Bhupisuhag wrote: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?
Tricky Approach:
Let us see for how many values of n from 1 to 8, the product is multiple of 8.
n = 1 --> n(n + 1)(n + 2) = 1*2*3 --> NO
n = 2 --> n(n + 1)(n + 2) = 2*3*4 --> YES
n = 3 --> n(n + 1)(n + 2) = 3*4*5 --> NO
n = 4 --> n(n + 1)(n + 2) = 4*5*6 --> YES
n = 5 --> n(n + 1)(n + 2) = 5*6*7 --> NO
n = 6 --> n(n + 1)(n + 2) = 6*7*8 --> NO
n = 7 --> n(n + 1)(n + 2) = 7*8*9 --> YES
n = 8 --> n(n + 1)(n + 2) = 8*9*10 --> YES
Hence, 5 values of n satisfies the condition.
Now, in next 8 integers also we will have 5 such values as all and so on.
We have 96/8 = 12 such groups of 8 consecutive integers.
And in each group we will have 5 such values of n.
Hence, a total of 12*5 = 60 values for n
Required probability = 60/96 = 5/8
The correct answer is D.
