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M15-05

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M15-05 [#permalink]

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If for any pair of two positive integers \(M\) and \(N\), their arithmetic mean \(A(M, N)\) is defined as \(\frac{M + N}{2}\) while their geometric mean \(G(M, N)\) is defined as \(\sqrt{MN}\), is \(M\) larger than \(N\)?


(1) \(A(G(M, N), M) = M\)

(2) \(A(M, N) - G(M, N) = 0\)
[Reveal] Spoiler: OA

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Re M15-05 [#permalink]

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Official Solution:


Statement (1) by itself is sufficient. \(A(G(M, N), M) = \frac{G(M, N) + M}{2} = \frac{\sqrt{MN} + M}{2}\). From S1 it follows that \(\frac{\sqrt{MN} + M}{2} = M\) or \(\frac{\sqrt{MN}}{2} = \frac{M}{2}\) or \(\sqrt{M} \sqrt{N} = M\) or \(\sqrt{M} = \sqrt{N}\). Thus, \(M = N\) and the answer to the question is "no". \(A(M, N) - G(M, N) = \frac{M + N}{2} - \sqrt{MN}\).

Statement (2) by itself is sufficient. From S2 it follows that \(M + N - 2\sqrt{MN} = 0\) or \((\sqrt{M} - \sqrt{N})^2 = 0\). Thus, \(M = N\) and the answer to the question is "no".


Answer: D
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Re: M15-05 [#permalink]

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New post 16 Oct 2015, 23:36
Hi,
Could you tell me if the method I used for calculating statement 2's sufficiency is valid, hmm foolproof ?
A(M,N)−G(M,N)=0,
M+N/2=SqRt MN, square both sides
(M+N/2)(M+N/2)= MN
2(M+N)= 2MN
M+N=MN
Since M and N are positive integers only two situations satisfy this equation- either both M and N are 1 or they are 2. In both these cases M=N, hence sufficient.
Is this correct method?
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Re M15-05 [#permalink]

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New post 22 Aug 2016, 10:36
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. m+n-2(sqrt of mn)=0

Can you please explain how you got this from statement 2?
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Re: M15-05 [#permalink]

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New post 23 Aug 2016, 09:52
tae808 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. m+n-2(sqrt of mn)=0

Can you please explain how you got this from statement 2?


m + n - 2sqrt(mn) = 0

LHS follows the formula (a-b)^2 = a^2 + b^2 -2ab
So we compress that to get:
[sqrt(m) - sqrt(n)]^2 = 0

This translates to: sqrt(m) - sqrt(n) = 0
=> sqrt(m) = sqrt(n)
=> m = n

also, recall that both are positive integers...
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Re: M15-05 [#permalink]

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New post 24 Aug 2016, 04:04
tae808 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. m+n-2(sqrt of mn)=0

Can you please explain how you got this from statement 2?


\(A(M, N) = \frac{M + N}{2}\)

\(G(M, N)=\sqrt{MN}\)

\(A(M, N) - G(M, N)=0\)

\(\frac{M + N}{2}-\sqrt{MN}=0\)

Multiply by 2: \(M + N - 2\sqrt{MN} = 0\)
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Re: M15-05 [#permalink]

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New post 23 Mar 2017, 16:54
Bunuel wrote:
Official Solution:


Statement (1) by itself is sufficient. \(A(G(M, N), M) = \frac{G(M, N) + M}{2} = \frac{\sqrt{MN} + M}{2}\). From S1 it follows that \(\frac{\sqrt{MN} + M}{2} = M\) or \(\frac{\sqrt{MN}}{2} = \frac{M}{2}\) or \(\sqrt{M} \sqrt{N} = M\) or \(\sqrt{M} = \sqrt{N}\). Thus, \(M = N\) and the answer to the question is "no". \(A(M, N) - G(M, N) = \frac{M + N}{2} - \sqrt{MN}\).

Statement (2) by itself is sufficient. From S2 it follows that \(M + N - 2\sqrt{MN} = 0\) or \((\sqrt{M} - \sqrt{N})^2 = 0\). Thus, \(M = N\) and the answer to the question is "no".


