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M30-09

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M30-09  [#permalink]

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New post 16 Sep 2014, 01:45
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A
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C
D
E

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  55% (hard)

Question Stats:

57% (01:20) correct 43% (01:14) wrong based on 151 sessions

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Re M30-09  [#permalink]

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New post 16 Sep 2014, 01:45
4
Official Solution:


(1) \(|k| + k = 0\). Re-arrange: \(|k|=-k\), This implies that \(k\leq{0}\). Since we are told that \(k\) is a nonzero integer, then we have that \(k < 0\). Not sufficient.

(2) \(|k^k| = k^0\). Any nonzero number to the power of 0 is 1, hence \(k^0=1\). So, we have that \(|k^k| = 1\). This implies that \(k=1\) or \(k=-1\). Not sufficient.

(1)+(2) Since from (1) \(k < 0\), then from (2) \(k=-1\). Sufficient.


Answer: C
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New post 23 Apr 2015, 04:09
1
MBAinSCM wrote:
Bunuel wrote:
Official Solution:


(1) \(|k| + k = 0\). Re-arrange: \(|k|=-k\), This implies that \(k\leq{0}\). Since we are told that \(k\) is a nonzero integer, then we have that \(k < 0\). Not sufficient.

(2) \(|k^k| = k^0\). Any nonzero number to the power of 0 is 1, hence \(k^0=1\). So, we have that \(|k^k| = 1\). This implies that \(k=1\) or \(k=-1\). Not sufficient.

(1)+(2) Since from (1) \(k < 0\), then from (2) \(k=-1\). Sufficient.


Answer: C



Hi Bunuel,

Can you please explain how you got \(k\leq{0}\) in 1 and \(k=1\) or \(k=-1\). in 2.

Thanks in advance!!


For 1:
Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\).

Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: absolute-value-tips-and-hints-175002.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html


For 2: can you please tell me what is unclear there?
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M30-09  [#permalink]

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New post 23 Apr 2015, 04:05
Bunuel wrote:
Official Solution:


(1) \(|k| + k = 0\). Re-arrange: \(|k|=-k\), This implies that \(k\leq{0}\). Since we are told that \(k\) is a nonzero integer, then we have that \(k < 0\). Not sufficient.

(2) \(|k^k| = k^0\). Any nonzero number to the power of 0 is 1, hence \(k^0=1\). So, we have that \(|k^k| = 1\). This implies that \(k=1\) or \(k=-1\). Not sufficient.

(1)+(2) Since from (1) \(k < 0\), then from (2) \(k=-1\). Sufficient.


Answer: C



Hi Bunuel,

Can you please explain how you got \(k\leq{0}\) from \(|k|=-k\) in (1) and \(k=1\) or \(k=-1\) from in (2).

Thanks in advance!!
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Re: M30-09  [#permalink]

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New post 26 Apr 2015, 02:45
Hi Bunuel,

Thanks you for the explanation and the link. I now understand that step clearly. :-D

Got the second one too. I think.
|k^k| = 1 can be represented by both -> k > 1 , 1^1 and k <-1 (-1)^-1

Thanks!!!
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Re: M30-09  [#permalink]

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New post 25 Aug 2016, 20:46
Just wondering ---- what is -1^0? I noticed in the solution, it said that any nonzero number raised to zero is 1, but is this true for negative numbers as well? Can't seem to find anything on the web about this...

Thanks.
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Re: M30-09  [#permalink]

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New post 05 May 2017, 09:19
Does the GMAT acknowledge 0^0 as 1 or indeterminate?
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New post 05 May 2017, 10:03
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Re: M30-09  [#permalink]

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New post 28 Sep 2018, 02:58
Bunuel wrote:
pierce514 wrote:
Does the GMAT acknowledge 0^0 as 1 or indeterminate?


0^0, in some sources equals to 1 (not 0), some mathematicians say it's undefined. But you won't need this for the GMAT because the case of 0^0 is not tested on the GMAT.



Please see the attached! Bunuel could you please explain this?

Thanks
>> !!!

