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If an integer n is to be chosen at random from the integer 1

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If an integer n is to be chosen at random from the integer 1 [#permalink] New post 02 Sep 2007, 00:08
If an integer n is to be chosen at random from the integer 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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 [#permalink] New post 02 Sep 2007, 04:47
I am getting 48/96 i.e. 1/2.
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 [#permalink] New post 02 Sep 2007, 05:32
I guess we have more than 50% of the outcomes. Hence the answer should be E.

1. n can be neither 95 nor 96 as n+1 and N+2 cannot be computed
2. If n is an odd number, then the this is valid only if n+1 is divisible by 8 There are 11 such possibilities.

What is the OA?
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 [#permalink] New post 02 Sep 2007, 06:19
In order for n(n+1)(n+2) to be divisible by 8, n has to be even.

lets pick the smallest even integer, 2. 2(3)(4) = 3x8
lets randomly pick another even integer, 20. 20(21)(22) = 220x21 divisible by 8.

The reason for that is if n is even, then n+2 is also even and we're sure that n(n+2) is divisible by 8.

desired outcomes = even integers from 1 to 96, inclusive = 48
total possible outcomes = 96

probability = 48/96 = 1/2

ANS: C
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 [#permalink] New post 02 Sep 2007, 08:24
1) in order for n(n+1)(n+2) to be divisible by 8, n has to be even.
as it was already discussed by Mishari i will not go into details
So there are 96 numbers and there are 48 even numbers

2) Also (n+1) makes it possible to accept some odd numbers (but not all of them as it is obvious). to be divisible by 8 those numbers should be one less than 8 so they are 8*m-1 = 7,15,23 etc. There are 12 those kind of numbers (last one is 95).

So 48+12 = 60

the probability is 60/96 = 5/8

Answer: D = 5/8
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Re: n(n+1)(n+2) [#permalink] New post 02 Sep 2007, 09:02
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solidcolor wrote:
If an integer n is to be chosen at random from the integer 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


for any number to be divisible by 8 , it has to be divisible by 2 and divisible by 4. (This requirement will not be satisfied only when the number is 4 itself. But 4 is not possible even for the lowest possible value of n 1*2*3.)

so in the series we have to find the probability of the product being divisble by 2 multiplied by product being divisible by 4.

so our probability is P(divisible by 2) * P(divisible by 4)

since the series will have at least one even number we know that it is always divisible by 2 or P(divisible by 2) = 1

for the series to be divisible by 4 we have 2 cases.
case 1) all the sequences which have 2 even numbers. obvious half of them would have 2 even numbers so P(sequence having 2 even numbers) = 1/2

case 2) Of the remaining sequences which have 2 odd numbers the lone even number should be divisble by 4. So out of 48 occurrences we will have 12 which are divisible by 4( every fourth one) so P(lone even number divisible by 4) = 12/48 = 1/4

So P(divisible by 4) = 1/2 + 1/4 = 5/8.

so P(divisible by 8) = 1*(5/8) = 5/8

answer is D
Re: n(n+1)(n+2)   [#permalink] 02 Sep 2007, 09:02
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