2000 AIME I Problems/Problem 7

Problem

Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Solution

Let $r = \frac{m}{n} = z + \frac {1}{y}$ .

$ $\begin{aligned} (5) (29) (r) & = (x + \frac{1}{z}) (y + \frac{1}{x}) (z + \frac{1}{y}) \\ = x y z + \frac{x y}{y} + \frac{x z}{x} + \frac{y z}{z} + \frac{x}{x y} + \frac{y}{y z} + \frac{z}{x z} + \frac{1}{x y z} \\ = 1 + x + z + y + \frac{1}{y} + \frac{1}{z} + \frac{1}{x} + \frac{1}{1} \\ = 2 + (x + \frac{1}{z}) + (y + \frac{1}{x}) + (z + \frac{1}{y}) \\ = 2 + 5 + 29 + r \\ = 36 + r \end{aligned}$ $ (Error compiling LaTeX. Unknown error_msg)

Thus $145r = 36+r \Rightarrow 144r = 36 \Rightarrow r = \frac{36}{144} = \frac{1}{4}$ . So $m + n = 1 + 4 = \boxed{5}$

2000 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2000 AIME I Problems/Problem 7

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