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New post 28 Sep 2018, 03:03
sunnyattri wrote:
Bunuel wrote:
pierce514 wrote:
Does the GMAT acknowledge 0^0 as 1 or indeterminate?


0^0, in some sources equals to 1 (not 0), some mathematicians say it's undefined. But you won't need this for the GMAT because the case of 0^0 is not tested on the GMAT.



Please see the attached! Bunuel could you please explain this?

Thanks


What is there to explain? Yes, -6^0 = -(6^0)= -1.
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New post 28 Sep 2018, 03:31
Please see the attached! Bunuel could you please explain this?

Thanks[/quote]

What is there to explain? Yes, -6^0 = -(6^0)= -1.[/quote]

if thats ture then .. shouldn't the ans B .. coz |X^X| has to positive integer, so the only way |X^X| = X^0 if X is 1.... please correct me if i'm wrong.
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New post 28 Sep 2018, 03:36
sunnyattri wrote:
Please see the attached! Bunuel could you please explain this?

Thanks


What is there to explain? Yes, -6^0 = -(6^0)= -1.[/quote]

if thats ture then .. shouldn't the ans B .. coz |X^X| has to positive integer, so the only way |X^X| = X^0 if X is 1.... please correct me if i'm wrong.[/quote]

Ah, I see what you mean. If k = -1, then we'd get (-1)^0 = 1 and not -1^0 = -1.
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Re: M30-09  [#permalink]

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New post 28 Sep 2018, 03:38
sunnyattri wrote:
Please see the attached! Bunuel could you please explain this?

Thanks


What is there to explain? Yes, -6^0 = -(6^0)= -1.[/quote]

if thats ture then .. shouldn't the ans B .. coz |X^X| has to positive integer, so the only way |X^X| = X^0 if X is 1.... please correct me if i'm wrong.[/quote]

Bunuel you quoted in your official solution that "Any nonzero number to the power of 0 is 1" and you agreed that .. "-6^0 is -1"... so, which one is correct?
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New post 28 Sep 2018, 03:42
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Re: M30-09  [#permalink]

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New post 28 Sep 2018, 03:44
Bunuel wrote:
sunnyattri wrote:
Please see the attached! Bunuel could you please explain this?

Thanks


What is there to explain? Yes, -6^0 = -(6^0)= -1.


if thats ture then .. shouldn't the ans B .. coz |X^X| has to positive integer, so the only way |X^X| = X^0 if X is 1.... please correct me if i'm wrong.[/quote]

Ah, I see what you mean. If k = -1, then we'd get (-1)^0 = 1 and not -1^0 = -1.[/quote]

if K =-1 the we'd get -1^0 = -1 not 1 ( see the attached)
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New post 28 Sep 2018, 03:46
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New post 28 Sep 2018, 03:49
Bunuel wrote:
sunnyattri wrote:
if K =-1 the we'd get -1^0 = -1 not 1 ( see the attached)


You are wrong.

If k = -1, then k^0 = (-1)^0 = 1.


Ah, it's clear now, Thanks ..
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Re: M30-09  [#permalink]

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New post 08 Jan 2019, 08:33
Bunuel wrote:
What is the value of nonzero integer \(k\)?



(1) \(|k| + k = 0\)

(2) \(|k^k| = k^0\)


(1)
\(|k|\) is always positive, so k is any negative integer.
For example \(|-1| + -1 = 0\) or \(|-2| + -2 = 0\) etc.
Not sufficient.

(2)
RHS will be 1 (any integer to the power of 0 is 1)
\(|k^k| = 1\)
\(k^k = 1\) ---> \(k = 1\)
OR
\(k^k = -1\) ---> \(k = -1\)
Not sufficient.

(+)
From (1) we determined k was negative and from (2) only one answer is negative
k = -1
Sufficient.

ANSWER C
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Re: M30-09  [#permalink]

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New post 09 Nov 2019, 05:24
Hi Bunuel Could you please clarify this?

Quote:
|k|=−k|k|=−k, This implies that k≤0.

According to the property isn't it just k<0 and not k≤0 ?

Please confirm.

Thank you,
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Re: M30-09   [#permalink] 09 Nov 2019, 05:24

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