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If k^(xy)=1, where k#0, which of the following must be true

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If k^(xy)=1, where k#0, which of the following must be true  [#permalink]

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New post Updated on: 22 May 2012, 00:44
1
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A
B
C
D
E

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  25% (medium)

Question Stats:

63% (00:56) correct 37% (01:11) wrong based on 190 sessions

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If \(k^{xy}=1\), where \(k\neq0\), which of the following must be true?

I. k=1
II. x=0
III. y=0

A. I only
B. II only
C. III only
D. I and II
E. None

Originally posted by LM on 26 Jan 2012, 06:46.
Last edited by Bunuel on 22 May 2012, 00:44, edited 2 times in total.
Edited the question
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Re: which of the following must be true  [#permalink]

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New post 26 Jan 2012, 06:52
2
3
LM wrote:
If \(k^{xy}=1\), where \(k\neq0\), which of the following must be true?

I. k=1
II. x=0
III. y=0

A. I only
B. II only
C. III only
D. I and II
E. None


Note that we are asked "which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

\(k^{xy}=1\) --> 3 cases:

\(xy=0\) and \(k=any \ integer\);
\(k=1\) and \(xy=any \ integer\);
\(k=-1\) and \(xy=even\).

So if we take for example the third case: \(k=-1\) and \(xy=2\) then none of option MUST be true.

Answer: E.
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Re: If k^(xy)=1, where k\neq0, which of the following must be  [#permalink]

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New post 26 Jan 2012, 07:00
1
LM wrote:
If \(k^{xy}=1\), where \(k\neq0\), which of the following must be true?

I. k=1
II. x=0
III. y=0

A. I only
B. II only
C. III only
D. I and II
E. None


Bunnel if we take all the 3 statements together then can't that be an answer??
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Re: If k^(xy)=1, where k\neq0, which of the following must be  [#permalink]

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New post 26 Jan 2012, 07:04
1
subhajeet wrote:
LM wrote:
If \(k^{xy}=1\), where \(k\neq0\), which of the following must be true?

I. k=1
II. x=0
III. y=0

A. I only
B. II only
C. III only
D. I and II
E. None


Bunnel if we take all the 3 statements together then can't that be an answer??


If k=1, x=0 and y=0 then k^(xy)=1^0=1, so this is certainly one of the possibilities (though not the only one: refer to my solution above). Also notice that we are asked which of the following MUST be true not COULD be true.

Hope it's clear.
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Re: If k^(xy)=1, where k\neq0, which of the following must be  [#permalink]

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New post 26 Jan 2012, 21:21
Bunuel wrote:
subhajeet wrote:
LM wrote:
If \(k^{xy}=1\), where \(k\neq0\), which of the following must be true?

I. k=1
II. x=0
III. y=0

A. I only
B. II only
C. III only
D. I and II
E. None


Bunnel if we take all the 3 statements together then can't that be an answer??


If k=1, x=0 and y=0 then k^(xy)=1^0=1, so this is certainly one of the possibilities (though not the only one: refer to my solution above). Also notice that we are asked which of the following MUST be true not COULD be true.

Hope it's clear.


Bunnel thanks for the explanation. Got confused between 'could be true' and 'must be true' :(
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Re: If k^(xy)=1, where k\neq0, which of the following must be  [#permalink]

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New post 22 May 2012, 00:40
a simpler way:

option 1) k=1 , not necessarily must be true as K can be 2,3,34, etc any number as long as either x or y is equal to 0 ( zero) ,

option 2)x=0 , doesn't have to be true because y = 0, and any number raised to 0 equals 1.so condition holds even if x is not equal to 0

option 3) y=0, doesn't have to be true as x=0 and the condition could still hold .

so we can see none of the options MUST be true , for the condition to hold .

Although Bunuel's solution is better a usual.
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Re: If k^(xy)=1, where k#0, which of the following must be true  [#permalink]

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New post 17 Jun 2019, 18:05
which of the following must be true? - This is a a very important part.

1) k = 1
Then definitely 1 ^ xy is always 1, but say k = 2 and xy = 0 even now K^xy = 1 . Hence K can actually take any integer other than 1. So k = 1 must be true is false, since K can also be 2,3,4 etc...

Following a similar logic the best option is E
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Re: If k^(xy)=1, where k#0, which of the following must be true   [#permalink] 17 Jun 2019, 18:05
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