first of all its easy to see that n(n+1)(n+2) can be divided by 8 if n is even
because one of n or n+2 will have to be divisible by 2 and the other by 4

so are we finished? no, but - > A and B are wrong

what happens if n is odd?

if n is odd, then n(n+1)(n+2) will be divisible be 8 if n+1 = 8k where k is from 1..12 (if k=13 then n = 103)

so we have another 12 numbers out of 96 -> 12.5%

so my answer is: 50%+12.5% = 62.5%

C

here's the OA:

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

E is divisible by 8, if n is even.
No of even numbers (between 1 and 96) = 48

E is divisible by 8, also when n = 8k - 1 (k = 1,2,3,.....)
Such numbers total = 12(7,15,....)

Favorable cases = 48+12 = 60.
Total cases = 96
P = 60/96 = 62.5

Method 2:
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 = 62.5

can someone explain this statement in the context of this question
Method 2:
From 1 to 10, there are 5 sets, which are divisible by 8.
(2*3*4) (4*7*6) (6*7*Cool(7*8*9)(8*9*10)

I think i can add another set i.e (10*11*12) which is also divisible by 8 so we have 6 sets between 1 to 10 hence

9*6(between 1 to 90)+3(between 91 & 96)=57 favorable events

so probability= 57/60 but doesn't coincide with any answer can someone suggest what is wrong

I also do not understand this statement

E is divisible by 8, also when n = 8k - 1 (k = 1,2,3,.....)
Such numbers total = 12(7,15,....)

how do we arrive at this magic number 12?? :roll:

ashish1354 wrote:can someone explain this statement in the context of this question
Method 2:
From 1 to 10, there are 5 sets, which are divisible by 8.
(2*3*4) (4*7*6) (6*7*Cool(7*8*9)(8*9*10)

I think i can add another set i.e (10*11*12) which is also divisible by 8 so we have 6 sets between 1 to 10 hence

9*6(between 1 to 90)+3(between 91 & 96)=57 favorable events

so probability= 57/60 but doesn't coincide with any answer can someone suggest what is wrong

I also do not understand this statement

E is divisible by 8, also when n = 8k - 1 (k = 1,2,3,.....)
Such numbers total = 12(7,15,....)

how do we arrive at this magic number 12?? :roll:

Here is the process

n(n+1)(n+2) divisible by 8

We know that when n is even, the product of three consecutive integers will be divisible by 8

[96-2]/2 = 47 + 1 = 48

When n is odd there could only be following cases
7,8,9
15,16,17
23,24,25...so on and so forth

Looking at above cases we just have find the multiples of 8

[96-8]/8 = 11+1 = 12

48+12 = 60

probability = 60/96 * 100 = 62.5%

Hope this helps.

ashish1354 wrote:can someone explain this statement in the context of this question
Method 2:
From 1 to 10, there are 5 sets, which are divisible by 8.
(2*3*4) (4*7*6) (6*7*Cool(7*8*9)(8*9*10)

Small correction:
In Method 2:
from n - 1 to 8 there are the following number divisible by 8
(2*3*4),(4*5*6), (6*7*8), (7*8*9) AND (8*9*10) - five diff nos.

1 to 96 can be divided into 12 sets of 8 (96/8) each of which will repeat similar calculation.
hence total nos divisible by 8 will be 12 * 5 = 60

I took the scenic route, which is like anju´s method 2:

1.2.3 = 6 not divisible by 8
2.3.4 = 24 divisible by 8
3.4.5 = 60 not divisible by 8
4.5.6 = 120 divisible by 8
5.6.7 = 210 not divisible by 8
6.7.8 = 336 divisible by 8 (of course)
7.8.9 = 504 divisible by 8 (of course)
8.9.10 = 720 divisible by 8 (of course)

Therefore, if we assume a repeating sequence up to 96, the answer is then 5/8 or 62.5% of products that are divisible by 8.

For all even values of n, n(n+1)(n+2) is divisible by 8
Eg: 2,3,4
16,17,18
22,23,24
and so on......
these constitute to a total of 96/2=48

and we do have 12 multiple of 8 between 1 and 96(8,16,24,32,.........96)
these come up when n is odd
Eg: 7,8,9
15,16,17
23,24,25............
.........
95,96,97

So a total of 48+12 =60 will be divisible

Now Probability is 60/96=62.5%