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# M24-06

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Math Expert
Joined: 02 Sep 2009
Posts: 50058

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16 Sep 2014, 01:20
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Difficulty:

45% (medium)

Question Stats:

61% (00:54) correct 39% (00:59) wrong based on 148 sessions

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If $$a$$ and $$b$$ are negative integers, is $$a^b$$ negative?

(1) $$ab$$ is even

(2) $$a + b$$ is even

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Math Expert
Joined: 02 Sep 2009
Posts: 50058

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16 Sep 2014, 01:20
1
2
Official Solution:

(1) $$ab$$ is even. If $$a=b=-2$$, then $$a^b=\frac{1}{4} \gt 0$$ but if $$a=-2$$ and $$b=-1$$, then $$a^b=-\frac{1}{2} \lt 0$$. Not sufficient.

(2) $$a+b$$ is even. If $$a=b=-2$$, then $$a^b=\frac{1}{4} \gt 0$$ but if $$a=-1$$ and $$b=-1$$, then $$a^b=-1 \lt 0$$. Not sufficient.

(1)+(2) From $$a+b=even$$ it follows that $$a$$ and $$b$$ are either both odd or both even. Now, if both are odd then $$ab=even$$ won't hold true, so both $$a$$ and $$b$$ must be even. Hence $$a^b=negative^{even}=positive$$. Sufficient.

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Joined: 08 Feb 2014
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22 Nov 2014, 19:55
We know that both a and b are negative integers. Therefore, a is being raised to the power of a negative number. This means that for a^b to be positive, b must be even (because a is neg. ex: a=(-2), (-2)^2 =4, while (-2)^3=(-8).

stmt 1: all we know is that at least one of a or b is even, because you need a min. of one even number in a set, for the product of a set of numbers to be even. a, or b, or both could be even. Insuff.

stmt 2: all we know is that a and b are either both odd or both even, because o+o=e and e+e=e but o+e=o.

stmts 1&2: stmt 1 tells us that at least a or b is even. combined with stmt 2, we know that both terms are even, therefore, b is even.
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13 Apr 2018, 04:16
+1 for option C.

The question requires us to find out if b is even or not.

St 1 : ie a is even , b is even or both are even. We can't be sure of b is even or not. NS
St 2 : i.e both are even or both are odd. NS

Both together : Yes , sufficient ! We now know that both are even. Hence the answer is C
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Re: M24-06 &nbs [#permalink] 13 Apr 2018, 04:16
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# M24-06

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