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# If a and b are integers, and a not= b, is |a|b > 0? (1)

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If a and b are integers, and a not= b, is |a|b > 0? (1) [#permalink]

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20 Dec 2006, 15:21
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If a and b are integers, and a not= b, is |a|b > 0?

(1) |a^b| > 0

(2) |a|^b is a non-zero integer
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20 Dec 2006, 15:37
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(E) for me.

|a|*b > 0 ?
<=> b > 0 ?

From (1)
|a^b| > 0

o If b=-1 and a=1, 1 > 0 and b < 0
o If b=1 and a=-1, 1 > 0 and b > 0

INSUFF.

From (2)
|a|^b is a non zero integer implies that b must be positive or equal to 0 in order to not create a real number such as 2^-1.

So, we remains with the cases of b = 0 and b > 0.

INSUFF.

Both (1) and (2)
It brings nothing more.

INSUFF.

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20 Dec 2006, 18:10
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Given: a , b are ints. and a is different from b
asking: |a| * b > 0

what the question is really asking if b > 0 [ |a| is always >0 ]

(1) |a^b| >0
----------------
says nothing, cuz |x| is always > 0
statement 1 is insufficient

(2) |a|^b is not zero
-------------------------
also says nothing .. we know |a|^b > 0
b could be -ve or +ve
statement 2 is insufficient

(1) and (2) together
------------------------
both statements really say nothing about b

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16 Jan 2007, 21:49
is |a|b > 0 is same as, is b > 0

stat 1 is not sufficient since all it says is that a <> 0

from stat 2 b>= 0 since a only negative value of b can cause |a|^b to be not a non zero integer

so E ?

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Re: DS_If a and b.... [#permalink]

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16 Jan 2007, 22:39
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mm007 wrote:
If a and b are integers, and a not= b, is |a|b > 0?
(1) |a^b| > 0
(2) |a|^b is a non-zero integer

First of all, the question is equivalent to: is a<>0 AND b>0?

(1) For |a^b| > 0, a could be <>0 but b could be <0 OR b>0 => insuff => B, C, or E.

(2) This is a similar case to (1): b, similarly to a, could be <>0 => insuff => C or E.

(1&2) Again, similar case to (1) and (2): b could be +ve or -ve => insuff => E.

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17 Jan 2007, 00:03
Quote:
If a and b are integers, and a not= b, is |a|b > 0?
(1) |a^b| > 0
(2) |a|^b is a non-zero integer

E2

I rephrased the stem to "is |a|>0 and is b -ve, 0 or +ve?"

1) |a^b| will always be +ve. But here, b can be greater than or equal to 0 and |a^b| will still be +ve
INS

2) |a|^b. Here, |a| cannot be 0 but B can be 0 or 1. b can't be -ve b/c |a|^b will become a non-integer, 1/a^b.
INS

Taken together, we're still not told anything about b. If we can't figure what b is we can't do much with this prob.
INS

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10 Sep 2009, 04:55
Can anybody chime in? Although I agree with the answer, I don't think the bold statement below is correct. B does not have to be
0 or greater. It can be -4 if A is 1.

If b= -4 and a=1. this satisfiies the conditions: 1) a,b are integers 2) a is not equal to b 3) statement 2; 1^(-4)=1. that is a non-zero integer. So you can't say B is definitely positive or 0. Combining statement 1 and 2. You can't determine if A is a 1 or not so E.

Please correct me if I am wrong. Thanks.

Fig wrote:
(E) for me.

|a|*b > 0 ?
<=> b > 0 ?

From (1)
|a^b| > 0

o If b=-1 and a=1, 1 > 0 and b < 0
o If b=1 and a=-1, 1 > 0 and b > 0

INSUFF.

[b]From (2)
|a|^b is a non zero integer implies that b must be positive or equal to 0 in order to not create a real number such as 2^-1.

So, we remains with the cases of b = 0 and b > 0.[/b]
INSUFF.

Both (1) and (2)
It brings nothing more.

INSUFF.

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Re: DS_If a and b.... [#permalink]

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26 Nov 2009, 20:34
mm007 wrote:
If a and b are integers, and a not= b, is |a|b > 0?

(1) |a^b| > 0

(2) |a|^b is a non-zero integer

Clearly E.
|a^b| and |a|^b are always >0, no matter what b is, because absolute value is always greater than 0.
Thus, we can't know if b>0 or not.
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Re: DS_If a and b.... [#permalink]

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21 Dec 2009, 09:22
mm007 wrote:
If a and b are integers, and a not= b, is |a|b > 0?

(1) |a^b| > 0

(2) |a|^b is a non-zero integer

Question is |a|b>0 this can be proved if we can prove that a not=0 and b>0.
1. |a^b| > 0 implies that a not = 0. a can be +ve or -ve, and be can be 0, +ve or -ve....but we are sure that a not=0 else |a^b| = 0.
Statement 1 itself is insuff.

2. |a|^b is a non zero integer.
we already know a,b both are integers....|a| is > 0, so b >= 0.
No clear value of b...

So ans E.

