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Difference between revisions of "2007 iTest Problems/Problem 9"

(New page: ==Problem== Suppose that <math>m</math> and <math>n</math> are positive integers such that <math>m < n</math>, the geometric mean of <math>m</math> and <math>n</math> is greater than <math...)
 
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==Solution==
 
==Solution==
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Since the arithmetic mean is less than 2007 and the geometric mean is greater than 2007, the arithmetic mean must be less than the geometric mean. But by the AM-GM inequality, this is impossible. Therefore no such pairs <math>(m, n)</math> exist, and the answer is <math>0\Rightarrow\boxed{A}</math>.
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Since the arithmetic mean is less than 2007 and the geometric mean is greater than 2007, the arithmetic mean must be less than the geometric mean. But by the [[AM-GM inequality]], this is impossible. Therefore no such pairs <math>(m, n)</math> exist, and the answer is <math>0\Rightarrow\boxed{A}</math>.

Revision as of 20:51, 26 September 2008

Problem

Suppose that $m$ and $n$ are positive integers such that $m < n$, the geometric mean of $m$ and $n$ is greater than $2007$, and the arithmetic mean of $m$ and $n$ is less than $2007$. How many pairs $(m, n)$ satisfy these conditions?

$\mathrm{(A)}\,0\quad\mathrm{(B)}\,1\quad\mathrm{(C)}\,2\quad\mathrm{(D)}\,3\quad\mathrm{(E)}\,4\quad\mathrm{(F)}\,5\quad\mathrm{(G)}\,6\quad\mathrm{(H)}\,7\quad\mathrm{(I)}\,2007$

Solution

Since the arithmetic mean is less than 2007 and the geometric mean is greater than 2007, the arithmetic mean must be less than the geometric mean. But by the AM-GM inequality, this is impossible. Therefore no such pairs $(m, n)$ exist, and the answer is $0\Rightarrow\boxed{A}$.

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