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 12?
Pick one
A) 1/48 B) 234/8 C) 1/24 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   E O E / 12 = Integer, therefore we have 48 / 96 numbers divisible by 8   O E O / 12= Not Integer   We cannot forget multiples of 12 from 1 to 96   We have 8 numbers that are multiple of 12   Therefore, 48/96 + 8/96 = 23/48   answer : 234/8
<result>B</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 12?