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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]

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Updated on: 15 Jul 2013, 23:20
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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 ?

Originally posted by tatane90 on 22 Sep 2010, 13:37.
Last edited by Bunuel on 15 Jul 2013, 23:20, edited 2 times in total.
Edited the OA.
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22 Sep 2010, 14:08
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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Updated on: 22 Sep 2010, 14:43
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

Originally posted by yezz on 22 Sep 2010, 14:35.
Last edited by yezz on 22 Sep 2010, 14:43, edited 1 time in total.
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22 Sep 2010, 14:43
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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22 Sep 2010, 14:44
2
1
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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05 Oct 2011, 07:43
If m and n are positive integers, is $$m^n$$ < $$n^m$$?
(1) m = $$\sqrt{n}$$
(2) n > 5

Let's discuss the relation between $$m^n$$ and $$n^m$$ when m and n are positive integers. To differentiate m from n, I say m is the smaller of the two except where they are equal.

Look at the pattern now:
Say m = 1
n = 1: $$1^1 = 1^1$$
n = 2: $$1^2 < 2^1$$
n = 3: $$1^3 < 3^1$$
and so on...

Say m = 2
n = 2: $$2^2 = 2^2$$
n = 3: $$2^3 < 3^2$$
n = 4: $$2^4 = 4^2$$ (m^n was less than n^m above but here they are equal so m^n should become greater from now on)
n = 5: $$2^5 > 5^2$$
n = 6: $$2^6 > 6^2$$
You see that the difference between m^n and n^m is increasing in every step.)

Say m = 3
n = 4: $$3^4 > 4^3$$ (m^n is already greater than n^m)
n = 5: $$3^5 > 5^3$$
n = 6: $$3^6 > 6^3$$
You see that the difference between m^n and n^m is increasing in every step.)

Say m = 4
n = 5: $$4^5 > 5^4$$ (m^n is already greater than n^m)
n = 6: $$4^6 > 6^4$$
n = 7: $$4^7 > 7^4$$
You see that the difference between m^n and n^m is increasing in every step.)

I hope you see the pattern. Let's forget about the cases where m = 1 and m = n.
In every case except m = 2, $$m^n > n^m$$
When m = 2, $$2^3 < 3^2$$ and $$2^4 = 4^2$$. Thereafter, $$m^n > n^m$$

It is a good idea to remember these relations.

Let's look at the question now.
Is $$m^n$$ < $$n^m$$?

(1) m = $$\sqrt{n}$$
First of all, both m and n are integers so n must be a perfect square. Also, m will be smaller than n except when both are equal to 1.
m = 1/2/3/4/5/6 etc
n = 1/4/9/16/25/36 etc

If m = 1 and n = 1, we know $$m^n = n^m$$
If m = 2 and n = 4, we know $$m^n = n^m$$
For all other cases, we know that $$m^n > n^m$$

Hence the answer to the question is 'No'. Sufficient.

(2) n > 5
If m = 1, 1^6 < 6^1
If m = 2, 2^6 > 6^2
Not sufficient.

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05 Oct 2011, 08:27
2
m^n < n^m

1) m=sqrt(n) => m^2 = n , putting this in above.

m^m^2 < m^2^m => m^2m < m^2m

which are equal, so NO. Sufficient.

2) n > 5 not sufficient

Ans: A. ( Am I missing something...?)
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12 Sep 2012, 16:28
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]

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13 Sep 2012, 04:56
1
3
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.

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Re: If m and n are positive integers, is m^n < n^m?  [#permalink]

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13 Sep 2012, 06:50
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.

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

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16 Jan 2015, 12:52
2
Hi All,

This DS question can be solved by TESTing VALUES.

We're told that M and N are POSITIVE INTEGERS. We're asked if M^N < N^M. This is a YES/NO question.

Fact 1: M = \sqrt{N}

IF....
N = 1
M = 1
1^1 is NOT < 1^1 and the answer to the question is NO.

N = 4
M = 2
2^4 is NOT < 4^2 and the answer to the question is NO.

N = 9
M = 3
3^9 is NOT < 9^3 and the answer to the question is NO.
This pattern will continue; the answer to the question is ALWAYS NO.
Fact 1 is SUFFICIENT.

Fact 2: N > 5

This tells us NOTHING about the value of M, so this is probably insufficient. Here's the proof.

IF...
M = 1
N = 6
1^6 < 6^1 and the answer to the question is YES.

IF....
M = 6
N = 6
6^6 is NOT < 6^6 and the answer to the question is NO.
Fact 2 is INSUFFICIENT

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Re: If m and n are positive integers, is m^n < n^m?  [#permalink]

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25 Jan 2015, 15:34
Very good question indeed for practice.

its given M= sqrt(N)

to show is whether M^n < N^m
substitute N = M^2 in the above equation we get

M^n < M^2m

it means n< 2m (As bases are same)

which means n < 2sqt(n)

above equation can be reduced to sqrt(n) < 1.

