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Re: If a, b, and c are consecutive positive even integers and a > b > c [#permalink]
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: If a, b, and c are consecutive positive even integers and a > b > c [#permalink]
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: If a, b, and c are consecutive positive even integers and a > b > c [#permalink]
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bmwhype2 wrote:
i have no idea what you did but i shifted the chain until it reached 10 8 6 and got E :-D


I have no idea why you shifted the chain to 10 8 6 but both of us solved the problem correctly. :-D
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Re: If a, b, and c are consecutive positive even integers and a > b > c [#permalink]
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)
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Re: If a, b, and c are consecutive positive even integers and a > b > c [#permalink]
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 :-D


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


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: If a, b, and c are consecutive positive even integers and a > b > c [#permalink]
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 :-D


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


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: If a, b, and c are consecutive positive even integers and a > b > c [#permalink]
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)
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Re: If a, b, and c are consecutive positive even integers and a > b > c [#permalink]
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since a,b,c are consecutive a-b is always 2 (even)
(a-b) - c =<0 since least positive even integer that c can assume is 2.

and since even - even is always even. -4 is only suitable choice
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Re: If a, b, and c are consecutive positive even integers and a > b > c [#permalink]
A simple way of doing this is to just pick three positive consecutive even integers.

1) 6-4-2=0 In the order of a-b-c. Since 0 is not in the options, we go up the order.

2) 8-6-4=-2. Again -2 is not in the option.

3) 10-8-6=-4. This is in the option and we cant go any further up the order as the next solution is -6 which is irrelevant considering the options given.

Hope this helps.
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Re: If a, b, and c are consecutive positive even integers and a > b > c [#permalink]
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


since n >=3 , even -> E
I understand all the substitutions,but how is the answer arrived as E? someone please explain.
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Re: If a, b, and c are consecutive positive even integers and a > b > c [#permalink]
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theperfectgentleman wrote:
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


since n >=3 , even -> E
I understand all the substitutions,but how is the answer arrived as E? someone please explain.


We can represent 3 consecutive integers as 2x - 2, 2x, and 2x + 2 (c, b, and a respectively).

Since the integers are positive then x must be more than 1.

a - b - c = (2x + 2) - 2x - (2x - 2) = 2(2 - x), which will be even non-positive integer. Only E fits.
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Re: If a, b, and c are consecutive positive even integers and a > b > c [#permalink]
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


why can it not be C? consider A=4, B=3 and C=2
you will end up with -1
Can somebody please clarify
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Re: If a, b, and c are consecutive positive even integers and a > b > c [#permalink]
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