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

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Manager
Joined: 08 Oct 2005
Posts: 95
If an integer n is to be chosen at random from the integers [#permalink]

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05 Aug 2006, 08:06
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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
Senior Manager
Joined: 22 May 2006
Posts: 368
Location: Rancho Palos Verdes

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05 Aug 2006, 09:58
1
KUDOS
ong 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

Total = 96

n(n+1)(n+2) => consecutive integer.. thus mutiples of 6

if n = even.. then n+2 is even.. =>n(n+1)(n+2) = multiples of 8
the number of even integers in given interval = 96/2 = 48
if n+1 = multiples of 8 also satisfy the condition.
the number of multiples of 8 in given interval = 96/8 = 12

Thus, (48+12)/96 = 5/8

D it is.

If u not sure why even consecutive intergers are multiples of 8..
An even integer = 2m
n*(n+2) = 2m(2m+2) = 4m(m+1) : either m or m+1 is even. Thus, always multiples of 8.
_________________

The only thing that matters is what you believe.

Last edited by freetheking on 05 Aug 2006, 10:25, edited 6 times in total.
Manager
Joined: 20 Mar 2006
Posts: 200

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05 Aug 2006, 12:33
freetheking wrote:
ong 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

Total = 96

n(n+1)(n+2) => consecutive integer.. thus mutiples of 6

if n = even.. then n+2 is even.. =>n(n+1)(n+2) = multiples of 8
the number of even integers in given interval = 96/2 = 48
if n+1 = multiples of 8 also satisfy the condition.
the number of multiples of 8 in given interval = 96/8 = 12

Thus, (48+12)/96 = 5/8

D it is.

If u not sure why even consecutive intergers are multiples of 8..
An even integer = 2m
n*(n+2) = 2m(2m+2) = 4m(m+1) : either m or m+1 is even. Thus, always multiples of 8.

Aren't you double counting multiples of 6 that are also multiples of 8 Such as (24,48,96) ?

Heman
Senior Manager
Joined: 22 May 2006
Posts: 368
Location: Rancho Palos Verdes

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05 Aug 2006, 12:36
I didn't count multiples of 6... just counted multiples of 8.

if n = even.. then n+2 is even.. =>n(n+1)(n+2) = multiples of 8
the number of even integers in given interval = 96/2 = 48

if n+1 = multiples of 8 also satisfy the condition.
the number of multiples of 8 in given interval = 96/8 = 12
_________________

The only thing that matters is what you believe.

Manager
Joined: 08 Oct 2005
Posts: 95

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05 Aug 2006, 20:25
freetheking wrote:
ong 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

Total = 96

n(n+1)(n+2) => consecutive integer.. thus mutiples of 6

if n = even.. then n+2 is even.. =>n(n+1)(n+2) = multiples of 8
the number of even integers in given interval = 96/2 = 48
if n+1 = multiples of 8 also satisfy the condition.
the number of multiples of 8 in given interval = 96/8 = 12

Thus, (48+12)/96 = 5/8

D it is.

If u not sure why even consecutive intergers are multiples of 8..
An even integer = 2m
n*(n+2) = 2m(2m+2) = 4m(m+1) : either m or m+1 is even. Thus, always multiples of 8.

OA is D
excellent explanation
Re: integer   [#permalink] 05 Aug 2006, 20:25
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# If an integer n is to be chosen at random from the integers

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