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> | + | 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>41\cdot39=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 16:27, 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 , 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 |