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

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Question Stats:

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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! _________________