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\ | + | 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 
with 
is the harmonic mean of 
and 
equal to 
?
Solution
The harmonic mean of and is equal to 
, so we have 
, and 
. 
has 
factors, one of which is the square root. Since $x\< y$ (Error compiling LaTeX. Unknown error_msg), the answer is half of the rest of them, which is 
.
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 | ||
