If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 16? Options
A) 1 / 4	B) 3 / 8	C) 1 / 2	D) 9 / 16	E) 3 / 4

Rationale: "i get 5 / 8 as well 1 to 96 inclusive means we have 48 odd and 48 even integers e o e / 16 = integer , therefore we have 48 / 96 numbers divisible by 8 o e o / 16 = not integer we can not forget multiples of 16 from 1 to 96 we have 6 numbers that are multiple of 16 therefore , 48 / 96 + 6 / 96 = 54 / 96 = 9 / 16 answer : d"
Answer: D