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Is the integer m an even number? (1) |m| = -m (2) (m)(m - 1)(m + 2) = 0

I'm happy to help with this.

Statement #1 is equivalent to the statement that m =< 0, that m is zero or a negative number. For more explanation of this, see this blog post: http://magoosh.com/gmat/2012/gmat-math- ... -of-minus/ We know from this that m is zero or negative, but not necessarily even or odd, so this statement, alone and by itself, is not sufficient.

Statement #2: We solve this with the Zero Product Property. The ZPP says: If A*B = 0, then A = 0 or B = 0 Notice that, in this statement, the word "or" is no garnish, but rather an essential piece of mathematical equipment. By extension, If A*B*C = 0, then A = 0 or B = 0 or C = 0.

Here, we have: (m)(m - 1)(m + 2) = 0 ===> m = 0 OR (m - 1) = 0 OR (m + 2) = 0 ===> m = 0 OR m = +1 or m = -2 From this statement, m could be any of these three, so it could be even or odd. This statement, alone and by itself, is not sufficient.

Combined statements We look at our set from the second statement, m = {0, +1, -2}, and because of the constraint of the first statement, we know m must be zero or negative, so we can keep 0 and -2, but we have to exclude +1. Now we are down to m = {0, -2}. We don't know which one m equals, but since both of them are even, we now definitively know that m is even. Thus, combined, the statements are sufficient.

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2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself, (or, which is the same, zero is a multiple of every integer except zero itself).

Thank Bunnel for your prompt reply! I chose E because I narrowed down to 0 and 2 from the 2nd statement.

If from (2) we had that m is either 0 or 2, then the answer would be B. Because both 0 and 2 are even. But from (2) m is 0, 1, or -2. If m is 0 or -2, then the answer to the question is YES but if m is 1, then the answer to the question is NO. So, (2) is not sufficient.
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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