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If mn≠0, is m^n > n^n ?

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If mn≠0, is m^n > n^n ?  [#permalink]

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New post 09 Aug 2014, 03:08
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If mn≠0, is \(m^n > n^n\)?

(1) \(|m| = n\)
(2) \(m < n\)
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Re: If mn≠0, is m^n > n^n ?  [#permalink]

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New post 09 Aug 2014, 05:28
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Cannot confirm:)

First of all here, you should be careful with even power.

(1) Here we have two possibilities.
If \(m\geq 0\), then \(m=n\) and \(n^n=n^n\). The answer is No.
If \(m<0\), then \(m=-n\) and we have two cases again: if n-even, then \((-n)^n=n^n\) (the answer is again No); if n-odd, then \((-n)^n<0\) and \((-n)^n<n^n\) (the answer is again No). Sufficient.

(2) If m=-3 and n=2, then \((-3)^2>2^2\), but if m=-1, n=2, then \((-1)^2<2^2\). Insufficient.

The correct answer is A.
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Re: If mn≠0, is m^n > n^n ?  [#permalink]

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New post 09 Aug 2014, 06:09
Good Day :-D

Since you've 2 possible cases for Statement 1 >= 0 and <0 and your conclusion suffices only on basis of 2nd case, how can Statement 1 alone be sufficient?

can you please elaborate!

Thanks,
Rajeev.


smyarga wrote:
Cannot confirm:)

First of all here, you should be careful with even power.

(1) Here we have two possibilities.
If \(m\geq 0\), then \(m=n\) and \(n^n=n^n\). The answer is No.
If \(m<0\), then \(m=-n\) and we have two cases again: if n-even, then \((-n)^n=n^n\) (the answer is again No); if n-odd, then \((-n)^n<0\) and \((-n)^n<n^n\) (the answer is again No). Sufficient.

(2) If m=-3 and n=2, then \((-3)^2>2^2\), but if m=-1, n=2, then \((-1)^2<2^2\). Insufficient.

The correct answer is A.
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Re: If mn≠0, is m^n > n^n ?  [#permalink]

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New post 09 Aug 2014, 14:54
raj44 wrote:
Good Day :-D

Since you've 2 possible cases for Statement 1 >= 0 and <0 and your conclusion suffices only on basis of 2nd case, how can Statement 1 alone be sufficient?

can you please elaborate!

Thanks,
Rajeev.


smyarga wrote:
Cannot confirm:)

First of all here, you should be careful with even power.

(1) Here we have two possibilities.
If \(m\geq 0\), then \(m=n\) and \(n^n=n^n\). The answer is No.
If \(m<0\), then \(m=-n\) and we have two cases again: if n-even, then \((-n)^n=n^n\) (the answer is again No); if n-odd, then \((-n)^n<0\) and \((-n)^n<n^n\) (the answer is again No). Sufficient.

(2) If m=-3 and n=2, then \((-3)^2>2^2\), but if m=-1, n=2, then \((-1)^2<2^2\). Insufficient.

The correct answer is A.


The question here asked is \(m^n > n^n\). In all two cases ( when m>=0 and m<0), the answer is No, meaning m^n cannot be larger than n^n. So information in (1) is sufficient to answer the question.
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Re: If mn≠0, is m^n > n^n ?  [#permalink]

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New post 02 Nov 2014, 06:40
I think the answer is E,the reasoning is as follows:

is m^n > n^n?

from statement 1, when |m| = n, we have either m = n or m = -n, when m = n, then sufficient, but when m = -n, then is? (-n)^n>n^n, when n =1, then we have -1^1 is not greater then 1^1, and when n = -2, we have (-n)^n = n^n. therefore insufficient to answer the question.

from statement 2 itself we cannot tell, as we get different answers, by substituting values. so, insufficient

combining, we have only case m = -n as, m<n, whch ends up with the same argument as described in statement 1, case 2. so, insufficient.

