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# If an integer n is to be chosen at random from the integers

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If an integer n is to be chosen at random from the integers [#permalink]

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11 Nov 2007, 21:09
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Jul 2014, 09:24, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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11 Nov 2007, 21:22
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plaza202 wrote:
If an integer n is to be chosen at randon 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.

I will go with D 5/8

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?

total nos of ways in which we can choose n = 96

n(n + 1)(n + 2) will be divisible by 8?

case 1: n = odd then n+2 =odd & n+1 will be even i.e this needs get divided by 8, hence is a multiple of 8 so we have 8..96 = 12 multiples to fill the n+1 pos hence 12 ways

case 2: n is even then n+2 will be even & the product will be divisible by 24 & thus 8

so nos of values that can be used for n= 2....96 (all even nos) i.e 48 nos

total = 48+12 =60 ways

so reqd P =60/96 =5/8
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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12 Nov 2007, 10:57
The question is asking if you select n what is the probability that the the product of n and the next two consecutive numbers is divisible by 8. For that to happen n would have to be an even number because if n is even then the third number would have to be an even number greater than 2.

ex. 2,3,4

The prime roots of 2 are of course 2 and the prime roots of 4 are 2 and 2. So any 3 consecutive numbers starting with an even number will be divisible by 8(2*2*2).

There are 48 even numbers and 96 possibilites. So 48/96 = 1/2.

Ans C
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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12 Nov 2007, 11:01
gixxer1000 wrote:
The question is asking if you select n what is the probability that the the product of n and the next two consecutive numbers is divisible by 8. For that to happen n would have to be an even number because if n is even then the third number would have to be an even number greater than 2.

ex. 2,3,4

The prime roots of 2 are of course 2 and the prime roots of 4 are 2 and 2. So any 3 consecutive numbers starting with an even number will be divisible by 8(2*2*2).

There are 48 even numbers and 96 possibilites. So 48/96 = 1/2.

Ans C

What abt 7,8,9

starts with an odd integer but is still divisible by 8
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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12 Nov 2007, 13:00
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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26 Aug 2008, 14:27
48 numbers are even for n
12 triplets have multiples of 8 in them --- this implies that those odd numbers that are in those triplets are all game, e.g., 7-8-9 ,15-16-17, 23-24-25 etc. Therefore, 7, 15, 23 would also be counted as possible n. There are 12 such nos. Therefore, the total should be 48 +12 = 60. Hence, prob. = 60/96 = 5/8
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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28 Sep 2009, 05:06
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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.

Soln:

n(n+1)(n+2) will be divisible by 8 when n is a multiple of 2 or when (n+1) is a multiple of 8.

Thus when n is even, this whole expression will be divisible by 8.
from 1 to 96, there are 48 even integers.

Now when (n+1) is multiple by 8, we have 12 such values for (n+1)

probability that n(n+1)(n+2) will be divisible by 8
= (48 + 12)/96
= 60/96
= 5/8

Ans is D
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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02 May 2011, 23:13
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choosing the first 8 numbers as sample period
(1,2,3),(2,3,4),(3,4,5),(4,5,6),(5,6,7),(6,7,8),(7,8,9),(8,9,10) for n= 1 to 8 respectively.

5 of the sets are divisible by 8
hence 5/8.
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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26 Jul 2014, 06:02
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If an integer n is to be chosen at random from the integers [#permalink]

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26 Jul 2014, 09:25
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plaza202 wrote:
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

$$n(n + 1)(n + 2)$$ is divisible by 8 in two cases:

A. $$n=even$$, in this case $$n+2=even$$ too and as $$n$$ and $$n+2$$ are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. $$n+1$$ is itself divisible by 8;

(Notice that these two sets have no overlaps, as when $$n$$ and $$n+2$$ are even then $$n+1$$ is odd and when $$n+1$$ is divisible by 8 (so even) then $$n$$ and $$n+2$$ are odd.)

Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8.

Similar questions to practice:
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OPEN DISCUSSION OF THIS QUESTION IS HERE: if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html
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If an integer n is to be chosen at random from the integers   [#permalink] 26 Jul 2014, 09:25
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