If k is a positive integer, which of the following must be divisible by 22?  Choose the correct choice:	A) (k – 4)(k)(k + 3)(k + 7)	B) (k – 4)(k – 2)(k + 3)(k + 5)	C) (k – 2)(k + 3)(k + 5)(k + 6)	D) (k + 1)(k + 3)(k + 5)(k + 7)	E) (k – 3)(k + 1)(k + 4)(k + 6)
22=11*2.   Note that the product of two consecutive even integers is always divisible by 8 (since one of them is divisible by 4 and another by 2). Only option B offers two consecutive even numbers for any integer value of k: k-4 and k-2, if k=even or k+3 and k+5 if k=odd.   Also from the following 3 consecutive integers:(k-4), (k-3),(k-2)one must be divisible by 3, if it's not k-4 or k-2 then it must be k-3 (if it's k-4 or k-2 option B is divisible by 3 right away). But if it's k-3 then (k-3)+6=k+3must also be divisible by 3.   So, option B: (k – 4)(k – 2)(k + 3)(k + 5) is divisible by 8 and 3 in any case.   Answer: (k + 1)(k + 3)(k + 5)(k + 7).
<result>D</result>
If k is a positive integer, which of the following must be divisible by 22?