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If m and n are positive integers, is m^n < n^m? [#permalink]
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22 Sep 2010, 13:37
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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 ?
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Last edited by Bunuel on 15 Jul 2013, 23:20, edited 2 times in total.
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Re: Exponentsroots DS [#permalink]
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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 \(m^2/2 < m\) \(m(m2) < 0\) 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 AThe answer would be (c) if the original question was m^n <= n^m
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Re: Exponentsroots DS [#permalink]
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22 Sep 2010, 14:35
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(m2) < 1 is m(m2)<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, 14:43, edited 1 time in total.



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Re: Exponentsroots DS [#permalink]
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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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Re: Exponentsroots DS [#permalink]
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22 Sep 2010, 14:44
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: Exponentsroots DS [#permalink]
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22 Sep 2010, 16:28
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: Exponentsroots DS [#permalink]
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22 Sep 2010, 16:39
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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: Exponentsroots DS [#permalink]
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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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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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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.
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]
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Re: If m and n are positive integers, is m^n < n^m? [#permalink]
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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
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 Final Answer: GMAT assassins aren't born, they're made, Rich
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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.
The overall answer is A.
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.
Hence answer is A.



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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.
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 : nofor 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
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 : yesfor n=4 : nofor 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=1Hence 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, 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.
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 : yesfor n=4 : nofor 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=1Hence 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



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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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