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# If z^n = 1, what is the value of z?

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If z^n = 1, what is the value of z? [#permalink]  29 Jun 2012, 04:50
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If z^n = 1, what is the value of z?

(1) n is a nonzero integer.
(2) z > 0

Zero raised to any power is zero and any number raised to the power of 0 equals one?
Is that the rule of it is reversed?
[Reveal] Spoiler: OA

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Re: If z^n = 1, what is the value of z? [#permalink]  29 Jun 2012, 04:59
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Stiv wrote:
If z^n = 1, what is the value of z?

(1) n is a nonzero integer.
(2) z > 0

Zero raised to any power is zero and any number raised to the power of 0 equals one?
Is that the rule of it is reversed?

I'm not sure understand the red part in your post above.

If z^n = 1, what is the value of z?

(1) n is a non zero integer --> $$1^{any \ integer}=1$$ and also $$(-1)^{even}=1$$, so $$z$$ can be 1 or -1. Not sufficient.

(2) z > 0 --> any nonzero number to the power of 0 is 1, so if $$n=0$$ then $$z$$ can be any non-zero number (any positive number in our case as given that $$z>0$$). Not sufficient.

(1)+(2) $$n$$ is a nonzero integer and $$z>0$$ implies that $$z$$ can equal to 1 only. Sufficient.

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PS questions on exponents: search.php?search_id=tag&tag_id=60

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Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html

Hope it helps.
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Re: If z^n = 1, what is the value of z? [#permalink]  04 Feb 2013, 19:50
Statement 1 is insufficient here are some examples

For z^n = 1, and n being a non zero integer, there are 3 possible ways.
a. 1^1 = 1
b. 1^- 1 = 1
c. - 1^2 = 1

Statement 1 not conclusive. Z could be 1 or -1.

Statement 2: z > 0.

Examples 2^0 = 1 , 3^0 = 1

not sufficient

together n can't be 0 and Z has to be greater than 1 only one case

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Re: If z^n = 1, what is the value of z? [#permalink]  07 Jan 2015, 16:13
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Re: If z^n = 1, what is the value of z? [#permalink]  08 Jan 2015, 04:48
Expert's post
Stiv wrote:
If z^n = 1, what is the value of z?

(1) n is a nonzero integer.
(2) z > 0

Zero raised to any power is zero and any number raised to the power of 0 equals one?
Is that the rule of it is reversed?

This is not a difficult question by itself, but clarity of approach matters a lot in getting it right. The low accuracy for this question suggests that most students could not think through this question clearly.

At eGMAT, we strongly advocate that in DS Questions, the student should first analyze the question statement thoroughly and only then move on to analyzing the two statements. You'll see how elegantly this question will simplify with this approach.

We are given that z^n = 1. So, what cases are possible for the value of z and n?

Case 1: z = 1; n has any integral value
Case 2: z = -1; n is an even integer
Case 3: z has any non-zero value; n = 0

Please note that only after this analysis are we going to the first Statement.

As per the first statement,
n is a non-zero integer
This rules out Case 3.
However, this still leaves out Case 1 and 2. So, z can either be equal to 1 or z can be equal to -1. So, Statement 1 alone is not sufficient.

As per the second statement,
z > 0
This rules out Case 2. However, Case 1 and 3 still remain. Again, we have not been able to determine a unique value of z. So, Statement 2 alone is not sufficient either.

Combining both the Statements,
From Statement 1, z could either be 1 or -1
From Statement 2, z > 0
Therefore, only possible value of z is 1.

Thus, by combining both the statements together, we have been able to determine a unique value of z. So, the correct answer is Choice C.

Takeaway: The correct answer is only a byproduct of a clear approach.

Hope this helps.

Japinder
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If z^n = 1, what is the value of z? [#permalink]  21 Mar 2015, 21:11
Bunuel wrote:
(1) n is a non zero integer --> $$1^{any \ integer}=1$$ and also $$(-1)^{even}=1$$, so $$z$$ can be 1 or -1. Not sufficient.

(2) z > 0 --> any nonzero number to the power of 0 is 1, so if $$n=0$$ then $$z$$ can be any non-zero number (any positive number in our case as given that $$z>0$$). Not sufficient.

(1)+(2) $$n$$ is a nonzero integer and $$z>0$$ implies that $$z$$ can equal to 1 only. Sufficient.

Can we take $$(\sqrt{1})^2$$ as a plug in value to check these statements..
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Re: If z^n = 1, what is the value of z? [#permalink]  22 Mar 2015, 05:13
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suhasancd wrote:
Bunuel wrote:
(1) n is a non zero integer --> $$1^{any \ integer}=1$$ and also $$(-1)^{even}=1$$, so $$z$$ can be 1 or -1. Not sufficient.

(2) z > 0 --> any nonzero number to the power of 0 is 1, so if $$n=0$$ then $$z$$ can be any non-zero number (any positive number in our case as given that $$z>0$$). Not sufficient.

(1)+(2) $$n$$ is a nonzero integer and $$z>0$$ implies that $$z$$ can equal to 1 only. Sufficient.

Can we take $$(\sqrt{1})^2$$ as a plug in value to check these statements..

$$\sqrt{1}=1$$, so are you asking whether we can plug 1 for n? Well, yes we can...
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Re: If z^n = 1, what is the value of z? [#permalink]  22 Mar 2015, 05:23
Bunuel wrote:

$$\sqrt{1}=1$$, so are you asking whether we can plug 1 for n? Well, yes we can...

Oh.. ya.. I forgot that the square root of a perfect number is always +ve.

$$\sqrt{36}$$ = 6 (not -6)

So $$\sqrt{1}$$ = 1 (not -1)

I had +/- 1 in my head while combining statements I and II together.

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Re: If z^n = 1, what is the value of z? [#permalink]  22 Mar 2015, 05:24
Expert's post
suhasancd wrote:
Bunuel wrote:

$$\sqrt{1}=1$$, so are you asking whether we can plug 1 for n? Well, yes we can...

Oh.. ya.. I forgot that the square root of a perfect number is always +ve.

$$\sqrt{36}$$ = 6 (not -6)

So $$\sqrt{1}$$ = 1 (not -1)

I had +/- 1 in my head while combining statements I and II together.

Any even root from any postie number is positive.
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Re: If z^n = 1, what is the value of z?   [#permalink] 22 Mar 2015, 05:24
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