duongthang wrote:if an integer n is to be chosen at random from the intergers 1 to 96, inclusive, what is the probability that n(n+1)(n+2) will be divisible by 8


1/4
3/8
1/2
5/8
3/4

pls, help with this

) Find number that are divisible by 8 btw 1 and 96 inclusive: 8 x 12 =96. So 12 numbers.

Examine n(n+1)(n+2)

Notice if n is divisible by 8, then n+1 and n+2 will not be and when n+1 is divisible by 8, n+2 and n will not be and vice versa. So you have a case of mutually exclusive possibilities.

By the Fundamental Counting Principle in such cases we add the number of outcomes

For N we have 12 from above

For N+1: (N+1)/8=q if N = 8q-1. There are q =12 numbers that satisfy this.
For N+2: (N+2)/8=p if N = 8p-2 There are p=12 numbers that satisfy this also.

Since these are the only possibilities ( see that 8 x8 x 8 not possible) we add:
12 +12+12=36 numbers that are divisible by 8 between 1 and 96 inclusive

Prob 36/96=3/8

Choose B

Tools Needed for quick solution

1) An understanding that 8 x 12 = 96 means there are 12 numbers divisible by 8 between 1 and 96 inclusive

2)An understanding of Fundamental Counting Principle when Mutually Exclusive Outcomes are involved

3) Transforming the concept of divisibility into math to easily get how many of n+1 or n+2 are divisble by 8 within a specified range: (N+1)/8=q if N = 8q-1. This is not too diffrent from 8 x 12, except that we can just easily substitute 1, 2, 3....12 to try to get to 96.

4) And of course plain old probability which is really trivial here.

If these can occur easily to you without putting pen on paper, more power to YOu!!! Mind you these beasts would still appear in another formidable guise.

if n is even, n(n+1)(n+2) is alway diviedec by 8

1. If n = even, then n(n+1)(n+2) is divisible by 8. So there are 48 even numbers from 1 to 96.

2. If n = 8x - 1, where x is some integer 1, 2, 3.., then n(n+1)(n+2) is also divisible by 8.

for eg. n = 7, 15, 23, etc. there are 8 numbers that satisfies this criteria.

So all together there are 48 + 8 different values for n (from 1 to 96 inclusive) which will make n(n+1)(n+2) divisible by 8. So probability of picking one of these numbers is:

(48 + 12) / 96
60/96
5/8

Choose D.

-BM-

n can be 6, 7 or 8 because (6)(7)(8), (7)(8)(9), (8)(9)(10) are all divisible by 8. since every 3 out of 8 numbers are divisible by 8. I think it s 3/8.


96 is exactly divisible by 8 and since the set from 88-96 is divisible 3/8 times as well it doesn't change the probability.

ssmiles08 wrote:n can be 6, 7 or 8 because (6)(7)(8), (7)(8)(9), (8)(9)(10) are all divisible by 8. since every 3 out of 8 numbers are divisible by 8. I think it s 3/8.

Not true. You did not include the following possibilites:

(2)(3)(4)
(4)(5)(6)

both of these are also divisible by 8. So you will have 5 out of any consecutive 8 numbers that will make n(n+1)(n+2) divisible by 8.

-BM-

IMO it's plain that each even number gives us divisibility by 8, so there 1/2 possible outcomes

I would go with 1/2. It is a tricky problem that could easily bog you down with work and would lead you to believe that the answer is 3/8. However we can not forget to look at 2(3)(4) or 4*5*6. The key here is to recognize that if n is an even number it is divisible by 2 and therefore n+2 is divisible by 4 and by multiplying them together you get a multipule of 8.

[2*(2+1)*(2+2)]/2*2*2

whats the answer finally. i got as 1/2. cos see the examples below

2,3,4
4,5,6
6,7,8
8,9,10
10,11,12

so u see tha every even no followed by the next 2 consecutive no is divisible by 8

hence between 1 & 96 there are 48 such nos i.e 96/2

hence prob= 48/96= 1/2

pls through some light :roll:

https://www.beatthegmat.com/ps-probabili ... =divisible

Hope this helps!

thks kanha it helped