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If m and n are positive integers, is m^n < n^m?

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If m and n are positive integers, is m^n < n^m? [#permalink] New post 22 Sep 2010, 12:37
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If m and n are positive integers, is m^n < n^m?

(1) m = \sqrt{n}
(2) n > 5

[Reveal] Spoiler:
i disagree with the OA.
My reasoning below
1) if m = 1 & n = 1 then m=sqrt(n) and m^n=n^m
if m=2 & n=4 then m=sqrt(n) and m^n=n^m
if m=3 & n=9 then m=sqrt(n) and m^n>n^m
So if 1) then m^n<n^m always false ==> sufficient

2) we do not know anything about m so insufficient

I assume answer is A. Do you agree ?
[Reveal] Spoiler: OA

Last edited by Bunuel on 15 Jul 2013, 22:20, edited 2 times in total.
Edited the OA.
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Re: Exponents-roots DS [#permalink] New post 22 Sep 2010, 13:08
tatane90 wrote:
If m and n are positive integers, is m^n < n^m?
(1) m = sqrt(n)
(2) n > 5

i disagree with the OA.
My reasoning below
1) if m = 1 & n = 1 then m=sqrt(n) and m^n=n^m
if m=2 & n=4 then m=sqrt(n) and m^n=n^m
if m=3 & n=9 then m=sqrt(n) and m^n>n^m
So if 1) then m^n<n^m always false ==> sufficient

2) we do not know anything about m so insufficient

I assume answer is A. Do you agree ?


(1) : m^n < n^m
m^{m^2} < n^m
(m^2)^{m^2/2} < n^m
n^{m^2/2} < n^m

For n,m integers and both greater than 1, this implies

m^2/2 < m
m(m-2) < 0

This expression is false for all m>2

Also we know for m=1 and m=2 that m^n=n^m ... so again the expression is false

So (1) is sufficient

(2) : Not sufficient as only condition on n


So I agree answer is A

The answer would be (c) if the original question was m^n <= n^m
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Re: Exponents-roots DS [#permalink] New post 22 Sep 2010, 13:35
If m and n are positive integers, is m^n < n^m?
(1) m = sqrt(n)
(2) n > 5


from 1

m = sqrt n ie: n = m^2

is m^(m^2)< m^2m , is m^m(m-2) < 1 is m(m-2)<0 is m>2.... insuff




from 2

obviously insuff

both suff

n>5, n = m^2 ( try worst case scenario n = 9 ) thus m = 3 >2...suff

Last edited by yezz on 22 Sep 2010, 13:43, edited 1 time in total.
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Re: Exponents-roots DS [#permalink] New post 22 Sep 2010, 13:43
yezz wrote:
If m and n are positive integers, is m^n < n^m?
(1) m = sqrt(n)
(2) n > 5


from 1

m = sqrt n ie: n = m^2

is m^(m^2)< (m^2)^m

plug m = 1 or 2 (no), plug m = 7 (yes) insuff

from 2

obviously insuff

both suff

C


With m=7, LHS is 7^49 or 49^(24.5)
RHS is 49^(7)
So LHS > RHS
So answer is NO not YES

A is sufficient !
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Re: Exponents-roots DS [#permalink] New post 22 Sep 2010, 13:44
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tatane90 wrote:
If m and n are positive integers, is m^n < n^m?
(1) m = sqrt(n)
(2) n > 5

i disagree with the OA.
My reasoning below
1) if m = 1 & n = 1 then m=sqrt(n) and m^n=n^m
if m=2 & n=4 then m=sqrt(n) and m^n=n^m
if m=3 & n=9 then m=sqrt(n) and m^n>n^m
So if 1) then m^n<n^m always false ==> sufficient

2) we do not know anything about m so insufficient

I assume answer is A. Do you agree ?


It seems that you are right, answer should be A.

Answer to be C question shouldn't say that m and n are integers. In this case if m=\sqrt{n}=\sqrt{2} then m^n=\sqrt{2}^2=2<2^{\sqrt{2}}=n^m, so (1) wouldn't be sufficient.

