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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) \(A(M, N)  G(M, N) = 0\)
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Re M1505 [#permalink]
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16 Sep 2014, 00:55
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: M1505 [#permalink]
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17 Oct 2015, 00:36
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 M1505 [#permalink]
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22 Aug 2016, 11:36
I think this is a highquality question and the explanation isn't clear enough, please elaborate. m+n2(sqrt of mn)=0
Can you please explain how you got this from statement 2?



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Re: M1505 [#permalink]
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23 Aug 2016, 10:52
tae808 wrote: I think this is a highquality question and the explanation isn't clear enough, please elaborate. m+n2(sqrt of mn)=0
Can you please explain how you got this from statement 2? m + n  2sqrt(mn) = 0 LHS follows the formula (ab)^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: M1505 [#permalink]
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24 Aug 2016, 05:04



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Re: M1505 [#permalink]
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23 Mar 2017, 17:54
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: M1505 [#permalink]
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24 Mar 2017, 04:57
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: M1505 [#permalink]
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24 Mar 2017, 09:54
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: M1505 [#permalink]
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24 Mar 2017, 10:51
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 therefore m^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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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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