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16 Sep 2014, 00:55
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For any pair of positive integers \(x\) and \(y\), their arithmetic mean \(A(x, y)\) is defined as \(\frac{x + y}{2}\) while their geometric mean \(G(x, y)\) is defined as \(\sqrt{xy}\). If \(m\) and \(n\) are positive integers, is \(m > n\)?
(1) \(A(G(m, n), m) = m\)
(2) \(A(m, n) - G(m, n) = 0\)
(1) \(A(G(m, n), m) = m\)
(2) \(A(m, n) - G(m, n) = 0\)
16 Sep 2014, 00:55
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Official Solution:
For any pair of positive integers \(x\) and \(y\), their arithmetic mean \(A(x, y)\) is defined as \(\frac{x + y}{2}\) while their geometric mean \(G(x, y)\) is defined as \(\sqrt{xy}\). If \(m\) and \(n\) are positive integers, is \(m > n\)?
(1) \(A(G(m, n), m) = m\).
\(A(\sqrt{mn}, m) = m\);
\(\frac{\sqrt{mn} + m}{2}= m\);
\(\sqrt{mn} + m= 2m\);
\(\sqrt{mn} = m\);
\(mn = m^2\);
Since \(m\) is positive we can safely reduce the above by it:
\(n = m\)
Therefore, the answer to the question is NO. Sufficient.
(2) \(A(m, n) - G(m, n) = 0\).
\(A(m, n) - G(m, n) =0\);
\(\frac{m + n}{2}-\sqrt{mn}=0\);
\(m + n-2\sqrt{mn}=0\);
\(m -2\sqrt{mn}+ n=0\);
Recognize that the above is the square of the difference:
\((\sqrt{m} - \sqrt{n})^2=0\);
\(\sqrt{m} = \sqrt{n}\);
\(m=n\)
Therefore, the answer to the question is NO. Sufficient.
Answer: D
For any pair of positive integers \(x\) and \(y\), their arithmetic mean \(A(x, y)\) is defined as \(\frac{x + y}{2}\) while their geometric mean \(G(x, y)\) is defined as \(\sqrt{xy}\). If \(m\) and \(n\) are positive integers, is \(m > n\)?
(1) \(A(G(m, n), m) = m\).
\(A(\sqrt{mn}, m) = m\);
\(\frac{\sqrt{mn} + m}{2}= m\);
\(\sqrt{mn} + m= 2m\);
\(\sqrt{mn} = m\);
\(mn = m^2\);
Since \(m\) is positive we can safely reduce the above by it:
\(n = m\)
Therefore, the answer to the question is NO. Sufficient.
(2) \(A(m, n) - G(m, n) = 0\).
\(A(m, n) - G(m, n) =0\);
\(\frac{m + n}{2}-\sqrt{mn}=0\);
\(m + n-2\sqrt{mn}=0\);
\(m -2\sqrt{mn}+ n=0\);
Recognize that the above is the square of the difference:
\((\sqrt{m} - \sqrt{n})^2=0\);
\(\sqrt{m} = \sqrt{n}\);
\(m=n\)
Therefore, the answer to the question is NO. Sufficient.
Answer: D
24 Nov 2017, 22:44
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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.
