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 1

We can rewrite $xyz=1$ as $\frac{1}{z}=xy$ .

Substituting into one of the given equations, we have $x+xy=5$ $x(1+y)=5$ $\frac{1}{x}=\frac{1+y}{5}.$

We can substitute back into $y+\frac{1}{x}=29$ to obtain $y+\frac{1+y}{5}=29$ $5y+1+y=145$ $y=24.$

We can then substitute once again to get $x=\frac15$ $y=\frac{5}{24}.$ Thus, $z+\frac1y=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}$ , so $m+n=\boxed{005}$ .

Solution 2

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

$\begin{align*} (5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\ &=xyz + \frac{xy}{y} + \frac{xz}{x} + \frac{yz}{z} + \frac{x}{xy} + \frac{y}{yz} + \frac{z}{xz} + \frac{1}{xyz}\\ &=1 + x + z + y + \frac{1}{y} + \frac{1}{z} + \frac{1}{x} + \frac{1}{1}\\ &=2 + \left(x + \frac {1}{z}\right) + \left(y + \frac {1}{x}\right) + \left(z + \frac {1}{y}\right)\\ &=2 + 5 + 29 + r\\ &=36 + r \end{align*}$

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

Solution 2

Since $x+(1/z)=5, 1=z(5-x)=xyz$ , so $5-x=xy$ . Also, $y=29-(1/x)$ by the second equation. Substitution gives $x=1/5$ , $y=24$ , and $z=5/24$ , so the answer is 4+1 which is equal to $5$ .

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

Contents

Problem

Solution 1

Solution 2

Solution 2

See also