Answer: D


How do we reduce step by step from statement 1? I am particularly confused with - > \(\sqrt{M} \sqrt{N} = M\) or \(\sqrt{M} = \sqrt{N}\). Thus, \(M = N\)
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Re: M15-05 [#permalink]

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New post 24 Mar 2017, 03:57
Prostar wrote:
Bunuel wrote:
Official Solution:


Statement (1) by itself is sufficient. \(A(G(M, N), M) = \frac{G(M, N) + M}{2} = \frac{\sqrt{MN} + M}{2}\). From S1 it follows that \(\frac{\sqrt{MN} + M}{2} = M\) or \(\frac{\sqrt{MN}}{2} = \frac{M}{2}\) or \(\sqrt{M} \sqrt{N} = M\) or \(\sqrt{M} = \sqrt{N}\). Thus, \(M = N\) and the answer to the question is "no". \(A(M, N) - G(M, N) = \frac{M + N}{2} - \sqrt{MN}\).

Statement (2) by itself is sufficient. From S2 it follows that \(M + N - 2\sqrt{MN} = 0\) or \((\sqrt{M} - \sqrt{N})^2 = 0\). Thus, \(M = N\) and the answer to the question is "no".


Answer: D


How do we reduce step by step from statement 1? I am particularly confused with - > \(\sqrt{M} \sqrt{N} = M\) or \(\sqrt{M} = \sqrt{N}\). Thus, \(M = N\)


\(\sqrt{M} \sqrt{N} = M\)

\(\sqrt{M} \sqrt{N} = \sqrt{M}*\sqrt{M}\)

\(\sqrt{N} = \sqrt{M}\)
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Re: M15-05 [#permalink]

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New post 24 Mar 2017, 08:54
Bunuel wrote:
Official Solution:


Statement (1) by itself is sufficient. \(A(G(M, N), M) = \frac{G(M, N) + M}{2} = \frac{\sqrt{MN} + M}{2}\). From S1 it follows that \(\frac{\sqrt{MN} + M}{2} = M\) or \(\frac{\sqrt{MN}}{2} = \frac{M}{2}\) or \(\sqrt{M} \sqrt{N} = M\) or \(\sqrt{M} = \sqrt{N}\). Thus, \(M = N\) and the answer to the question is "no". \(A(M, N) - G(M, N) = \frac{M + N}{2} - \sqrt{MN}\).

Statement (2) by itself is sufficient. From S2 it follows that \(M + N - 2\sqrt{MN} = 0\) or \((\sqrt{M} - \sqrt{N})^2 = 0\). Thus, \(M = N\) and the answer to the question is "no".


Answer: D


Bunuel For Statement 1 , I did this m=(MN)^1/2;Thus M-(MN)^1/2=0 and therefore m^1/2 = 0 or/and m=n.Please help me in understanding why this is wrong
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Re: M15-05 [#permalink]

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New post 24 Mar 2017, 09:51
KARISHMA315 wrote:
Bunuel wrote:
Official Solution:


Statement (1) by itself is sufficient. \(A(G(M, N), M) = \frac{G(M, N) + M}{2} = \frac{\sqrt{MN} + M}{2}\). From S1 it follows that \(\frac{\sqrt{MN} + M}{2} = M\) or \(\frac{\sqrt{MN}}{2} = \frac{M}{2}\) or \(\sqrt{M} \sqrt{N} = M\) or \(\sqrt{M} = \sqrt{N}\). Thus, \(M = N\) and the answer to the question is "no". \(A(M, N) - G(M, N) = \frac{M + N}{2} - \sqrt{MN}\).

Statement (2) by itself is sufficient. From S2 it follows that \(M + N - 2\sqrt{MN} = 0\) or \((\sqrt{M} - \sqrt{N})^2 = 0\). Thus, \(M = N\) and the answer to the question is "no".


Answer: D


Bunuel For Statement 1 , I did this m=(MN)^1/2;Thus M-(MN)^1/2=0 and thereforem^1/2 = 0 or/and m=n.Please help me in understanding why this is wrong


If for any pair of two positive integers M and N...
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M15-05 [#permalink]

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New post 24 Nov 2017, 21:44
This is a high quality question. Thanks Bunuel for the amazing solution, it took me quite sometime to see how statement 2 is sufficient.
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M15-05   [#permalink] 24 Nov 2017, 21:44
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