Problem Solving

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

What's the answer? The other linked post does not give the OA.

I think it's 3/8.

The n(n+1)(n+2) will be divisible by 8 for three consecutive numbers, ie 6,7,8. Between 1 and 96, there are 12 numbers that are a multiple of 8. Given the expression, there will be 12*3 numbers that will be divisible by 8, so we get 36/96=3/8.

Can u pls confirm the answer. I feel the answer should be 1/2

n(n+1)(n+2) will be divisible for any value of n, where n is even.

Since there are 48 even numbers from 1 to 96 and a total of 96 numbers ans is 48/96 = 1/2

Can u pls confirm the answer. I feel the answer should be 1/2

n(n+1)(n+2) will be divisible for any value of n, where n is even.

Since there are 48 even numbers from 1 to 96 and a total of 96 numbers ans is 48/96 = 1/2