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2000 AIME I Problems/Problem 7

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Contents

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

Solution 1

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)\\&=...

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 solution is 5.

See also

2000 AIME I (ProblemsResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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