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?
please explain, thank you.
For n(n+1)(n+2) to be divisible by 8, either n has to be even ( because if, n and n+2 are even, then n is divisible by 8) or n+1 as a whole should be divisible by 8
There are 96 values for n. The possibility of n to be even is (96/2) = 48 and the possibility of n to be divisible by 8 is (96/8)= 12
Therefore the total prob = (48+12)/96 = 60/96 = 5/8
The OA is D
Kudos if I helped