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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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25 Apr 2006, 12:24
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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

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25 Apr 2006, 19:23
1/2?

Probabiltity that n is chosen even.. is 1/2..
as long as n is even, the product is divisible by 8?

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Manager
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25 Apr 2006, 19:45
Can someone shed some light on this one and provide a detailed explanation?

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Director
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25 Apr 2006, 21:10
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

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Director
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25 Apr 2006, 21:18
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?

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Senior Manager
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26 Apr 2006, 10:49
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?

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Intern
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26 Apr 2006, 11:27
5/8

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

Probability = 48+12 / 96

Ans: 5/8

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Senior Manager
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30 Apr 2006, 04:19
OA is 'D', but can anybody explain the fast solution?

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VP
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30 Apr 2006, 06:25
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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30 Apr 2006, 06:25
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