Also the question would be a little bit trickier in this case.


yezz wrote:
If m and n are positive integers, is m^n < n^m?
(1) m = sqrt(n)
(2) n > 5


from 1

m = sqrt n ie: n = m^2

is m^(m^2)< (m^2)^m

plug m = 1 or 2 (no), plug m = 7 (yes) insuff
from 2

obviously insuff

both suff

C


If m=7 then n=49 and 7^{49}>49^7, so answer is still NO.
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Re: Exponents-roots DS [#permalink] New post 22 Sep 2010, 15:28
I will go with C...OA seems correct

1. With one alone I get

Is m^n > n^m
Is (\sqrt{n})^n >n^(\sqrt{n}
Is n^(\frac{n}{2}) = n^(\sqrt{n}). Since I made bases same I have to answer is n/2 greater than squreroot n....cannot answer

2. knowing n alone will not tell me about expression

If I combine both I can answer my question - Is N/2 greater than \sqrt{n}
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Re: Exponents-roots DS [#permalink] New post 22 Sep 2010, 15:39
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I think I answered the question by reversing the signs of equality. Is the question really correct?
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Re: Exponents-roots DS [#permalink] New post 12 Sep 2012, 15:28
Bunuel wrote:
tatane90 wrote:
If m and n are positive integers, is m^n < n^m?
(1) m = sqrt(n)
(2) n > 5

i disagree with the OA.
My reasoning below
1) if m = 1 & n = 1 then m=sqrt(n) and m^n=n^m
if m=2 & n=4 then m=sqrt(n) and m^n=n^m
if m=3 & n=9 then m=sqrt(n) and m^n>n^m
So if 1) then m^n<n^m always false ==> sufficient

2) we do not know anything about m so insufficient

I assume answer is A. Do you agree ?


It seems that you are right, answer should be A.

Answer to be C question shouldn't say that m and n are integers. In this case if m=\sqrt{n}=\sqrt{2} then m^n=\sqrt{2}^2=2<2^{\sqrt{2}}=n^m, so (1) wouldn't be sufficient.

Also the question would be a little bit trickier in this case.


yezz wrote:
If m and n are positive integers, is m^n < n^m?
(1) m = sqrt(n)
(2) n > 5


from 1

m = sqrt n ie: n = m^2

is m^(m^2)< (m^2)^m

plug m = 1 or 2 (no), plug m = 7 (yes) insuff
from 2

obviously insuff

both suff

C


If m=7 then n=49 and 7^{49}>49^7, so answer is still NO.



Hi Bunuel,

Can you discuss this question from scratch as a new question. As per me answer is A.
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Re: If m and n are positive integers, is m^n < n^m? [#permalink] New post 13 Sep 2012, 03:56
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If m and n are positive integers, is m^n < n^m?

(1) m = sqrt(n) --> m^2=n. Substitute n in the question: is m^{m^2}<(m^2)^m? --> is m^{m^2}<m^{2m}? Now, if m is 1 or 2, then m^{m^2}=m^{2m}, so the answer is NO and if m is an integer greater than 2, then m^{m^2}>m^{2m}, so the answer is still NO. Sufficient.

(2) n > 5. If m=1, then the answer is YES but if m=2, then the answer is NO. Not sufficient.

Answer: A.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If m and n are positive integers, is m^n < n^m? [#permalink] New post 13 Sep 2012, 05:50
Bunuel wrote:
If m and n are positive integers, is m^n < n^m?

(1) m = sqrt(n) --> m^2=n. Substitute n in the question: is m^{m^2}<(m^2)^m? --> is m^{m^2}<m^{2m}? Now, if m is 1 or 2, then m^{m^2}=m^{2m}, so the answer is NO and if m is an integer greater than 2, then m^{m^2}>m^{2m}, so the answer is still NO. Sufficient.

(2) n > 5. If m=1, then the answer is YES but if m=2, then the answer is NO. Not sufficient.

Answer: A.



Thanks Bunuel for the quick reply
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Re: If m and n are positive integers, is m^n < n^m? [#permalink] New post 15 Jul 2013, 22:22
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Re: If m and n are positive integers, is m^n < n^m?   [#permalink] 15 Jul 2013, 22:22
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