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# a b c are integers, is their product abc = 0 1 a^2 = 2a 2

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CEO
Joined: 17 May 2007
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a b c are integers, is their product abc = 0 1 a^2 = 2a 2 [#permalink]

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06 Jun 2007, 16:48
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a b c are integers, is their product abc = 0

1 a^2 = 2a
2 b/c = [(a+b)^2 / (a^2 + 2ab + b^2)] - 1
But I reckon its B

My logic is :
if we look at statement 2
b/c = 0 when the equation is simplified
which means b must be zero therefore abc = 0

However the OA is not B, It is E. Can someone explain why ?

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Intern
Joined: 02 Apr 2007
Posts: 40

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06 Jun 2007, 17:35
doesnt say a,b,c are finite integers. c could be infinity, hence E.

i doubt that this question will appear in gmat, seems advanced.

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Manager
Joined: 18 Apr 2007
Posts: 120

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06 Jun 2007, 17:42
The way I understand it, you cannot conclude that (a+b)^2/(a+b)^2 = 1 because a, b, and c are not defined as positive integers. Therefore, if a = -b, then (a+b)^2 would = 0 and the equation would be undefined.

Does this make sense?

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CEO
Joined: 17 May 2007
Posts: 2947

Kudos [?]: 667 [0], given: 210

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06 Jun 2007, 17:44
Hi bluebird, you are right , it makes sense.

cheers.

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Intern
Joined: 04 Feb 2007
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07 Jun 2007, 12:22
2) b/c = [(a+b)^2 / (a^2 + 2ab + b^2)] - 1----- If this is given to be true then does not this imply that (a^2 + 2ab + b^2) =/= 0 i.e. a is never equal to -b.

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VP
Joined: 08 Jun 2005
Posts: 1144

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07 Jun 2007, 14:38
b/c = [(a+b)^2 / (a^2 + 2ab + b^2)] - 1

lets simplify:

b/c = {(a+b)(a+b)/(a+b)(a+b)} - 1
b/c + 1 = {(a+b)(a+b)/(a+b)(a+b)}
b/c + 1 = 1

one way to solve:

b/c + c/c = 1
(b+c)/c = 1
b+c = c
b = 0

another way to solve:

b/c + b/b = 1
b^2 + bc = bc
b^2 = 0
b = 0

now we know that:

a+b <> 0

has to be ! otherwise the expression is undefined !!

a <> 0

Bluebird's solution assumes an undefined expression (i.e a+b=0). This is not the official GMAT approach ! the official GMAT will not assume undefined solutions. So I second (B).

What is the source of this problem ?

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Manager
Joined: 18 Apr 2007
Posts: 120

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07 Jun 2007, 14:48
I saw this problem on one of the challenges....is this where you saw it bsd_lover?

I admit that I answered B on the challenge when I did it, but got it wrong...official answer on the challenge is E.

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Manager
Joined: 12 Apr 2007
Posts: 166

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07 Jun 2007, 15:05
has anyone heard of the actual gmat throwing these kinds of problems at people?

cus thats a tough one...im thinking that if i got this problem on the gmat and noticed that a could equal b, i might still choose B because i'd be like.."naww they wouldn't do this on the real thing"

then id get it wrong even when i knew the concept they were testing..

eh?

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CEO
Joined: 17 May 2007
Posts: 2947

Kudos [?]: 667 [0], given: 210

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07 Jun 2007, 23:43
Thanks guys. Yes thats exactly where I saw it. I just couldn't agree with the answer at that time. But now its clearer. If I see this kind of problem in real GMAT I will be pretty happy, since it would imply that I am doing pretty well so far

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Director
Joined: 25 Oct 2006
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08 Jun 2007, 15:18
b/c = [(a+b)^2 / (a^2 + 2ab + b^2)] - 1

I can put it in this way,
b/c = (a+b)^2/(a+b)^2 - 1
hence a and b is not defined as +ve or -ve, but still we'll always get square term as +ve, if it's not integer, then also we can make it 0, becoz, 2/3*3/2 is always 1.
Hence b/c = 1-1 = 0, so B=0.
Hence B should be the soluition.

Let me know if I am wrong. Thanks

Kudos [?]: 637 [0], given: 6

08 Jun 2007, 15:18
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