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

Why do we have to start at 8?
if n=2 then
2(3)(4) is divisible by 8?

I get every even number from 1-96 when n(n+1)(n+2) is divisible by 8.

plus integers like (7,8,9) or (15,16,17)....so 4/8+1/8=5/8
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C 1/2 is my answer.
A no can be divided by eight if we can divide that no by both 2 and 4.
n(n+1)(n+2) , three consecutive nos can always be divided by 2,
and if n is even, we can divide by 4 also
Therefore for all even n we can divide n(n+1)(n+2) by 8
we have 48 even and 96 total nos
so, prob = 48/96= 1/2
whats the OA pls?????

walker can you help me with this approach?

96=2^5 * 3=2^3 * 2^2 *3= so we have a total of 12 factors. to find multiples of 8 we must find all multiples of 2^3, so 6 factors, 1/2. what do I miss?

Total numbers 8*12
There are 12 numbers divisible by 8 -> 3*12 (if 8 is an example - (6,7,8), (7,8,9), (8,9,10))
and 12 numbers divisible by 4 but not divisible by 8 -> 2*12 (if 4 is an example (2,3,4) and (4,5,6))

The answer 5/8 -> D

that's a great job really. quick and easy. kudo for you

OK, here is a long explanation for a problem that took me less than a minute.

First and most important, we know that for n(n+1)(n+2) to be divisible by 8, either n or n+1 or n+2 must be divisible by 8

Second we have to realize that we are dealing with consecutive integers.

We also realize that as we have consecutive integers and multiplication, every even integer n from 2 to 96 satisfy n(n+1)(n+2) to be divisible by 8, allowing us to quickly count all even integers from 2 to 96 to be 47, when we look when n or n+1 or n+2 is divisible by 8.

Realizing the above three in a snap, we simply check when n or n+1 or n+2 is divisible by 8:

1. n is divisible by 8 when n is 8,16,32,36,40,48, 56,64,72,80,88,96, but these even integers are already included in our conclusion that every even integer n from 2 to 96 satisfy the condition that n(n+1)(n+2) to be divisible by 8.

2. n+1 is divisible by 8, when n+1 is 8,16,32,36,40,48, 56,64,72,80,88,96, but as we deal with consecutive integers, this means that in such case, n will be odd integer 7,31,35,39,47,55,63,71,79,87,95, therefore we can pick another 13 integers that make n(n+1)(n+2) divisible by eight. We have to add those 13 odds to our 47 evens.

3. n+2 is divisible by 8, when n+2 is 8,16,32,36,40,48, 56,64,72,80,88,96, but these even integers are already included in our conclusion that every even integer n from 2 to 96 satisfy the condition that n(n+1)(n+2) to be divisible by 8.

we have 47+13 integers satisfying n(n+1)(n+2) is divisible by 8, thus the probability of picking one of these integers is simply P(E) = n of event occuring/total number, that is 60/96, which is equal to 5/8.

The OA is wrong here, it should be D.
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