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Re: If a, b, and c are all integers divisible by 3 and a > b > c > 0, then [#permalink]

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21 Mar 2017, 16:02

1

This post received KUDOS

I'm having difficulty understanding why answer is E and not D.

The question does not specify that negative numbers can't be used, in which case if we use a=6 b=3 and C=-3 , statement II would be insufficient since 0 is not divisible by 3. Can someone clarify?

Re: If a, b, and c are all integers divisible by 3 and a > b > c > 0, then [#permalink]

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25 Mar 2017, 09:42

mogarza08 wrote:

I'm having difficulty understanding why answer is E and not D.

The question does not specify that negative numbers can't be used, in which case if we use a=6 b=3 and C=-3 , statement II would be insufficient since 0 is not divisible by 3. Can someone clarify?

Thanks in advance.

As per the question, a>b>C>0 so c=-3 is not greater than 0. Also, 0 is divisible by every integer except 0 itself.

Re: If a, b, and c are all integers divisible by 3 and a > b > c > 0, then [#permalink]

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25 Apr 2017, 15:36

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This post received KUDOS

mogarza08 wrote:

If a, b, and c are all integers divisible by 3 and a > b > c > 0, then which of the following must be divisible by 3?

I. a+b+c

II. a-b+c

III. abc/9

A) I only

B) III only

C) I and II

D) I and III

E) I, II, and III

I. a+b+c Addition and subtraction between multiples of a number result in a multiple of that number. Therefore, 3 multiples of 3 added together give another multiple of 3.

II. a-b+c Addition and subtraction between multiples of a number result in a multiple of that number. Note that the resulting multiple must be +ve, since a>b. However, even if the resulting multiple was -ve, it would still be divisible by 3. E.g. a=9, b=18, c=3, a-b+c=-6. -6/2=-3.

III. abc/9 Here, we're multiplying at least 3 powers of 3 and dividing by 2 powers of 3. This leaves at least one power of 3 and therefore the result will be divisible by 3.

Agree, give kudos! Have an improvement, please comment.