The greatest common divisor of (b-2)!,(b+1)!, and (b+4)! is 5040. what is the value of b?  Pick:
A) 8	B) 10	C) 12	D) 9	E) 6
GCD of (b-2)!,(b+1)!, and (b+4)! is 5040. Find x.   Best way to find n is substitute each of the answer and find out the correct one.   Only option D gave the correct answer.   Option D: 9 and find GCD or 7!, 10! and 13!. 7! itself is 5040 and 7! which is contained in the other two factorials. This satisfies the condition, hence answer is 9.   Answer is 9.
<result>D</result>
The greatest common divisor of (b-2)!,(b+1)!, and (b+4)! is 5040. what is the value of b?