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2007 iTest Problems/Problem 9: Difference between revisions

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==Solution==
==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 <math>(m, n)</math> exist, and the answer is <math>0\Rightarrow\boxed{A}</math>.
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>.
==See Also==
{{iTest box|year=2007|num-b=8|num-a=10}}
[[Category:Introductory Algebra Problems]]

Latest revision as of 19:24, 10 January 2019

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 $0\Rightarrow\boxed{A}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 8 		Followed by:
Problem 10
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