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# a, b, and c are integers. Is abc = 0?

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a, b, and c are integers. Is abc = 0? [#permalink]

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10 Jan 2008, 00:13
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a, b, and c are integers. Is abc = 0?

(1) a^2 = 2a
(2) (b/c) = [[(a+b)^2] / [a^2 + 2ab + b^2]] - 1; where a does not equal -b and c does not equal 0;
[Reveal] Spoiler: OA
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Re: DS_ M03, question 19 [#permalink]

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10 Jan 2008, 02:18
From S1 we can conclude that a=0 or a=2 so we can not conclude anything
From S2 we can know that b/c=0 so b=0 and we know that a·b·c=0.

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Re: DS_ M03, question 19 [#permalink]

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10 Jan 2008, 06:20
1. tells us that a = 0 or 2. INSUFFICIENT
2. since (a+b)^2 = (a^2+2ab+b^2) the whole fraction = 1. 1-1 = 0 and (b/c) = 0 so b must be 0. as long as one of the 3 integers is 0 abc will = 0 as well

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Re: DS_ M03, question 19 [#permalink]

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10 Jan 2008, 09:58
Thanks. I understood why statement 1 didn't work but couldn't figure out how statement 2 lead to B=0.

Now I get it. [b/c] = 0, where c cannot = 0 (given), so b must =0. I kept thinking [b/c] must =1 and forgot to take into account the last part that says [b/c] actually equals 0.
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Re: DS_ M03, question 19 [#permalink]

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10 Jan 2008, 10:26
misterJJ2u wrote:
a , b , and c are integers. Is abc = 0?

1) a^2 = 2a
2) (b/c) = [[(a+b)^2] / [a^2 + 2ab + b^2]] - 1; where a does not equal -b and c does not equal 0;

1. tells us only that x could be 0 or 2...second degree equation
2. says that b=0, thus abc=0.

OA is B
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Re: DS_ M03, question 19 [#permalink]

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10 Jan 2008, 10:59
misterJJ2u wrote:
a , b , and c are integers. Is abc = 0?

1) a^2 = 2a
2) (b/c) = [[(a+b)^2] / [a^2 + 2ab + b^2]] - 1; where a does not equal -b and c does not equal 0;

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient

Statement 1: does not give any information about a.b.c---> Insufficient
Statement 2:
Can be rewrite as
b/c = [(a+b)^2/(a+b)^2]-1
b/c = 1-1 = 0
which gives product of a,b and c is zero. Hence Sufficient
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Re: a , b , and c are integers. Is abc = 0? 1) a^2 = 2a 2) (b/c) [#permalink]

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05 Jul 2013, 11:58
I saw this question in a gmat club test, and I still have some doubts about the answer.

How can i reduce the expresion [(a+b)^2/(a+b)^2], if we don't know whether (a+b)^2 is 0 or not. ¿??

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Re: a , b , and c are integers. Is abc = 0? 1) a^2 = 2a 2) (b/c) [#permalink]

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05 Jul 2013, 12:16
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jacg20 wrote:
I saw this question in a gmat club test, and I still have some doubts about the answer.

How can i reduce the expresion [(a+b)^2/(a+b)^2], if we don't know whether (a+b)^2 is 0 or not. ¿??

BELOW IS REVISED VERSION OF THIS QUESTION:

Is $$abc = 0$$ ?

In order $$abc = 0$$ to be true at least one of the unknowns must be zero.

(1) $$a^2 = 2a$$ --> $$a^2-2a=0$$ --> $$a(a-2)=0$$ --> $$a=0$$ or $$a=2$$. If $$a=0$$ then the answer is YES but if $$a=2$$ then $$abc$$ may not be equal to zero (for example consider: $$a=2$$, $$b=3$$ and $$c=4$$). Not sufficient.

(2) $$b= \frac{(a*\sqrt{c}+b*\sqrt{c})^2}{a^2+2ab+b^2} - c$$ --> $$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$ --> $$b=c-c$$ --> $$b=0$$. Sufficient.

If a+b were equal 0, then $$\frac{c*(a+b)^2}{(a+b)^2}$$ would be undefined and $$b= \frac{c*(a+b)^2}{(a+b)^2} - c$$ (which is given as a true statement) wouldn't make sense.

Hope it's clear.
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Re: a , b , and c are integers. Is abc = 0? 1) a^2 = 2a 2) (b/c)   [#permalink] 05 Jul 2013, 12:16
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