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We want to have n(n+1)(n+2) divisible by 8
There are 2 distinct cases:
A: If n is even: In this case, (n+2) is also divisible by 2 and one of n and (n+2) is divisible by 4.
Hence, the product is definitely divisible by 2 x 4 = 8
Think of cases like: 2,3,4 | 4,5,6 | ... | 10,11,12 | 16,17,18 | etc.
Thus, there are 96/2 = 48 possible values of n
B: If n is odd and hence, (n+2) is also odd, but (n+1) is even and a multiple of 8 itself.
Hence, the product is definitely divisible by 2 x 4 = 8
Think of cases like: 7,8,9 | 15,16,17 | ... 95,96,97 | etc.
Thus, there are 96/8 = 12 possible values of (n+1), i.e. 12 values of n
Note: n must be less than or equal to 96, but (n+1) and (n+2) can exceed 96
Total favourable values of n = 48+12 = 60
=> Probability = 60/96 = 5/8
Alternative: Since we are interested in multiples of 8, we look at the first 8 numbers and scale up 12 times (each set of 8 numbers behaves in an identical manner)
Possible vaues of n from 1 to 8 are: n = 2, 4, 6, 7 and 8 i.e. 5 out of 8
=> Probability = 5/8
Answer D
