integers and probabilities

[demoskopico]

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


Please help

[kevin]

For this product to be a multiple of 8=2^3, n must be either odd or a multiple of 8. Find the probability of each and then simply add these probabilities, as n cannot satisfy both conditions

[demoskopico]

For this product to be a multiple of 8=2^3, n must be either odd or a multiple of 8. Find the probability of each and then simply add these probabilities, as n cannot satisfy both conditions

why do you say that? if n is odd we won't have multiples of 8 always. ie n=1, product=1*2*3. same problem with n=3,5,9....

maybe you would have said even: 2*3*4 is amultiple of 8, thus:

prob even= 1/2
prob n+1 is a multiple of 8=12/96

sum=5/8