Difference between revisions of "1996 AIME Problems/Problem 8"
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== Problem == | == 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 <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 == | ||
+ | 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 == | ||
− | + | {{AIME box|year=1996|num-b=7|num-a=9}} | |
− | {{ | + | [[Category:Intermediate Algebra Problems]] |
+ | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{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 with is the harmonic mean of and equal to ?
Solution
The harmonic mean of and is equal to , so we have , and by SFFT, . Now, has factors, one of which is the square root (). Since , the answer is half of the remaining number of factors, 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 |
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