I think its E
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Re: If mn≠0, is m^n > n^n ?  [#permalink]

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New post 02 Nov 2014, 06:47
santorasantu wrote:
I think the answer is E,the reasoning is as follows:

is m^n > n^n?

from statement 1, when |m| = n, we have either m = n or m = -n, when m = n, then sufficient, but when m = -n, then is? (-n)^n>n^n, when n =1, then we have -1^1 is not greater then 1^1, and when n = -2, we have (-n)^n = n^n. therefore insufficient to answer the question.

from statement 2 itself we cannot tell, as we get different answers, by substituting values. so, insufficient

combining, we have only case m = -n as, m<n, whch ends up with the same argument as described in statement 1, case 2. so, insufficient.

I think its E


Let me ask you a question: don't you have a NO answer in statement (1) for both of your examples?
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Re: If mn≠0, is m^n > n^n ?  [#permalink]

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New post 02 Nov 2014, 13:30
Bunuel wrote:
santorasantu wrote:
I think the answer is E,the reasoning is as follows:

is m^n > n^n?

from statement 1, when |m| = n, we have either m = n or m = -n, when m = n, then sufficient, but when m = -n, then is? (-n)^n>n^n, when n =1, then we have -1^1 is not greater then 1^1, and when n = -2, we have (-n)^n = n^n. therefore insufficient to answer the question.

from statement 2 itself we cannot tell, as we get different answers, by substituting values. so, insufficient

combining, we have only case m = -n as, m<n, whch ends up with the same argument as described in statement 1, case 2. so, insufficient.

I think its E


Let me ask you a question: don't you have a NO answer in statement (1) for both of your examples?



aah sorry, true, A is the answer. Thanks for the tip.
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Re: If mn≠0, is m^n > n^n ?  [#permalink]

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Re: If mn≠0, is m^n > n^n ?  [#permalink]

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If m and n are non-zero integers, is m^n > n^n?  [#permalink]

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New post Updated on: 10 Sep 2015, 21:06
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If m and n are non-zero integers, is m^n > n^n?

(1) |m| = n
(2) m < n

Originally posted by Hemavakade on 10 Sep 2015, 12:47.
Last edited by Bunuel on 10 Sep 2015, 21:06, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If m and n are non-zero integers, is m^n > n^n?  [#permalink]

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New post 10 Sep 2015, 13:57
Stmt 1. |m|=n if, m=-2,n will be 2 or m=2.n will be 2. The answer is a definite no! A
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Re: If m and n are non-zero integers, is m^n > n^n?  [#permalink]

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New post 10 Sep 2015, 14:16
Went with E at first, but changed my answer to A after working it out. Good question, +1.

If m and n are non-zero integers, is m^n>n^n?

Statement 1: |m| = n

Tells us the absolute value of m equals n, so if we pick numbers we can come up with multiple solutions. For example, if m = -2 and n = 2. In this case, -2^2 = 4 and 2^2 = 4 so the answer would be no.

But, if m = -3 and n = 3. -3^3 = -27 and 3^3 = 27. In this case the answer is still no.

There isn't any more possible ways to get different values for m and n because of question stem tells us they are integers and no-zero. So there for m^n will always equal or be less than n^n. Therefore statement 1 is sufficient.

Statement 2: m < n

Picking numbers will work for this again.

Example 1. m = -2 and n =2

-2^2=2^2

Example 2. m = -3 and n =2

-3^2>2^2

Since you can have two solutions that give you two different outcomes, statement 2 is insufficient.

Ans A
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If m and n are non-zero integers, is m^n > n^n?  [#permalink]

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Re: If m and n are non-zero integers, is m^n > n^n?  [#permalink]

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New post 29 Jan 2016, 22:04
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Bunuel wrote:
If m and n are non-zero integers, is m^n > n^n?