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15 Sep 2010, 09:15
Mishari wrote:

Given: a , b are ints. and a is different from b
asking: |a| * b > 0

what the question is really asking if b > 0 [ |a| is always >0 ]

(1) |a^b| >0
----------------
says nothing, cuz |x| is always > 0
statement 1 is insufficient

(2) |a|^b is not zero
-------------------------
also says nothing .. we know |a|^b > 0
b could be -ve or +ve
statement 2 is insufficient

(1) and (2) together
------------------------
both statements really say nothing about b

>>Please tell me how can |a| be taken as positive in the above steps without knowing its sign..I mean if a is negative,then |a| wil be negative..Right?Am i missing anythin badly?

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15 Sep 2010, 09:39
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ravitejapandiri wrote:
>>Please tell me how can |a| be taken as positive in the above steps without knowing its sign..I mean if a is negative,then |a| wil be negative..Right?Am i missing anythin badly?

Absolute value of of an expression is alway non-negative: $$|some \ expression|\geq{0}$$. Please check Walker's post on Absolute Value at: math-absolute-value-modulus-86462.html

As for the question:

If a and b are integers, and a does not equal to b, is |a|*b > 0?
(1) |a^b| > 0
(2) |a|^b is a non-zero integer.

$$|a|*b>0$$ is true when $$b>0$$ and $$a$$ does not equal to zero.

(1) $$|a^b| > 0$$ --> $$a$$ does not equal to zero, but we don't know about $$b$$, it can be any value, positive or negative. Not sufficient.

(2) $$|a|^b$$ is a non-zero integer --> $$a$$ can be 1 and $$b$$ any integer, positive or negative. Not sufficient.

(1)+(2) If $$a=1$$ and $$b=2$$, then $$|a|*b>0$$, but if $$a=1$$ and $$b=-2$$, then $$|a|*b<0$$. Not sufficient.

Other discussion of this question at: good-set-of-ds-85413.html
Similar question: the-power-of-absolutes-manhattan-challenge-problem-101661.html

Hope it helps.
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Re: DS question : need help [#permalink]

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28 Oct 2010, 19:20
Basically just need to find out if b is positive or negative, since a will always be positive as it is inside of the | |.

1) Doesn't give you anything because everything is inside of the | |, so you can't tell if b is positive or negative, so insufficient.

2) Tells you that b is not negative since that would result in a non-integer. However, b could be 0 since a number raised to the 0 is 1, which is a non-integer number. Also insufficient.

Using both statements still doesn't provide anything because b can still be equal to 0, in which case |a|b > 0 is false. However, b can also be any positive number which would make |a|b > 0 true. Thus E.

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Re: If a and b are integers, and a not= b, is |a|b > 0? (1) [#permalink]

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04 Jun 2014, 19:35
The reason why s2 alone or taken together with s1 is not sufficient bcos we need info on the signs that is, a is +ve or -ve and whether b is +ve or -ve & not wether they are zero or non-zero integers.

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Re: If a and b are integers, and a not= b, is |a|b > 0? (1) [#permalink]

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02 Sep 2014, 04:43
Bunuel wrote:
ravitejapandiri wrote:
>>Please tell me how can |a| be taken as positive in the above steps without knowing its sign..I mean if a is negative,then |a| wil be negative..Right?Am i missing anythin badly?

Absolute value of of an expression is alway non-negative: $$|some \ expression|\geq{0}$$. Please check Walker's post on Absolute Value at: math-absolute-value-modulus-86462.html

As for the question:

If a and b are integers, and a does not equal to b, is |a|*b > 0?
(1) |a^b| > 0
(2) |a|^b is a non-zero integer.

$$|a|*b>0$$ is true when $$b>0$$ and $$a$$ does not equal to zero.

(1) $$|a^b| > 0$$ --> $$a$$ does not equal to zero, but we don't know about $$b$$, it can be any value, positive or negative. Not sufficient.

(2) $$|a|^b$$ is a non-zero integer --> $$a$$ can be 1 and $$b$$ any integer, positive or negative. Not sufficient.

(1)+(2) If $$a=1$$ and $$b=2$$, then $$|a|*b>0$$, but if $$a=1$$ and $$b=-2$$, then $$|a|*b<0$$. Not sufficient.

Other discussion of this question at: good-set-of-ds-85413.html
Similar question: the-power-of-absolutes-manhattan-challenge-problem-101661.html

Hope it helps.

Hi Bunuel,
I've always struggled when to consider 0 as an integer and when not. Is there any concept that you can share? Appreciate your help!!

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If a and b are integers, and a not= b, is |a|b > 0? (1) [#permalink]

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02 Sep 2014, 04:49
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shahPranay14 wrote:
Bunuel wrote:
ravitejapandiri wrote:
>>Please tell me how can |a| be taken as positive in the above steps without knowing its sign..I mean if a is negative,then |a| wil be negative..Right?Am i missing anythin badly?

