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M28-37

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M28-37  [#permalink]

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New post 16 Sep 2014, 01:31
1
9
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

30% (00:35) correct 70% (00:41) wrong based on 244 sessions

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Re M28-37  [#permalink]

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New post 16 Sep 2014, 01:31
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Re: M28-37  [#permalink]

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New post 16 Jul 2015, 04:25
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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New post 16 Jul 2015, 04:38
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 12 Feb 2016, 00:22
x=1, y=1 discards 3d option
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Re: M28-37  [#permalink]

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New post 18 Jun 2017, 03:33
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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New post 18 Jun 2017, 03:37
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Re: M28-37  [#permalink]

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New post 05 Oct 2017, 07:58
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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New post 05 Oct 2017, 08:00
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Re: M28-37  [#permalink]

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New post 28 Nov 2018, 06:34
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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New post 28 Nov 2018, 06:40
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Re: M28-37  [#permalink]

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New post 28 Nov 2018, 08:20
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, 08:22
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, 10:38
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, 18:25
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, 18:27
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  [#permalink]

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New post 18 Dec 2018, 16:30
I think this is a poor-quality question and I don't agree with the explanation. the statements that are given clearly say that x=1 and not x<1 or x>1. I don't understand why are we considering x=-1 when each statement clearly gives a value to x and/or y and not a range of values
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New post 18 Dec 2018, 22:11
FrankMartin wrote:
I think this is a poor-quality question and I don't agree with the explanation. the statements that are given clearly say that x=1 and not x<1 or x>1. I don't understand why are we considering x=-1 when each statement clearly gives a value to x and/or y and not a range of values


I think you did not understand the question.

The stem says: \(x\) is not equal to 0 and \(x^y=1\).

Then we are given three options, and are asked to determine which of them MUST be true.

The solution shows that none of the options MUST be true (none of the options is NECESSARILY true).

The questions is 100% correct.
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Re: M28-37   [#permalink] 18 Dec 2018, 22:11
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