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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.
_________________

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.

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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(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]

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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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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]

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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]

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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

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, 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

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

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30 Jun 2015, 22:41

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.

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

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22 Dec 2017, 10:24

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..