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M28-37

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M28-37  [#permalink]

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New post 16 Sep 2014, 00:31
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A
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Question Stats:

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If \(x\) is not equal to 0 and \(x^y=1\), then which of the following must be true?
I. \(x=1\)

II. \(x=1\) and \(y=0\)

III. \(x=1\) or \(y=0\)


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

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Re M28-37  [#permalink]

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New post 16 Sep 2014, 00:31
1
Official Solution:

If \(x\) is not equal to 0 and \(x^y=1\), then which of the following must be true?
I. \(x=1\)

II. \(x=1\) and \(y=0\)

III. \(x=1\) or \(y=0\)


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


Notice that if \(x=-1\) and \(y\) is any even number, then \((-1)^{even}=1\), thus none of the options must be true.


Answer: E
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Re: M28-37  [#permalink]

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New post 16 Jul 2015, 03:25
Bunuel wrote:
Official Solution:

If \(x\) is not equal to 0 and \(x^y=1\), then which of the following must be true?
I. \(x=1\)

II. \(x=1\) and \(y=0\)

III. \(x=1\) or \(y=0\)


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


Notice that if \(x=-1\) and \(y\) is any even number, then \((-1)^{even}=1\), thus none of the options must be true.


Answer: E

Dear Bunuel,

Please explain why C is incorrect. Even that satisfies the must condition.
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Re: M28-37  [#permalink]

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New post 16 Jul 2015, 03:38
shanti47 wrote:
Bunuel wrote:
Official Solution:

If \(x\) is not equal to 0 and \(x^y=1\), then which of the following must be true?
I. \(x=1\)

II. \(x=1\) and \(y=0\)

III. \(x=1\) or \(y=0\)


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


Notice that if \(x=-1\) and \(y\) is any even number, then \((-1)^{even}=1\), thus none of the options must be true.


Answer: E

Dear Bunuel,

Please explain why C is incorrect. Even that satisfies the must condition.


III says \(x=1\) or \(y=0\). This is NOT necessarily true: consider x = -1 and y = 2 --> (-1)^2 = 1.
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Re: M28-37  [#permalink]

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New post 11 Feb 2016, 23:22
x=1, y=1 discards 3d option
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Re: M28-37  [#permalink]

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New post 18 Jun 2017, 02:33
Hi Bunuel, I didn't understand why B is incorrect. When x=1 and y=0, then the answer is always 1. Please clarify.
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Re: M28-37  [#permalink]

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New post 18 Jun 2017, 02:37
Romannepal wrote:
Hi Bunuel, I didn't understand why B is incorrect. When x=1 and y=0, then the answer is always 1. Please clarify.


The question asks which of the following MUST be true? Not COULD be true.

II says: \(x=1\) and \(y=0\).

This is not necessarily true. Consider x = -1 and y = 0.
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Re: M28-37  [#permalink]

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New post 05 Oct 2017, 06:58
Hi Bunuel, why is C incorrect. It says X=1 or Y=0. Now if X=1, x^y will always be 1 irrespective of Y and if Y = 0, then x^y will be 1 as well since X cannot be 0. Can you please explain.
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Re: M28-37  [#permalink]

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New post 05 Oct 2017, 07:00
mansi20081990 wrote:
Hi Bunuel, why is C incorrect. It says X=1 or Y=0. Now if X=1, x^y will always be 1 irrespective of Y and if Y = 0, then x^y will be 1 as well since X cannot be 0. Can you please explain.


This is explained above:

III says \(x=1\) or \(y=0\). This is NOT necessarily true: consider x = -1 and y = 2 --> (-1)^2 = 1.
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Re: M28-37  [#permalink]

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New post 28 Nov 2018, 05:34
Hi,
Can someone explain why A isn't correct. When X=1, then 1 to the power of any except 0 is 1 right?
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Re: M28-37  [#permalink]

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New post 28 Nov 2018, 05:40
swathi27 wrote:
Hi,
Can someone explain why A isn't correct. When X=1, then 1 to the power of any except 0 is 1 right?


The question asks: which of the following must be true? NOT which of the following could be true?

I says \(x=1\). This is NOT necessarily true: consider x = -1 and y = 2 --> (-1)^2 = 1.
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Re: M28-37  [#permalink]

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New post 28 Nov 2018, 07:20
Bunuel wrote:
Official Solution:

If \(x\) is not equal to 0 and \(x^y=1\), then which of the following must be true?
I. \(x=1\)

II. \(x=1\) and \(y=0\)

III. \(x=1\) or \(y=0\)


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


Notice that if \(x=-1\) and \(y\) is any even number, then \((-1)^{even}=1\), thus none of the options must be true.


Answer: E








In the question, its already mentioned that x=1, so why are we taking x=-1 as other possibility ?
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Re: M28-37  [#permalink]

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New post 28 Nov 2018, 07:22
bhateja wrote:
Bunuel wrote:
Official Solution:

If \(x\) is not equal to 0 and \(x^y=1\), then which of the following must be true?
I. \(x=1\)

II. \(x=1\) and \(y=0\)

III. \(x=1\) or \(y=0\)


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


Notice that if \(x=-1\) and \(y\) is any even number, then \((-1)^{even}=1\), thus none of the options must be true.


Answer: E








In the question, its already mentioned that x=1, so why are we taking x=-1 as other possibility ?


The stem says that \(x\) is not equal to 0. Where does it say that x = 1?
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Re: M28-37  [#permalink]

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New post 28 Nov 2018, 09:38
Bunuel, why we are considering x=-1 when its nowhere mentioned in options.
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Re: M28-37  [#permalink]

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New post 28 Nov 2018, 17:25
dine5207 wrote:
Bunuel, why we are considering x=-1 when its nowhere mentioned in options.


I'm not Bunuel but going to answer this one.

The question asks "which of the following MUST be true?"
In order to show that the answer is "E. None", we must present a solution in which we can have \(x,y\) such that \(x^y = 1\),and \(x\neq 0\) and all of I, II, and III are all not true.

The example Bunuel gives here is \(x=-1\) and \(y={\text{any even number}}\). Suppose \(y = 2\). Then \(x^y = (-1)^2 = 1\) and yet all of I, II, and III are all not true.
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Re: M28-37  [#permalink]

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New post 28 Nov 2018, 17:27
Bunuel wrote:
If \(x\) is not equal to 0 and \(x^y=1\), then which of the following must be true?
I. \(x=1\)

II. \(x=1\) and \(y=0\)

III. \(x=1\) or \(y=0\)


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


Could someone confirm that the set specified by III also includes where \(x=1 \text{ AND } y = 0\)?
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Re: M28-37 &nbs [#permalink] 28 Nov 2018, 17:27
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