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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. _________________

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. _________________

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Re: If m and n are positive integers, is m^n < n^m? [#permalink]
13 Sep 2012, 03:56

Expert's post

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.

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