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Difference between revisions of "1996 AIME Problems/Problem 8"

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== Problem ==
 
== Problem ==
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The harmonic mean of two positive integers is the reciprocal of the arithmetic mean of their reciprocals. For how many ordered pairs of positive integers <math>(x,y)</math> with <math>x<y</math> is the harmonic mean of <math>x</math> and <math>y</math> equal to <math>6^{20}</math>?
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The [[harmonic mean]] of two positive integers is the reciprocal of the [[arithmetic mean]] of their reciprocals. For how many ordered pairs of positive integers <math>(x,y)</math> with <math>x<y</math> is the harmonic mean of <math>x</math> and <math>y</math> equal to <math>6^{20}</math>?
  
 
== Solution ==
 
== Solution ==
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The harmonic mean of <math>x</math> and <math>y</math> is equal to <math>2xy/(x+y)</math>, so we have <math>xy=(x+y)(3^{20}\cdot2^{19})</math>, and <math>(x-3^{20}\cdot2^{19})(y-3^{20}\cdot2^{19})=3^{40}\cdot2^{19}</math>. <math>3^{40}\cdot2^{19}</math> has <math>39\cdot41=1599</math> factors, one of which is the square root, so the answer is half of the rest of them, which is <math>799</math>.
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The harmonic mean of <math>x</math> and <math>y</math> is equal to <math>\frac{1}{\frac{\frac{1}{x}+\frac{1}{y}}2} = \frac{2xy}{x+y}</math>, so we have <math>xy=(x+y)(3^{20}\cdot2^{19})</math>, and by [[SFFT]], <math>(x-3^{20}\cdot2^{19})(y-3^{20}\cdot2^{19})=3^{40}\cdot2^{38}</math>. Now, <math>3^{40}\cdot2^{38}</math> has <math>41\cdot39=1599</math> factors, one of which is the square root (<math>3^{20}2^{19}</math>). Since <math>x<y</math>, the answer is half of the remaining number of factors, which is <math>\frac{1599-1}{2}= \boxed{799}</math>.
  
 
== See also ==
 
== See also ==
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*[[1996 AIME Problems]]
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{{AIME box|year=1996|num-b=7|num-a=9}}
  
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{{AIME box|year=1996|num-b=7|num-a=9}}
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[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 18:32, 4 July 2013

Problem

The harmonic mean of two positive integers is the reciprocal of the arithmetic mean of their reciprocals. For how many ordered pairs of positive integers $(x,y)$ with $x<y$ is the harmonic mean of $x$ and $y$ equal to $6^{20}$?

Solution

The harmonic mean of $x$ and $y$ is equal to $\frac{1}{\frac{\frac{1}{x}+\frac{1}{y}}2} = \frac{2xy}{x+y}$, so we have $xy=(x+y)(3^{20}\cdot2^{19})$, and by SFFT, $(x-3^{20}\cdot2^{19})(y-3^{20}\cdot2^{19})=3^{40}\cdot2^{38}$. Now, $3^{40}\cdot2^{38}$ has $41\cdot39=1599$ factors, one of which is the square root ($3^{20}2^{19}$). Since $x<y$, the answer is half of the remaining number of factors, which is $\frac{1599-1}{2}= \boxed{799}$.

See also

1996 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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