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If m and n are positive integers, is m^n < n^m? [#permalink]
 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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Re: Exponents-roots DS [#permalink]
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^mFor n,m integers and both greater than 1, this implies m^2/2 < mm(m-2) < 0This 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: Exponents-roots DS [#permalink]
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(m-2) < 1 is m(m-2)<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: Exponents-roots DS [#permalink]
22 Sep 2010, 14:37
It seems like I've seen this problem before. Would you mind sharing the source?
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Re: Exponents-roots DS [#permalink]
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: Exponents-roots DS [#permalink]
22 Sep 2010, 14:44
1
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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: Exponents-roots DS [#permalink]
22 Sep 2010, 14:45
shrouded1 wrote: 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 ! Apologies i have edited my post
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Re: Exponents-roots DS [#permalink]
22 Sep 2010, 16:28
I will go with C...OA seems correct 1. With one alone I get Is m^n > n^mIs (\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: Exponents-roots DS [#permalink]
22 Sep 2010, 16:39
I think I answered the question by reversing the signs of equality. Is the question really correct?
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Re: Exponents-roots DS [#permalink]
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]
13 Sep 2012, 04:56
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Re: If m and n are positive integers, is m^n < n^m? [#permalink]
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?
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13 Sep 2012, 06:50
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