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

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Senior Manager
Joined: 08 Jun 2004
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If an integer n is to be chosen at random from the integers [#permalink]  25 Apr 2006, 11:24
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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

Director
Joined: 04 Jan 2006
Posts: 926
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1/2?

Probabiltity that n is chosen even.. is 1/2..
as long as n is even, the product is divisible by 8?
Manager
Joined: 21 Mar 2006
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Kudos [?]: 1 [0], given: 0

Can someone shed some light on this one and provide a detailed explanation?
Director
Joined: 04 Jan 2006
Posts: 926
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Kudos [?]: 16 [0], given: 0

if n is even, n*n+1*n+2 is always going to be divisible by 8
example:
n = 2,n+1=3,n+2 = 4
since we have two consecutive even numbers in this, the product will be always divisible by 8..

try out numbers

4,5,6
6,7,8
8,9,10
14,15,16
18,19,20

2 will always be repeated three times..

so then the probability to get n as even would be :
48/96(half of them are even)
1/2

Does that help? maybe someone else can explain better
Director
Joined: 04 Jan 2006
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ARGHHHHHHHHH while explaining to u i realised i was wrong..

p(n is even) or p(n+1 is div by 8,i.e n is odd , but n = 7,15 etc..)

1/2 + 1/8 = 5/8..

Is that OA?
Senior Manager
Joined: 08 Jun 2004
Posts: 499
Location: Europe
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willget800 wrote:
ARGHHHHHHHHH while explaining to u i realised i was wrong..

p(n is even) or p(n+1 is div by 8,i.e n is odd , but n = 7,15 etc..)

1/2 + 1/8 = 5/8..

Is that OA?

Intern
Joined: 12 Mar 2006
Posts: 26
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5/8

Even: 48 possibilities
Odd: every 4 hits, so 48/4 = 12

Probability = 48+12 / 96

Ans: 5/8
Senior Manager
Joined: 08 Jun 2004
Posts: 499
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Kudos [?]: 20 [0], given: 0

OA is 'D', but can anybody explain the fast solution?
VP
Joined: 21 Sep 2003
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M8 wrote:
OA is 'D', but can anybody explain the fast solution?

Here is a link to the previous discussion:
http://www.gmatclub.com/phpbb/viewtopic.php?p=189499
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