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# Is m^n a perfect square?

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Is m^n a perfect square?  [#permalink]

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Updated on: 01 Dec 2018, 06:24
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Difficulty:

75% (hard)

Question Stats:

23% (01:13) correct 77% (01:10) wrong based on 86 sessions

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Is $$m^n$$ a perfect square?

(1) m is a perfect square

(2) n is a perfect square

Originally posted by barryseal on 29 Nov 2018, 05:01.
Last edited by barryseal on 01 Dec 2018, 06:24, edited 2 times in total.
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Re: Is m^n a perfect square?  [#permalink]

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Updated on: 30 Nov 2018, 06:24
barryseal wrote:
Is $$m^n$$ a perfect square?

(1) m is a perfect square

(2) n is a perfect square

$${m^n}\,\,\mathop = \limits^? \,\,{K^2}\,\,\,,\,\,\,K\,\,{\mathop{\rm int}}$$

$$\left( 1 \right)\,\,m = {J^2},\,\,J\mathop \ge \limits^{{\rm{WLOG}}} 0\,\,\,{\mathop{\rm int}} \,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {0,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {4,0.5} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$

$$\left( 2 \right)\,\,n = {W^2},\,\,W\mathop \ge \limits^{{\rm{WLOG}}} 0\,\,\,\,{\mathop{\rm int}} \,\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {0,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$

$$\left( {1 + 2} \right)\,\,{m^n} = {\left( {{J^2}} \right)^{{W^2}}} = {\left( {{J^{{W^2}}}} \right)^2}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\left( {{J^{{W^2}}} = K} \right)\,\,\,$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

POST-MORTEM (important to the most careful students): $$0^0$$ is an undefined expression in Math (and also in the GMAT, of course),
hence the case $$m=0$$ is implicitly associated with $$n \neq 0$$. (The question stem implicitly implies that $$m^n$$ exists!)
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Originally posted by fskilnik on 29 Nov 2018, 09:03.
Last edited by fskilnik on 30 Nov 2018, 06:24, edited 1 time in total.
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Re: Is m^n a perfect square?  [#permalink]

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29 Nov 2018, 17:38
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barryseal wrote:
Is $$m^n$$ a perfect square?

(1) m is a perfect square

(2) n is a perfect square

Since there's no information to suggest that m and n are integers, the correct answer is not A.

Statement 1: m is a perfect square
Consider these conflicting cases that satisfy statement 1:

CASE A) m = 9 and n = 2. In this case, m^n = 9^2 = 81. So, the answer to the target question is "YES m^n IS a perfect square."
CASE B) m = 9 and n = 0.5. In this case, m^n = 9^0.5 = 3. So, the answer to the target question is "NO m^n is NOT a perfect square."

Cheers,
Brent
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Re: Is m^n a perfect square?  [#permalink]

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30 Nov 2018, 05:36
GMATPrepNow wrote:
barryseal wrote:
Is $$m^n$$ a perfect square?

(1) m is a perfect square

(2) n is a perfect square

Since there's no information to suggest that m and n are integers, the correct answer is not A.

Statement 1: m is a perfect square
Consider these conflicting cases that satisfy statement 1:

CASE A) m = 9 and n = 2. In this case, m^n = 9^2 = 81. So, the answer to the target question is "YES m^n IS a perfect square."
CASE B) m = 9 and n = 0.5. In this case, m^n = 9^0.5 = 3. So, the answer to the target question is "NO m^n is NOT a perfect square."

Cheers,
Brent

Hi Brent,

The correct answer is C here. If we know both m and n are perfect squares, then m^n is not a perfect square.
m = 25
n = 4
m^n is not a perfect square.

Is this correct?

Posted from my mobile device
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Re: Is m^n a perfect square?  [#permalink]

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30 Nov 2018, 06:25
Hi, Brent!

You are right. I have corrected my solution accordingly.

Regards,
Fabio.
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Re: Is m^n a perfect square?  [#permalink]

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Updated on: 30 Nov 2018, 10:13
Top Contributor
dollytaneja51 wrote:
Hi Brent,

The correct answer is C here. If we know both m and n are perfect squares, then m^n is not a perfect square.
m = 25
n = 4
m^n is not a perfect square.

Is this correct?

Posted from my mobile device

EDIT: I first read the above as m^n IS a perfect square.
See Fabio's response below.

Cheers,
Brent
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Originally posted by GMATPrepNow on 30 Nov 2018, 09:45.
Last edited by GMATPrepNow on 30 Nov 2018, 10:13, edited 2 times in total.
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Re: Is m^n a perfect square?  [#permalink]

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Updated on: 30 Nov 2018, 18:00
dollytaneja51 wrote:

If we know both m and n are perfect squares, then m^n is not a perfect square.
m = 25
n = 4
m^n is not a perfect square.

Is this correct?

This is NOT correct.

First of all: when both m and n are perfect squares, we are sure both are nonnegative integers.
(Each one is the square of an integer, therefore a nonnegative integer.)

