pesfunk wrote:

Hello Fluke,

I understand your question. However, I have a small doubt.

In this case when we are counting the even numbers, if they have a lot of multiples of 8 already included.

In that case, aren't we over counting the possibilities ?

Regards - Manish

Not Really!!!

f(n) = n*(n+1)*(n+2)

Now, the two cases I have mentioned above are mutually exclusive.

f(6) = 6*7*8. Divisible by 8 because 6 is even i.e. n is even

f(7) = 7*8*9. Divisible by 8 because 7 is odd and the middle term is a multiple of 8 i.e. (n+1) is multiple of 8.

f(8) = 8*9*10. Divisible by 8 because 8 is even i.e. n=even

f(6) and f(8) will be counted in the even numbers case.

f(7) will be counted in the case where we are finding the (n+1) or the middle number divisibility by 8.

Let's consider a smaller set.

If n is an integer from 1 to

10 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)

n(1) = 1*2*3

n(2) = 2*3*4

n(3) = 3*4*5

n(4) = 4*5*6

n(5) = 5*6*7

n(6) = 6*7*8

n(7) = 7*8*9

n(8) = 8*9*10

n(9) = 9*10*11

n(10) = 10*11*12

Here there are total 10 cases.

n(1) not divisible by 8 because n=1(odd) and the middle number i.e. (n+1)=2 is not divisible by 8.

n(2) is divisible by 8 because n=even=2

n(3) not divisible by 8 because n=3(odd) and the middle number i.e. (n+1)=4 is not divisible by 8.

n(4) is divisible by 8 because n=even=4

n(5) not divisible by 8 because n=5(odd) and the middle number i.e. (n+1)=6 is not divisible by 8.

n(6) is divisible by 8 because n=even=6

n(7) is divisible by 8 because n=odd=7 and (n+1)=8 is divisible by 8

n(8) is divisible by 8 because n=even=8

n(9) not divisible by 8 because n=9(odd) but the middle number i.e. (n+1)=10 is not divisible by 8.

n(10) is divisible by 8 because n=even=10

How many cases are divisible by 8:

n(2) - counted for the n=even case

n(4) - counted for the n=even case

n(6) - counted for the n=even case

n(7) - counted for the n+1 divisible by 8 case

n(8) - counted for the n=even case

n(10) - counted for the n=even case

Count=6 out of total 10 possibilities. Do you see any repetitions here? No, right.

With formula, if we just find the number of even numbers from 1 to 10 and numbers divisible by 8. We should have our result.

Even numbers from 1 to 10 = ((10-2)/2)+1 = 5

Numbers from 1 to 10 that are divisible by 8= (8-8)/8+1=0+1=1

Total=5+1=6