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

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Director
Joined: 03 Sep 2006
Posts: 780
If k^(xy)=1, where k#0, which of the following must be true  [#permalink]

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Updated on: 21 May 2012, 23:44
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4
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Difficulty:

25% (medium)

Question Stats:

64% (00:55) correct 36% (01:07) wrong based on 205 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, 05:46.
Last edited by Bunuel on 21 May 2012, 23:44, edited 2 times in total.
Edited the question
Math Expert
Joined: 02 Sep 2009
Posts: 52290
Re: which of the following must be true  [#permalink]

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26 Jan 2012, 05:52
1
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.

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Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 139
Location: India
WE: Information Technology (Investment Banking)
Re: If k^(xy)=1, where k\neq0, which of the following must be  [#permalink]

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26 Jan 2012, 06:00
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??
Math Expert
Joined: 02 Sep 2009
Posts: 52290
Re: If k^(xy)=1, where k\neq0, which of the following must be  [#permalink]

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26 Jan 2012, 06:04
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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Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 139
Location: India
WE: Information Technology (Investment Banking)
Re: If k^(xy)=1, where k\neq0, which of the following must be  [#permalink]

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26 Jan 2012, 20:21
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'
Manager
Joined: 17 Oct 2010
Posts: 75
Re: If k^(xy)=1, where k\neq0, which of the following must be  [#permalink]

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21 May 2012, 23:40
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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19 Sep 2018, 08:49
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Re: If k^(xy)=1, where k#0, which of the following must be true &nbs [#permalink] 19 Sep 2018, 08:49
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