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 6?
Select: A) 2/3 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   E O E / 6 = Integer, therefore we have 48 / 96 numbers divisible by 6   O E O / 6 = Not Integer   We cannot forget multiples of 6 from 1 to 96   We have 16 numbers that are multiple of 8   Therefore, 48/96 + 16/96 = 64/96 = 2/3   answer : 2/3
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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 6?