(1) |m| = n
(2) m < n


Hi,
Again what can we make out of the Q stem..
m and n are non-zero integers, is m^n > n^n..
Just "there are various possiblities of value of m and n..." so lets move on to the statements

(1) |m| = n
in the given condition that m and n are non-zero integers..
m^n > n^n will never be true.. so its suff


But lets see why?
we know that n>0..
but we do not know anything about m, just that it is equal to n in numeric value..
so lets see the possiblities..

both m and n are positive..
is m^n > n^n?... NO, as both will be equal..

m is negative and n is positive...
we will have two scenarios here too when n is even integer and when n is positive integer..
when n is even, m, a -ive number to even power will make m^n = n^n..
when n is odd, LHS,m^n, will be negative, so m^n < n^n..

In each case our answer is NO..
so Sufficient A

(2) m < n
if m is -4 and n is 2.. ans will be yes..
if m is -1 and n is 1.. ans will be No..
different answers..
insuff..

ans A
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Re: If m and n are non-zero integers, is m^n > n^n?  [#permalink]

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New post 31 Jan 2016, 02:22
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St1: |m| = n --> n is positive
If m = -3, n = 3 --> m^n = -27; n^n = 27... m^n > n^n? No
If m = 3, n = 3 --> m^n = 27; n^n = 27... m^n > n^n? No
If m = -2, n = 2 --> m^n = 4; n^n = 4... m^n > n^n? No
Thus m^n is not greater than n^n. St1 is sufficient

St2: m < n
Let m = 1; n = 2 --> m^n = 1; n^n = 4... m^n > n^n? No
Let m = -2, n = -1 --> m^n = -(1/2); n^n = -1.. m^n > n^n? Yes
St2 is not sufficient.

Answer: A
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Re: If mn≠0, is m^n > n^n ?  [#permalink]

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New post 03 Mar 2016, 03:59
Hi Bunuel,

is it correct to solve like usual the abs value function for both the cases and then put alltogether in a system?
In this case, I have that everything is above the root should be >= 0, |x^2-2|>=0. Then I solve the two cases:

1. x^2-2 for x^2>=2 I obtain: x<=-1 x>=2 (only x>=2 is acceptable
2.- x^2+2 for x^2<=2 I obtain: x<=-1 -2<x<=1 acceptable range

Now if I pu all togheter to have a solution that is >=0: x<-2 and 1<x<2 so I take only the second range?
Correct me on my reasoning .

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Re: If mn≠0, is m^n > n^n ?  [#permalink]

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New post 03 Mar 2016, 21:51
mahendru1992 wrote:
If mn≠0, is \(m^n > n^n\)?

(1) \(|m| = n\)
(2) \(m < n\)


1st statement:
mod(m) = n.
i.e. n>0.
now if m<0,-m=n.
m>0, m=n.
In both cases, m^n is either equal or less than n^n (equal in case of even powers and less than in case of odd powers)
Sufficient!

2nd statement:
m<n
no way to break down this statement and be able to use it.
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Re: If mn≠0, is m^n > n^n ?  [#permalink]

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New post 08 Mar 2016, 20:48
mahendru1992 wrote:
If mn≠0, is \(m^n > n^n\)?

(1) \(|m| = n\)
(2) \(m < n\)


Took me 5 minute to solve this one...
But in the correct answer made the effort worth it..
Here is my approach
In statement 1 => m=n and m=-n and n>0
both the cases give a Complete NO
as m=n => equal quantities
and m=-n => (-n)^n >n^n => they can be best equal for even powers and hence NO again
The third case we must check for proper fractions (THEY BEHAVE FUNNY WHEN SQUARED)
(-1/2)^1/2 >(1/2)^1/2 => LHS is some unreal number as negative is in the root making it Iota hence RHS is again NO is the answer again
SO A is sufficient

As for statement 2 we can plug in -4 & 3 and -4 & 10 to get YES and NO making it insufficient.
hence A
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