probability that n(n + 1)(n + 2)

[alivapriyada]

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 8?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

[Rich@VeritasPrep]

n(n+1)(n+2) will be divisible by 8 in two situations:

1. n is even. This is because n and n+2 would be consecutive even integers. Even integers always alternate between being multiples of 4 and NOT multiples of 4. So no matter what even integer you chose for n, there will always be a multiple of 4 between n and n+2, thus the entire product will be a multiple of 4*2 = 8.

2. n+1 is a multiple of 8, making the entire product a multiple of 8. This is because if n+1 is even, n and n+2 are both odd. So there's no way to get a factor of 8 from n or n+2. It has to come from n+1, thus n+1 has to be divisible by 8 if the entire product is to be divisble by 8.

For case 1, n could be any even number from 1 to 96 (i.e. 2, 4, 6, 8, ... , 96). There are 48 such numbers.

For case 2, n+1 could be any multiple of 8 from 1 to 96, which means that n could be any number from 1 to 96 that is one less than a multiple of 8 (e.g. 7, 15, 23, 31, ... , 95). There are 12 such numbers.

So there are a total of 48+12 = 60 numbers you're interested in.

Probability: [spoiler]60/96 = 5/8[/spoiler]