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

Question

a b c are integers, is their product abc = 0

1 a^2 = 2a
2 b/c =   - 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 ?

qxjmba · Answer

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.

Bluebird · Answer

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?

bsd_lover · Answer

Hi bluebird, you are right , it makes sense. 

cheers.

vidyasagar · Answer

2) b/c =   - 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.

KillerSquirrel · Answer

b/c = - 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 ? :shock:

Bluebird · Answer

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.

plaguerabbit · Answer

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?

bsd_lover · Answer

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

priyankurml · Answer

b/c =   - 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

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

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