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Difference between revisions of "2000 AIME I Problems/Problem 7"

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== Problem ==
 
== Problem ==
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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>.
  
 
== Solution ==
 
== Solution ==
 
Let <math>r = \frac{m}{n} = z + \frac {1}{y}</math>.
 
Let <math>r = \frac{m}{n} = z + \frac {1}{y}</math>.
  
−
<math>
+
<cmath>
 
\begin{align*}
 
\begin{align*}
 
(5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\
 
(5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\
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&=36 + r
 
&=36 + r
 
\end{align*}
 
\end{align*}
−
</math>
+
</cmath>
  
−
Thus <math>145r = 36+r \Rightarrow 144r = 36 \Rightarrow r = \frac{36}{144} = \frac{1}{4}</math>. So <math>m + n = 1 + 4 = \boxed{5}</math>
+
Thus <math>145r = 36+r \Rightarrow 144r = 36 \Rightarrow r = \frac{36}{144} = \frac{1}{4}</math>. So <math>m + n = 1 + 4 = \boxed{5}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=I|num-b=6|num-a=8}}
 
{{AIME box|year=2000|n=I|num-b=6|num-a=8}}
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 +
[[Category:Intermediate Algebra Problems]]

Revision as of 17:14, 31 December 2007

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{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 $m + n = 1 + 4 = \boxed{5}$.

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
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