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Re: If a, b, and c are consecutive positive even integers and a > b > c [#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 ?

Re: If a, b, and c are consecutive positive even integers and a > b > c [#permalink]
09 Jul 2015, 03:15

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.

gmatclubot

Re: If a, b, and c are consecutive positive even integers and a > b > c
[#permalink]
09 Jul 2015, 03:15

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...