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Re: Is a*b*c divisible by 32? [#permalink]
25 May 2007, 05:10

Juaz wrote:

Is a*b*c is divisible by 32?

1. a,b and c are consecutive even integers.

2. a*c < 0

*Question edited.

OA is correct. Its C.

Zero divided by a negative or positive number is either zero or is expressed as a fraction with zero as numerator and the finite quantity as denominator.

Re: Is a*b*c divisible by 32? [#permalink]
25 May 2007, 15:02

Statement I: insufficient but helpful

Statement two says that a*c <0 so one of the numbers must be negative, all could be negative as well:

-8 * -6 * -4 = -192 which is divisible by 32 as per GMAT definition y=xq+r when q and r are unique integers and 0<=r<x ( -192=32*-6 + 0)

-6*-4*-2 =-48 which is not divisible by 32.

insufficient.

Taken together we have that they are consecutive, and that one is negative and the other is not. Because of the restriction of being consecutive and even we have only one possible set of numbers: -2,0,2.

abc =0. 0 is divisable by any integer except zero. so the answer was C.

tricky problem and I would have no doubt gotten this wrong on the test if i didn't have 10 minutes to sit and think about it

Re: Is a*b*c divisible by 32? [#permalink]
21 Oct 2014, 07:14

Juaz wrote:

Is a*b*c is divisible by 32?

1. a,b and c are consecutive even integers.

2. a*c < 0

*Question edited.

What on earth question ?

Statements (1) and (2) combined are sufficient. From S1 + S2 it follows that either a or c is negative. As a , b, and C are consecutive even integers, one of these three numbers must be 0. Thus, abc=0 which is divisible by 32

and what on earth explanation ! Mind blown!!!

gmatclubot

Re: Is a*b*c divisible by 32?
[#permalink]
21 Oct 2014, 07:14

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