Absolute value of of an expression is alway non-negative: $$|some \ expression|\geq{0}$$. Please check Walker's post on Absolute Value at: math-absolute-value-modulus-86462.html

As for the question:

If a and b are integers, and a does not equal to b, is |a|*b > 0?
(1) |a^b| > 0
(2) |a|^b is a non-zero integer.

$$|a|*b>0$$ is true when $$b>0$$ and $$a$$ does not equal to zero.

(1) $$|a^b| > 0$$ --> $$a$$ does not equal to zero, but we don't know about $$b$$, it can be any value, positive or negative. Not sufficient.

(2) $$|a|^b$$ is a non-zero integer --> $$a$$ can be 1 and $$b$$ any integer, positive or negative. Not sufficient.

(1)+(2) If $$a=1$$ and $$b=2$$, then $$|a|*b>0$$, but if $$a=1$$ and $$b=-2$$, then $$|a|*b<0$$. Not sufficient.

Other discussion of this question at: good-set-of-ds-85413.html
Similar question: the-power-of-absolutes-manhattan-challenge-problem-101661.html

Hope it helps.

Hi Bunuel,
I've always struggled when to consider 0 as an integer and when not. Is there any concept that you can share? Appreciate your help!!

0 is neither positive nor negative even integer.

Check for more here: number-properties-tips-and-hints-174996.html
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Re: If a and b are integers, and a not= b, is |a|b > 0? (1) [#permalink]

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20 Oct 2014, 23:23
If a and b are integers, and a does not equal to b, is |a|*b > 0?
(1) |a^b| > 0
(2) |a|^b is a non-zero integer.
to check whether |a|*b > 0, we need to identify whether b>0

1 - This statement has to be positive, irrespective of the value of a and b.
This is insufficient

2. |a|^b is non-zero integer -

Possibilities -
a = -.5 or .5 and b = -1 The value of expression would be 2
a = any number and b = 0. The value will be 1
a = any positive / negative number and b = any positive number Thus the result will be a positive number.

Thus insufficient.

Combining two,

We'll get positive values for y, and zero.

Thus combining two will not give solution.

Thus Ans - E
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Re: If a and b are integers, and a not= b, is |a|b > 0? (1) [#permalink]

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If a and b are integers, and a not= b, is |a|b > 0? (1) [#permalink]

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20 Mar 2017, 20:51
mm007 wrote:
If a and b are integers, and a not= b, is $$|a|b > 0$$?

(1) $$|a^b| > 0$$

(2) $$|a|^b$$ is a non-zero integer

OFFICIAL SOLUTION

Let us start be examining the conditions necessary for $$|a|b > 0$$. Since |a| cannot be negative, both |a| and b must be positive. However, since |a| is positive whether a is negative or positive, the only condition for a is that it must be non-zero.

Hence, the question can be restated in terms of the necessary conditions for it to be answered "yes":

“Do both of the following conditions exist: a is non-zero AND b is positive?”

(1) INSUFFICIENT: In order for a = 0, $$|a^b|$$ would have to equal 0 since 0 raised to any power is always 0. Therefore (1) implies that a is non-zero. However, given that a is non-zero, b can be anything for $$|ab| > 0$$ so we cannot determine the sign of b.

(2) INSUFFICIENT: If a = 0, |a| = 0, and $$|a|^b = 0$$ for any b. Hence, a must be non-zero and the first condition (a is not equal to 0) of the restated question is met. We now need to test whether the second condition is met. (Note: If a had been zero, we would have been able to conclude right away that (2) is sufficient because we would answer "no" to the question is |a|b > 0?) Given that a is non-zero, |a| must be positive integer. At first glance, it seems that b must be positive because a positive integer raised to a negative integer is typically fractional (e.g., $$a^{-2} = \frac{1}{{a^2}}$$. Hence, it appears that b cannot be negative. However, there is a special case where this is not so:

If |a| = 1, then b could be anything (positive, negative, or zero) since $$|1|^b$$ is always equal to 1, which is a non-zero integer . In addition, there is also the possibility that b = 0. If |b| = 0, then $$|a|^0$$ is always 1, which is a non-zero integer.

Hence, based on (2) alone, we cannot determine whether b is positive and we cannot answer the question.

An alternative way to analyze this (or to confirm the above) is to create a chart using simple numbers as follows:

a b Is $$|a|^b$$ a non-zero integer? Is $$|a|b > 0$$?
1 2 Yes Yes
1 -2 Yes No
2 1 Yes Yes
2 0 Yes No

We can quickly confirm that (2) alone does not provide enough information to answer the question.

(1) AND (2) INSUFFICIENT: The analysis for (2) shows that (2) is insufficient because, while we can conclude that a is non-zero, we cannot determine whether b is positive. (1) also implies that a is non-zero, but does not provide any information about b other than that it could be anything. Consequently, (1) does not add any information to (2) regarding b to help answer the question and (1) and (2) together are still insufficient. (Note: As a quick check, the above chart can also be used to analyze (1) and (2) together since all of the values in column 1 are also consistent with (1)).

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If a and b are integers, and a not= b, is |a|b > 0? (1)   [#permalink] 20 Mar 2017, 20:51
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