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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? (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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Originally posted by tatane90 on 22 Sep 2010, 12:37.
Last edited by Bunuel on 15 Jul 2013, 22:20, edited 2 times in total.
Edited the OA.



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Re: Exponentsroots DS
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22 Sep 2010, 13:08
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
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Updated on: 22 Sep 2010, 13:43
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(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
Originally posted by yezz on 22 Sep 2010, 13:35.
Last edited by yezz on 22 Sep 2010, 13:43, edited 1 time in total.



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Re: Exponentsroots DS
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22 Sep 2010, 13:43
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
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22 Sep 2010, 13:44
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: Inequality
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05 Oct 2011, 06:43
kad wrote: 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. Answer (A)
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Re: Inequality
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05 Oct 2011, 07:27
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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Re: Exponentsroots DS
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12 Sep 2012, 15:28
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?
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13 Sep 2012, 03:56
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?
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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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Re: If m and n are positive integers, is m^n < n^m?
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16 Jan 2015, 11:52
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?
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25 Jan 2015, 14:34
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?
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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.



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If m and n are positive integers, is m^n < n^m?
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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 : 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?
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26 Jun 2015, 23:19
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?
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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 : 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?
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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.
A



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Re: If m and n are positive integers, is m^n < n^m?
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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..
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?
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22 Dec 2017, 11:52
Answer seems to be wrong.. 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?
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