Difference between revisions of "2000 AIME I Problems/Problem 7"
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Suppose that <math>x,</math> <math>y,</math> and <math>z</math> are three positive numbers that satisfy the equations <math>xyz = 1,</math> <math>x + \frac {1}{z} = 5,</math> and <math>y + \frac {1}{x} = 29.</math> Then <math>z + \frac {1}{y} = \frac {m}{n},</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>. | Suppose that <math>x,</math> <math>y,</math> and <math>z</math> are three positive numbers that satisfy the equations <math>xyz = 1,</math> <math>x + \frac {1}{z} = 5,</math> and <math>y + \frac {1}{x} = 29.</math> Then <math>z + \frac {1}{y} = \frac {m}{n},</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>. | ||
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===Solution 1=== | ===Solution 1=== | ||
Let <math>r = \frac{m}{n} = z + \frac {1}{y}</math>. | Let <math>r = \frac{m}{n} = z + \frac {1}{y}</math>. | ||
Revision as of 11:17, 2 July 2016
Contents
Problem
Suppose that 
and 
are three positive numbers that satisfy the equations 
and 
Then 
where 
and 
are relatively prime positive integers. Find 
.
Solution 1
Let 
.
Thus 
. So 
.
Solution 2
Since 
, so 
. Also, 
by the second equation. Substitution gives 
, 
, and 
, so the solution is 
.
See also
| 2000 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
