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 4?  Options: A) 1/4 B) 3/8 C) 1/2 D) 5/8 E) 3/4
I get 5/8 as well   1 to 96 inclusive means we have 48 odd and 48 even integers   3/4 O 3/4 / 4 = Integer, therefore we have 48 / 96 numbers divisible by 8   O 3/4 O / 4 = Not Integer   We cannot forget multiples of 8 from 1 to 96   We have 24 numbers that are multiple of 4   Therefore, 48/96 + 24/96 = 72/96 = 3/4   answer : 3/4
<result>E</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 4?