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If a, b, and c are consecutive positive even integers and a

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Senior Manager
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If a, b, and c are consecutive positive even integers and a [#permalink]  06 Feb 2008, 07:40
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If a, b, and c are consecutive positive even integers and a > b > c,
which of the following could be equal to a - b - c ?

A) 6
(B) 2
(C) -1
(D) -3
(E) -4
Director
Joined: 01 Jan 2008
Posts: 629
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Re: PS Question [#permalink]  06 Feb 2008, 07:48
1
KUDOS
lumone wrote:
If a, b, and c are consecutive positive even integers and a > b > c,
which of the following could be equal to a - b - c ?

A) 6
(B) 2
(C) -1
(D) -3
(E) -4

a = 2n (n>=3), b = 2n - 2, c = 2n - 4

a -b -c = 2n - (2n-2) - (2n-4) = 6 - 2n <= 0 since n >=3 , even -> E
CEO
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Re: PS Question [#permalink]  06 Feb 2008, 10:21
maratikus wrote:
lumone wrote:
If a, b, and c are consecutive positive even integers and a > b > c,
which of the following could be equal to a - b - c ?

A) 6
(B) 2
(C) -1
(D) -3
(E) -4

a = 2n (n>=3), b = 2n - 2, c = 2n - 4

a -b -c = 2n - (2n-2) - (2n-4) = 6 - 2n <= 0 since n >=3 , even -> E

i have no idea what you did but i shifted the chain until it reached 10 8 6 and got E
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Re: PS Question [#permalink]  06 Feb 2008, 11:15
maratikus,

a = 2n (n>=3), b = 2n - 2, c = 2n - 4

Why cant it be n>=2
If n =2 we get
a = 4 b= 2 c = 0

Not sure where I am going wrong?

-Jack
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Re: PS Question [#permalink]  06 Feb 2008, 12:00
jackychamp wrote:
maratikus,

a = 2n (n>=3), b = 2n - 2, c = 2n - 4

Why cant it be n>=2
If n =2 we get
a = 4 b= 2 c = 0

Not sure where I am going wrong?

-Jack

a,b,c - consecutive POSITIVE even integers, 0 is not a positive integer
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Re: PS Question [#permalink]  06 Feb 2008, 12:03
bmwhype2 wrote:
i have no idea what you did but i shifted the chain until it reached 10 8 6 and got E

I have no idea why you shifted the chain to 10 8 6 but both of us solved the problem correctly.
Director
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Re: PS Question [#permalink]  06 Feb 2008, 12:22
lumone wrote:
If a, b, and c are consecutive positive even integers and a > b > c,
which of the following could be equal to a - b - c ?

A) 6
(B) 2
(C) -1
(D) -3
(E) -4

a-b = 2

therefore 2-c = x
also note since even-even-even = even eliminate c),d)

if 2-c > 2 , c<= 0

hence x < 0

only E fits this,

ans = e)
Senior Manager
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Re: PS Question [#permalink]  14 Mar 2008, 07:13
maratikus wrote:
bmwhype2 wrote:
i have no idea what you did but i shifted the chain until it reached 10 8 6 and got E

I have no idea why you shifted the chain to 10 8 6 but both of us solved the problem correctly.

And I have no idea what you all did and still can't solve the problem.

Is anyone able to explain with more details?
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Re: PS Question [#permalink]  14 Mar 2008, 08:24
lumone wrote:
maratikus wrote:
bmwhype2 wrote:
i have no idea what you did but i shifted the chain until it reached 10 8 6 and got E

I have no idea why you shifted the chain to 10 8 6 but both of us solved the problem correctly.

And I have no idea what you all did and still can't solve the problem.

Is anyone able to explain with more details?

The long way is plugging in numbers for a > b > c,
Remember they are consecutive positive even integers
CBA could be 2,4,6 or 4, 6, 8 or 6, 8, 10 and so on. Then you plug into a - b - c.
The first two sets of numbers aren't in the answers so this is why I say it's the LONG way.

I would use the formula provided above by maratikus.
It says A (the largest number) is 2N, then B would be 2N-2 (b/c it's consecutive even),
Then that would make the smallest number C, 2N-4.
Then you'd plug in ABC into formula as seen below. Hope this helps.

a = 2n (n>=3), b = 2n - 2, c = 2n - 4

a -b -c = 2n - (2n-2) - (2n-4) = 6 - 2n <= 0 since n >=3 , even -> E

If a, b, and c are consecutive positive even integers and a > b > c,
which of the following could be equal to a - b - c ?

A) 6
(B) 2
(C) -1
(D) -3
(E) -4
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Re: PS Question [#permalink]  14 Mar 2008, 11:59
consectutive even , X , X + 2 , X + 4 where X = Even

now subtract X - (X+2) - (X+4)

doing this leaves us with -X + -2

what even number can we substitute in X to yield an answer choice

E works -4 (plug in 2 for X)
Re: PS Question   [#permalink] 14 Mar 2008, 11:59
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