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M24-06

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M24-06  [#permalink]

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New post 16 Sep 2014, 01:20
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

65% (01:23) correct 35% (01:24) wrong based on 109 sessions

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Re M24-06  [#permalink]

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New post 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.


Answer: C
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Re: M24-06  [#permalink]

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New post 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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Re: M24-06  [#permalink]

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New post 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  [#permalink]

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New post 10 Apr 2019, 18:22
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. In statement 2: When A=B=-2, that given A+B= -4 which is not even and when A=B=-1, then also A+B=-2 which is not even.
To answer the question, values of A and B are not sufficient.
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Re: M24-06  [#permalink]

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New post 10 Apr 2019, 21:19
oharsha12 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. In statement 2: When A=B=-2, that given A+B= -4 which is not even and when A=B=-1, then also A+B=-2 which is not even.
To answer the question, values of A and B are not sufficient.


Both -4 and -2 are even integers.b Even number is an integer, which is divisible by 2 without a remainder.
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Re: M24-06  [#permalink]

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New post 11 Apr 2019, 00:29
Bunuel wrote:
If \(a\) and \(b\) are negative integers, is \(a^b\) negative?


(1) \(ab\) is even

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


#1
\(ab\) is even
one of integers has to be even so we dont know which one of a or b is even ; insufficient
#2
\(a + b\) is even ;
both a & b odd or both even ; insufficient
from1 & 2
ab both are even
so [m]a^b[/m is not negative
IMO C
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Re: M24-06   [#permalink] 11 Apr 2019, 00:29
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