n-2 turns out to be less than 5! and hence cannot have a divisor of 120.   8nswer is 8.
<result>A</result>
Given the following factorials (x+3)!, (x-2)!, and (x+4)! what is the value of x if the greatest common divisor is 720?