Set Question : Probality

[simplythebest]

A random number N is chosen from 1 to 96 inclusive, what is the probability that the number N(N+1)(N+2) is divisible by 8?

[jayhawk2001]

A random number N is chosen from 1 to 96 inclusive, what is the probability that the number N(N+1)(N+2) is divisible by 8?

5/8 ?

If n is even, then n*(n+2) will yield a product which is a multiple of 8. So,
1/2 the numbers will satisfy this.

If n is odd, the condition will be satisfied when n+1 is a multiple of 8.
There is 1 such number for every 4 consecutive odd numbers (7,
15, 23 etc.).

So, total = 1/2 + 1/2*1/4 (where 1/2 is probability of odd and
1/4 is probability of n+1 being multiple of 8 in the list of odd integers).

So, 1/2 + 1/8 = 5/8

[800GMAT]

agree with 5/8

all even numbers will satisfy the given condition
i.e. 96/2 = 48
when n is odd, multiples of 8 will satisfy the given condition i.e. 12 multiples

(48+12)/96 = 5/8

[800GMAT]

what is the OA...........

[AMalik]

Hi,

From 1 to 10, there are 5 sets, which are divisible by 8.
(2*3*4) (4*7*6) (6*7*8)(7*8*9)(8*9*10)

So till 96, there will be 12 * 5 such sets = 60 sets

so probability will be 60/96=5/8

[simplythebest]

Perfect...