* divisible by 8

[sssddq]

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?

a. 1/4
b. 3/8
c. 1/2
d. 5/8
e. 3/4

[michael_shaunn]

I believe that the answer is 3/8.

[Guest]

Please explain. However, the official answer is 5/8

[michael_shaunn]

Actually I found the probabaility that n(n+1)(n+2) will not be divisible by 8 which gave me the answer as 3/8.
I am in the bad habit of day dreaming.
Sorry about that.
Here's my approach.
The number n is either even or odd.
If n is even then n+2 is also even.Now plz observe urself that if n is a multiple of 2 then (n+2) will be a multiple of 4 or if n is a multiple of 4 then (n+2) will be a multiple of 2.
Check it urself.Simple observation.Write down few numbers and you will be able to observe that.
This means that for n is even...the given expression will always be divisible by 8 which gives a total of 96/2=48 even numbers.

Lets come to 'n' being an odd number.
If n is odd then (n+2) is also odd.
The only chance of n(n+1)(n+2) being divisible by 8 is that (n+1) should be divisible by 8.Total number of numbers from 1 to 96 which are multiples of 8 are 96/8=12 and corresponding to these 12 numbers we get 12 odd values of n (subtract 1 from each multiple of 8..as multiple of 8 is n+1....though you are not required to do that )
Hence total values of 'n' are 48+12=60.

The probability is thus 60/96=5/8.

thanks!!

[Guest]

awesome explanation...:).

thanks!!

[JonathanSchneider]

Can you please post the source of this question? Otherwise we will have to delete it.