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

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If an integer n is to be chosen at randon from the integers [#permalink]  11 Nov 2007, 20:09
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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.

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Re: probability [#permalink]  11 Nov 2007, 20:22
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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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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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: probability [#permalink]  26 Aug 2008, 13:27
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: probability [#permalink]  28 Sep 2009, 04:06
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: probability [#permalink]  02 May 2011, 22:13
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: probability   [#permalink] 02 May 2011, 22:13
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