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^{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 hale of the rest of them, which is <math>799</math>. |
== See also == | == See also == |
Revision as of 19:14, 20 February 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, so the answer is hale 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 |