solidcolor wrote:

If an integer n is to be chosen at random from the integer 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

for any number to be divisible by 8 , it has to be divisible by 2 and divisible by 4. (This requirement will not be satisfied only when the number is 4 itself. But 4 is not possible even for the lowest possible value of n 1*2*3.)

so in the series we have to find the probability of the product being divisble by 2 multiplied by product being divisible by 4.

so our probability is P(divisible by 2) * P(divisible by 4)

since the series will have at least one even number we know that it is always divisible by 2 or P(divisible by 2) = 1

for the series to be divisible by 4 we have 2 cases.

case 1) all the sequences which have 2 even numbers. obvious half of them would have 2 even numbers so P(sequence having 2 even numbers) = 1/2

case 2) Of the remaining sequences which have 2 odd numbers the lone even number should be divisble by 4. So out of 48 occurrences we will have 12 which are divisible by 4( every fourth one) so P(lone even number divisible by 4) = 12/48 = 1/4

So P(divisible by 4) = 1/2 + 1/4 = 5/8.

so P(divisible by 8) = 1*(5/8) = 5/8

answer is D