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# Is the integer m an even number?

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Manager
Joined: 25 Jun 2012
Posts: 63
Is the integer m an even number?  [#permalink]

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15 Nov 2012, 11:03
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35% (medium)

Question Stats:

65% (01:00) correct 35% (00:56) wrong based on 435 sessions

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Is the integer m an even number?

(1) |m| = -m

(2) (m)(m - 1)(m + 2) = 0
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4485
Re: Is the integer m an even number?  [#permalink]

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15 Nov 2012, 15:07
5
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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.

Does all this make sense?
Mike
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Mike McGarry
Magoosh Test Prep

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Re: Is the integer m an even number?  [#permalink]

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15 May 2014, 02:18
1
Is the integer m an even number?

(1) |m| = -m

(2) (m)(m - 1)(m + 2) = 0

Sol: Given in St 1 |m|=-m or $$m+|m|=0$$ ----->this implies some a positive no + m= 0 and therefore m has to be a negative number.

Now m can be -2 then ans is Y but if m=-3 then no

So Option A and D are ruled out

St 2 tells you that possible values of m are 0,1 and -2
Again if m=0,-2 then m is even number but if m=1 then m is odd. Option B ruled out

Combining we get that m is a negative number so m=-2. Ans is C
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Manager
Joined: 07 May 2013
Posts: 97
Re: Is the integer m an even number?  [#permalink]

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06 Jun 2014, 19:07
Great explanation mike. I solved the problem but without understanding the importance of OR. Good job and please keep us enlightened
Manager
Joined: 20 Dec 2013
Posts: 122
Re: Is the integer m an even number?  [#permalink]

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06 Jun 2014, 23:31
Is the integer m an even number?

(1) |m| = -m

(2) (m)(m - 1)(m + 2) = 0

Statement I is insufficient:

m = 0, m = -1 would be suffice the statement and m can be even or odd. M can be any negative number or zero

Statement II is insufficient:

Either m = 0 OR m = 1 OR m = -2

Combining:

m = 0 or m = -2 which states that m will be even

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Manager
Joined: 04 Jan 2014
Posts: 103
Re: Is the integer m an even number?  [#permalink]

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18 Jun 2014, 01:00
Is zero considered an even integer?
Math Expert
Joined: 02 Sep 2009
Posts: 52386
Re: Is the integer m an even number?  [#permalink]

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18 Jun 2014, 01:02
1
1
pretzel wrote:
Is zero considered an even integer?

ZERO:

1. 0 is an integer.

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).

Check more here: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1371030

Hope it helps.
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Re: Is the integer m an even number?  [#permalink]

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18 Jun 2014, 01:04
Thank Bunnel for your prompt reply! I chose E because I narrowed down to 0 and 2 from the 2nd statement.
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Re: Is the integer m an even number?  [#permalink]

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18 Jun 2014, 01:29
pretzel wrote:
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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Re: Is the integer m an even number?  [#permalink]

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12 Aug 2016, 20:15
Is the integer m an even number?

(1) |m| = -m

(2) (m)(m - 1)(m + 2) = 0

From statement 1, we know LHS has absolute value.
Since absolute value can never be negative, so the only plausible solution was 0

I am sure am missing out something- can anyone help me out?
Math Expert
Joined: 02 Sep 2009
Posts: 52386
Re: Is the integer m an even number?  [#permalink]

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12 Aug 2016, 23:51
1
Anigr16 wrote:
Is the integer m an even number?

(1) |m| = -m

(2) (m)(m - 1)(m + 2) = 0

From statement 1, we know LHS has absolute value.
Since absolute value can never be negative, so the only plausible solution was 0

I am sure am missing out something- can anyone help me out?

This is explained above: from (1) $$-m\geq 0$$ --> $$m \leq 0$$ --> m can be any non-positive value to satisfy |m| = -m.
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Re: Is the integer m an even number?  [#permalink]

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20 Jun 2017, 19:42
Imo C
Very tricky question
From statement 1 we have m<=0 , m can take any value hence insufficient
From statement 2 we have m=0,1,-2 hence insufficient .
Combining we have 0 and -2 hence sufficient
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Re: Is the integer m an even number?  [#permalink]

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01 Nov 2018, 07:40
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Re: Is the integer m an even number? &nbs [#permalink] 01 Nov 2018, 07:40
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