The greatest common divisor of (b-2)!,(b+1)!, and (b+4)! is 5040. what is the value of b? Choices
A) 8
B) 10
C) 12
D) 9
E) 6

Rationale: 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 d .
Answer: D