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n(n+1)(n-1) is divisible by 8 if any or all of its terms are divisible by 8.
Lets assume X=[1,96];
The probability P that X is divisible by 8 is P = 1/8.
X could be equal to n or (n-1) or (n+1), so there are 3 possible favorable outcomes out of 8 possible outcomes.
I'd go with C 1/2
We need the expression n(n+1)(n-1) to contain 2^3 to be divisible by 8
Plug in 1, 2, 3, 4... and you will see that every other 2nd number starting with 1 will have 2^3 in it.
the total cases : 96
for all multiples of 8 we have 3 triplets i.e in total 12 * 3 = 36 triplets
for all odd multiples of 4 we have 2 triplets so 12 *2 = 24
OA is E.
the key here is to determine the favorable outcomes for the task: either 12x3, either 12x2, either 12 events
12 12 12
The total sum of favorable outcomes is 12x3 + 12x2 + 12 = 72
So it is 72/96 = 3/4.
we are choosing a number n from 1 to 96. If we choose n=1, then we have (n-1)(n)(n+1)=0. However, 0/8 is still 0. 0 is divisible by 8. then next odd number is 3 in which case (n-1)(n)(n+1) would be 2*3*4 = 24, which is divisble by 8. Same is the case with 5...so on - totalling to 48 of them that are divisible by 8.
I didn't quite get your reference to 2.
I feel that the question will have NO ambiguity if it reads n(n+1)(n+2)
gayathri, the result set is (1,2,0) meaning that when you take 1, you get 1*(1+1)*(1-1) = 1*2*0 = 0
0 IS a multiple of 8
Hence, take every odd number from the domain, 1 to 96, you will have a number divisible by 8 and there are 48 of those. But as Venksune explained, every even number that is a multiple of 8 will also give a result divisible by 8 and there are 12 of those.