Sqrt(n) can never be less than 1 as n is positive integer.

So using stmt 1 we can derive the answer as NO.

Stmnt2 does not have any reference of M. so not sufficient.

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If m and n are positive integers, is m^n < n^m?  [#permalink]

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09 Jun 2015, 23:06
It would have been interesting to see if the inequality had been reversed. => m^n > n^m

In this case.. there is a special value which will fail it... m = 2 & n = 4 ... i.e. .. m = n^1/2 & 2^4 = 4^2... In this case it doesn't satisfy the inequality but all other values do.. So then answer would have been C to ensure that m =! 2.
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If m and n are positive integers, is m^n < n^m?  [#permalink]

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26 Jun 2015, 22:09
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.

if I compare with base using "n" rather than "m"

I get
following equation after reduction :

m^n < n^m reduces to

n^(n/2) < (n)^sqrt (n)

means we need find if
n/2 < Sqrt (n), Given n is perfect square bcs m = sqrt(n) and m is integer.

for n=1 : yes
for n=4 : no
for n=64 : No
for n=9 : No

then how can we answer : A is sufficient.

on other side if n>5 then
Answer B: is always hold. So sufficient....

I am sure I must be wrong somewhere in logic!!!
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Re: If m and n are positive integers, is m^n < n^m?  [#permalink]

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27 Jun 2015, 00:19
1
Jam2014 wrote:
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.

if I compare with base using "n" rather than "m"

I get
following equation after reduction :

m^n < n^m reduces to

n^(n/2) < (n)^sqrt (n)

means we need find if
n/2 < Sqrt (n), Given n is perfect square bcs m = sqrt(n) and m is integer.

for n=1 : yes
for n=4 : no
for n=64 : No
for n=9 : No

then how can we answer : A is sufficient.

on other side if n>5 then
Answer B: is always hold. So sufficient....

I am sure I must be wrong somewhere in logic!!!

Look at the highlighted steps only

n^(n/2) < (n)^sqrt (n)

for n=1 : yes The answer is not Yes it's No here as well because n^(n/2) will NOT be less than (n)^sqrt (n) for n=1

Statement 2: n>5

@n=6, and m=1, m^n will be less than n^m
@n=6, and m=2, m^n will NOT be less than n^m
NOT SUFFICIENT

I hope it helps!
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Re: If m and n are positive integers, is m^n < n^m?  [#permalink]

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27 Jun 2015, 03:52
GMATinsight wrote:
Jam2014 wrote:
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.

if I compare with base using "n" rather than "m"

I get
following equation after reduction :

m^n < n^m reduces to

n^(n/2) < (n)^sqrt (n)

means we need find if
n/2 < Sqrt (n), Given n is perfect square bcs m = sqrt(n) and m is integer.

for n=1 : yes
for n=4 : no
for n=64 : No
for n=9 : No

then how can we answer : A is sufficient.

on other side if n>5 then
Answer B: is always hold. So sufficient....

I am sure I must be wrong somewhere in logic!!!

Look at the highlighted steps only

n^(n/2) < (n)^sqrt (n)

for n=1 : yes The answer is not Yes it's No here as well because n^(n/2) will NOT be less than (n)^sqrt (n) for n=1

Statement 2: n>5

@n=6, and m=1, m^n will be less than n^m
@n=6, and m=2, m^n will NOT be less than n^m
NOT SUFFICIENT

I hope it helps!

My take away is that I can't compare just power. Based on what Bases are, results can be different!!!!
Great concept and question.
Thnaks
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If m and n are positive integers, is m^n < n^m?  [#permalink]

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30 Jun 2015, 23:41
If m and n are positive integers, is m^n < n^m?

(1) m=√n
Squaring both sides yields m^2 = n
This can be substituted into the original equation m^(m^2) < (m^2)^m
This is sufficient (no). You can try one or two numbers to confirm.

(2) n > 5 Since we know nothing of m, this is not sufficient.

A
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Re: If m and n are positive integers, is m^n < n^m?  [#permalink]

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22 Dec 2017, 11:24
A bit of juggling but great question overall. Ans is A.
St 1 says --> n = M^2
Substituting it give --> m ^2m NOT greater than m ^2m. Since it answers the stem, A is suff..

St 2 - n can be 6,7,8,.. and no information on N.
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Re: If m and n are positive integers, is m^n < n^m?  [#permalink]

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22 Dec 2017, 12:52
Please check the condition for m=1 and n=1
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Re: If m and n are positive integers, is m^n < n^m?  [#permalink]

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22 Dec 2017, 12:57
akgulhane wrote:
Please check the condition for m=1 and n=1

For these numbers you'll still get the same NO answer as for any other. Please re-read the discussion above carefully.
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Re: If m and n are positive integers, is m^n < n^m?   [#permalink] 22 Dec 2017, 12:57

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