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M13-01

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M13-01 [#permalink]

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New post 16 Sep 2014, 00:48
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New post 16 Sep 2014, 00:48
Official Solution:


(1) \(m^n=1\). If \(m=1\) then \(n\) can be any positive number. Not sufficient.

(2) \(n^m=1\). If \(n=1\) then \(m\) can be any positive number. Not sufficient.

(1)+(2) From above we have that \(m=1\) and \(n=1\), hence \(mn=1\). Sufficient.


Answer: C
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Re: M13-01 [#permalink]

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What about n=m=0?

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New post 14 Mar 2015, 20:12
samcialek wrote:
What about n=m=0?


hi
since it is given m and n are positive, they cannot be 0..
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Re: M13-01 [#permalink]

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New post 06 Dec 2015, 12:49
Bunuel wrote:
Official Solution:


(1) \(m^n=1\). If \(m=1\) then \(n\) can be any positive number. Not sufficient.

(2) \(n^m=1\). If \(n=1\) then \(m\) can be any positive number. Not sufficient.

(1)+(2) From above we have that \(m=1\) and \(n=1\), hence \(mn=1\). Sufficient.


Answer: C


But the Stem doesn't mention m and n to be positive integers.The stem says they are positive numbers > 0

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Re: M13-01 [#permalink]

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New post 15 Jan 2016, 18:51
Question stem says: m & n are positive numbers. Note that 0 isn't a positive number, hence zero is ruled out

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New post 10 Jun 2016, 04:51
HI ... why have we ruled out the fact that a number raised to 0 is also equal to 1 and that 0 raised to 0 is also equal to one. m and n could both be equal to 0

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New post 10 Jun 2016, 06:17
sheetaldodani wrote:
HI ... why have we ruled out the fact that a number raised to 0 is also equal to 1 and that 0 raised to 0 is also equal to one. m and n could both be equal to 0


We are told that m and n are positive numbers, 0 is neither positive nor negative.
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M13-01 [#permalink]

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New post 06 Aug 2016, 08:23
I don't agree with the explanation. both the numbers m & n can be 0 . so it should be E.

* OOPS! My bad! Ignored vital point in question statement.

Last edited by Lastlap2016 on 07 Aug 2016, 05:09, edited 1 time in total.

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New post 06 Aug 2016, 11:08
Lastlap2016 wrote:
I think this is a poor-quality question and I don't agree with the explanation. both the numbers m & n can be 0 . so it should be E.


The question is correct. You should read questions more carefully: we are told that m and n are positive numbers, 0 is neither positive nor negative.
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New post 10 Sep 2016, 09:42
Hey - can m or n be decimals which happen to equals to both m and n being raised to the power of each other =1 ?

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New post 19 Sep 2016, 08:33
marshnaz wrote:
Hey - can m or n be decimals which happen to equals to both m and n being raised to the power of each other =1 ?



If m and n are positive numbers, what is the value of m∗nm∗n?

In GMAT positive number means integer greater than 0.
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New post 19 Sep 2016, 15:07
0ld wrote:
marshnaz wrote:
Hey - can m or n be decimals which happen to equals to both m and n being raised to the power of each other =1 ?



If m and n are positive numbers, what is the value of m∗nm∗n?

In GMAT positive number means integer greater than 0.


Is this true? If we see x is a positive number, we know that x is a positive integer?

Thanks!

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New post 20 Sep 2016, 04:25
ddb123 wrote:
0ld wrote:
marshnaz wrote:
Hey - can m or n be decimals which happen to equals to both m and n being raised to the power of each other =1 ?



If m and n are positive numbers, what is the value of m∗nm∗n?

In GMAT positive number means integer greater than 0.


Is this true? If we see x is a positive number, we know that x is a positive integer?

Thanks!


No, that's NOT true. Positive numbers are numbers greater than 0, not necessarily integers.
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M13-01 [#permalink]

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New post 14 Sep 2017, 06:04
Bunuel wrote:
If \(m\) and \(n\) are positive numbers, what is the value of \(m*n\)?


(1) \(m^n = 1\)

(2) \(n^m = 1\)



Hi Bunuel,

When we combine both statements, we get
m^n = n^m

2^4= 4^2 also satisfies this condition. Hence mn can be 8 as well as 1. So the answer should be E.

Is this incorrect?

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New post 14 Sep 2017, 06:07
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Yashodhan123 wrote:
Bunuel wrote:
If \(m\) and \(n\) are positive numbers, what is the value of \(m*n\)?


(1) \(m^n = 1\)

(2) \(n^m = 1\)



Hi Bunuel,

When we combine both statements, we get
m^n = n^m

2^4= 4^2 also satisfies this condition. Hence mn can be 8 as well as. So the answer should be E.

Is this incorrect?


No, because we now that \(m^n = 1\) and \(n^m = 1\), not 16...
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Collection of Questions:
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Re: M13-01   [#permalink] 14 Sep 2017, 06:07
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