Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

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

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.
_________________

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 \(\sqrt{n}\)
_________________

If you like my post, consider giving me some KUDOS !!!!! Like you I need them

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.
_________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

(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.

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

Show Tags

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
_________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

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

Show Tags

16 Jan 2015, 07:58

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

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

Show Tags

09 Jun 2015, 22:06

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.

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

Show Tags

26 Jun 2015, 21:09

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.

Answer: A.

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....

(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.

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

Hence consistent answer so SUFFICIENT

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

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

Show Tags

27 Jun 2015, 02:52

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.

Answer: A.

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

Hence consistent answer so SUFFICIENT

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

(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.

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

Show Tags

02 Nov 2016, 09:00

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...