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?  Choose the correct answer:
A) 1/4
B) 3/8
C) 1/2
D) 9/16
E) 3/4
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 cannot 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 : 9/16
<result>D</result>
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?