Sol:

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

The product of the three consecutive number will be divisible by 8 if the first of the 3 consecutive numbers is even.

f(2); the function becomes 2*3*4 will be divisible by 8 because 2 is even.
f(52), the function becomes 52*53*54 will be divisible by 8 because 52 is even
f(61), the function becomes 61*62*63 will NOT be divisible by 8 because 61 is odd

However, there is one more case where the function will be divisible by 8 even when first number of the three consecutive number is odd;
f(7): the function becomes 7*8*9; the middle term 8 is divisible by 8 and thus the entire function becomes divisible by 8.
f(63): the function becomes 63*64*65; the middle term 64 is divisible by 8 and thus the entire function becomes divisible by 8.


Probability = favorable/total

How many functions are there;
f(1) = 1*2*3
f(2) = 2*3*4
f(3) = 3*4*5
f(4) = 4*5*6
.
.
.
f(95) = 95*96*97
f(96) = 96*97*98

Total = 96

To find the favorable cases; we need to find the count of all even numbers from 1 to 96 and the count of all numbers that are divisible by 8.

Count of even numbers = (96-2)/2+1=47+1=48
Total numbers divisible by 8 = (96-8)/8+1 = 12

Probability = \frac{48+12}{96} = \frac{60}{96}=0.625=62.5%

Ans: "C"

Please see this as well:
http://gmatclub.com/forum/tough-tricky-set-of-problms-85211-40.html#p799275
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~fluke

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@fluke

However, there is one more case where the function will be divisible by 8 even when first number of the three consecutive number is odd;

3 * 4 * 5 is not divisible by 8, right, perhaps it's more accurate to say the odd number should be at least 7 ?

Also, just to find the number of integers "(n+1)" divisible by 8 from 1 - 96 in this case :

95 = 7 + (N-1)*8 (N is total such numbers, and it's an AP)

=> (N-1)*8 = 88

=> N = 12
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PROBABILITY OF INTEGER BEING DIVISIBLE BY 8:
If n is an integer from 1 to 96 (inclusive), what is the probability for N being divisible by 8 when N=n*(n+1)*(n+2)

Total number of multiples of 8 between 1 to 96, inclusive = 12
sum of even numbers = [(first even+last even) / 2] - 1 = 48

Probability = 48+12 / 96 = 60 / 96 = 62.5%

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
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~fluke

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Is this something that you derived, or is this a well-known property of 8?

Hi all, thanks so much for your replies, has been very helpful. I have a new problem though...

Question: For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n inclusive. If p is the smallest prime factor of h(100)+1, then p is
a) between 2 and 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 40

Any replies would be greatly appreciated - thanks!

Dear fluke
I did it by simple method as below

first i counted the no which are divisible by 8 through1 - 96 it was 96/8=12
so for each value there will be 3 pairs I.e. 12 x 3 =36
similarly i counted the nos which are divisible by 4 it was 96/4=24 now (as if n = multiple of 4 then n+2 will be even hence multiple of 8)

i added up 36+24=60
and hence the required probability = 60/96=.625

please correct me if i am wrong
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hey fluke
you are right man....Thanks for your help
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WarLocK
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The War is oNNNNNNNNNNNNN for 720+
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