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

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== Solution ==
 
== Solution ==
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^{38}</math>. <math>3^{40}\cdot2^{38}</math> has <math>39\cdot41=1599</math> factors, one of which is the square root. Since <math>x\geq y</math>, 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>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^{38}</math>. <math>3^{40}\cdot2^{38}</math> has <math>39\cdot41=1599</math> factors, one of which is the square root. Since <math>x\< y</math>, the answer is half of the rest of them, which is <math>799</math>.
  
 
== See also ==
 
== See also ==

Revision as of 16:25, 22 July 2008

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 $2xy/(x+y)$, so we have $xy=(x+y)(3^{20}\cdot2^{19})$, and $(x-3^{20}\cdot2^{19})(y-3^{20}\cdot2^{19})=3^{40}\cdot2^{38}$. $3^{40}\cdot2^{38}$ has $39\cdot41=1599$ factors, one of which is the square root. Since $x\< y$ (Error compiling LaTeX. ! Undefined control sequence.), the answer is half of the rest of them, which is $799$.

See also

1996 AIME (ProblemsAnswer KeyResources)
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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