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16 Sep 2014, 01:45



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Bunuel wrote: Official Solution:
(1) \(k + k = 0\). Rearrange: \(k=k\), 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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23 Apr 2015, 04:09
MBAinSCM wrote: Bunuel wrote: Official Solution:
(1) \(k + k = 0\). Rearrange: \(k=k\), 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: \(5=5\). For 2: can you please tell me what is unclear there?
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26 Apr 2015, 02:45
Hi Bunuel, Thanks you for the explanation and the link. I now understand that step clearly. 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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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...
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05 May 2017, 09:19
Does the GMAT acknowledge 0^0 as 1 or indeterminate?



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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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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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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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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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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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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










