Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If an integer n is to be chosen at random from the integers [#permalink]
04 May 2006, 11:19

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

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

Re: PS Probability [#permalink]
06 May 2006, 11:03

Dunno if this is right but here is how I approached this....
There are 12 integers between 1 and 96 that are divisible by 8. So the probability of finding an integer between 96 integers is (12/96) which is 1/8.
Now the product given is a product of three consecutive integers. If we map out the scenarios :
1. OEO : The EVEN integer in this sequence is divisible by 8 (eg.31,32,33), probability of which is 1/8.
2. EOE : This scenario has to be further broken down as:
2a. The first EVEN integer in this sequence ONLY is divisible by 8 (eg. 16,17,18), probability of which is 1/8.
2b. The second EVEN integer in this sequence ONLY is divisble by 8 (eg. 30,31,32), probability of which is 1/8.
2c. NEITHER of the EVEN integers in the sequence is divisible by 8 (eg. 2,3,4) but the product is, probability of which is 1/8.

Therefore the probability of having a random integer picked from a set of {1...96} such that its n(n+1)(n+2) product is divisible by 8 is 1/8+1/8+1/8+1/8 = 4/8 = 1/2.

n(n+1)(n+2) is always divisible by 8 if n = 2,4,6,8,....96 i.e. if n is an even number. Total # of such cases = 48

n(n+1)(n+2) is NOT divisible by 8 if n = 1,3,5,7,.....95.

There are no other cases. So in 48 cases out of 96, n(n+1)(n+2) is divisible by 8. The required probability = 48/96 = 0.5. Ans C.

Not true, If n = 7...then product is 7*8*9 ... it IS divisible by 8

Good catch . I got influenced by one of the explanations above!

Here is a revised explanation:

12 7
Π Π (8n â€“ j)(8n â€“ j+1)(8n-j+2)
n=1 j = 0

For three combinations of n and j the product would not be divisible by 8.
(8n-7)(8n-6)(8n-5), (8n-5)(8n-4)(8n-3), (8n-3)(8n-2)(8n-1). n varies from 1 to 12.

So the probability that it would NOT be divisible by 8 is 3/8.

So the probability that it would be divided by 8 is 5/8. So answer should be D. Has the official answer been correctly listed? For clearly there are more than 48 cases of the product being divisible out the possible 96 cases.
_________________

If u realize that the value will be divisble by 8 only if n is even..

if n is even. that is 96/2 = 48 even in 1-96

Thus, Prob = 48/96 = 1/2

shevy has given a counterexample against 48 cases. Starting with even numbers satisfy the 'divisibility by 8' condition. Since there are more cases such as 7*8*9, we need to count them as well.
_________________

every even number, plus every odd number that is 1 less than a multiple of 8.

took about 6 mins

I think the OA is wrong if it is written C. The correct answer should be 5/8 as given by kook44.
1) The number n(n+1)(n+2) is divisible by all even n. so there are total 48 even numbers between 1 and 96.
2) The expression is again divisible by all n which is 1 less than the multiple of 8 i.e, 7,15,23,31...etc total 12 numbers.
Therefore total numbers satifying the criteria is 60. Hence probability 60/96 = 5/8

question can be simply rephrase as what is prbability n is even and offcourse it is 1/2 because 1/2 time n will be even and 1/2 time it's odd.

Now how can I rephrase question n is even, simple

n(n+1)(n+2) => if we say n is even we can rewrite n=2k

so 2k(2k+1)(2k+2) => 2k(2k+1)2(k+1) => 4k(k+1)(2k+1)

now either k or k+1 is multiple of 2 => n(n+1)(n+2) is multiple of 8

Other simple way is to look at numbers 1,2,3,4,5,6,7,8,

so if you pick 2,3,4 than 2 and 4 will give 8 as multiple if you pick 4,5,6 than 4 and 6 will give 8 as multiple

Even for odd n, n(n+1)(n+2) can be divided by 8. So are we considering that?

I know what you are saying I think krishnosarkar explaination is correct and mine is wrong!

Actually I too got influenced by an earlier message. My original posting in this discussion is also wrong. But yes, 5/8 is the answer!
_________________