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Math Expert V
Joined: 02 Sep 2009
Posts: 55266
M28-37  [#permalink]

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9 00:00

Difficulty:   75% (hard)

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

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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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Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re M28-37  [#permalink]

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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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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M28-37  [#permalink]

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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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x=1, y=1 discards 3d option
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Re: M28-37  [#permalink]

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Hi Bunuel, I didn't understand why B is incorrect. When x=1 and y=0, then the answer is always 1. Please clarify.
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M28-37  [#permalink]

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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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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M28-37  [#permalink]

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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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Hi,
Can someone explain why A isn't correct. When X=1, then 1 to the power of any except 0 is 1 right?
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M28-37  [#permalink]

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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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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 ?
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M28-37  [#permalink]

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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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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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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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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 $$x=1 \text{ AND } y = 0$$?
Intern  Joined: 17 Oct 2018
Posts: 1
Re M28-37  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 55266
Re: M28-37  [#permalink]

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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.
_________________ Re: M28-37   [#permalink] 18 Dec 2018, 22:11
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