From that, when a perfect square (m) is put to the power of a nonnegative ineteger (n),
we may conclude that $$m^n$$ is also a perfect square:

$${m^n} = {\left( {{J^2}} \right)^n} = {\left( {{J^n}} \right)^2} = {\text{in}}{{\text{t}}^2}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {J,n \geqslant 0\,\,{\text{ints}}\,,\,{\text{not}}\,\,{\text{both}}\,\,{\text{zero}}} \right)$$

In the example provided, please note that:

$${25^4} = {\left( {{{25}^2}} \right)^2}\,\,\underline {{\rm{is}}} \,\,{\rm{a}}\,\,{\rm{perfect}}\,\,{\rm{square}}$$

If you prefer (equivalently),

$$\sqrt {{{25}^4}} = {25^2} = {\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,\,\,{25^4}\,\,\underline {{\rm{is}}} \,\,{\rm{a}}\,\,{\rm{perfect}}\,\,{\rm{square}}$$

Regards,
Fabio.
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Originally posted by fskilnik on 30 Nov 2018, 10:09.
Last edited by fskilnik on 30 Nov 2018, 18:00, edited 2 times in total.
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Re: Is m^n a perfect square?  [#permalink]

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30 Nov 2018, 11:00
Dear Moderator,

Looks like OA is incorrect for this question.

Posted from my mobile device
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Re: Is m^n a perfect square?  [#permalink]

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30 Nov 2018, 11:08
fskilnik wrote:
dollytaneja51 wrote:

If we know both m and n are perfect squares, then m^n is not a perfect square.
m = 25
n = 4
m^n is not a perfect square.

Is this correct?

This is NOT correct.

First of all: when both m and n are perfect squares, we are sure both are integers.
(Each one is the square of an integer, therefore an integer.)

From that, when a perfect square (m) is to the power of an INTEGER (n),
we may conclude that $$m^n$$ is also a perfect square:

$${m^n} = {\left( {{J^2}} \right)^n} = {\left( {{J^n}} \right)^2} = {{\mathop{\rm int}} ^2}\,\,\,\,\,\left( {J \ge 0\,\,{\mathop{\rm int}} ,\,\,J\,\,{\rm{and}}\,\,{\rm{n}}\,\,{\rm{not}}\,\,{\rm{both}}\,\,{\rm{zero}}} \right)$$

In the example provided, please note that:

$${25^4} = {\left( {{{25}^2}} \right)^2}\,\,\underline {{\rm{is}}} \,\,{\rm{a}}\,\,{\rm{perfect}}\,\,{\rm{square}}$$

If you prefer (equivalently),

$$\sqrt {{{25}^4}} = {25^2} = {\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,\,\,{25^4}\,\,\underline {{\rm{is}}} \,\,{\rm{a}}\,\,{\rm{perfect}}\,\,{\rm{square}}$$

Regards,
Fabio.

Thanks Fabio for the response. It will be a perfect square.

I dont know why I was thinking perfect square will have only three factors.

Posted from my mobile device
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Re: Is m^n a perfect square?  [#permalink]

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30 Nov 2018, 11:55
dollytaneja51 wrote:

Thanks Fabio for the response. It will be a perfect square.

I dont know why I was thinking perfect square will have only three factors.

Hi dollytaneja51 !

We all get things wrong sometimes. The best proof is the posts above (mine included, of course)!

Regards,
Fabio.
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Re: Is m^n a perfect square?  [#permalink]

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30 Nov 2018, 18:22
What is the correct answer of this question? Is it C or E?
In this cae what happened if 0^0?, because 0 is a perfect square.

Thank you.
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Re: Is m^n a perfect square?  [#permalink]

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30 Nov 2018, 19:07
jorgetomas9 wrote:
What is the correct answer of this question? Is it C or E?
In this cae what happened if 0^0?, because 0 is a perfect square.

Thank you.

Hi jorgetomas9 !

I explained in my first post that the case 0^0 must be implicitly considered excluded.

Reason: we were asked whether m^n is a perfect square, hence we may (and should) assume that m^n exists (=is defined).

Regards,
Fabio.
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Re: Is m^n a perfect square?  [#permalink]

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30 Nov 2018, 19:10
fskilnik wrote:
jorgetomas9 wrote:
What is the correct answer of this question? Is it C or E?
In this cae what happened if 0^0?, because 0 is a perfect square.

Thank you.

Hi jorgetomas9 !

I explained in my first post that the case 0^0 must be implicitly considered excluded.

Reason: we were asked whether m^n is a perfect square, hence we may (and should) assume that m^n exists (=is defined).

Regards,
Fabio.

Thank you, very much I marked C too, only I had a doubt related with 0^0
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Re: Is m^n a perfect square?  [#permalink]

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30 Nov 2018, 19:23
jorgetomas9 wrote:
Thank you, very much I marked C too, only I had a doubt related with 0^0

I am glad things are clear now.

See you in other posts and success in your studies!
Fabio.
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Re: Is m^n a perfect square?  [#permalink]

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02 Dec 2018, 23:36
Need help. How can C be the right answer?

Say m = 9 (perfect square)
n = 9 (perfect square)

m^n = 9^9 = 9*9*..... 9 times.. ths is not a perfect square.
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Re: Is m^n a perfect square?  [#permalink]

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03 Dec 2018, 00:11
agpamit wrote:
Need help. How can C be the right answer?

Say m = 9 (perfect square)
n = 9 (perfect square)

m^n = 9^9 = 9*9*..... 9 times.. ths is not a perfect square.

9 is not a prime factor -> 9^9 = 3^18 = perfect square

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Re: Is m^n a perfect square?   [#permalink] 03 Dec 2018, 00:11
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