Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2)
(Added Solution)
Line 4: Line 4:
  
 
===Solution 1===
 
===Solution 1===
 +
We can rewrite <math>xyz=1</math> as <math>\frac{1}{z}=xy</math>.
 +
 +
Substituting into one of the given equations, we have
 +
<cmath>x+xy=5</cmath>
 +
<cmath>x(1+y)=5</cmath>
 +
<cmath>\frac{1}{x}=\frac{1+y}{5}.</cmath>
 +
 +
We can substitute back into <math>y+\frac{1}{x}=29</math> to obtain
 +
<cmath>y+\frac{1+y}{5}=29</cmath>
 +
<cmath>5y+1+y=145</cmath>
 +
<cmath>y=24.</cmath>
 +
 +
We can then substitute once again to get
 +
<cmath>x=\frac15</cmath>
 +
<cmath>y=\frac{5}{24}.</cmath>
 +
Thus, <math>z+\frac1y=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}</math>, so <math>m+n=\boxed{005}</math>.
 +
 +
 +
===Solution 2===
 
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 10:23, 21 July 2018

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 $5$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_7&oldid=96391"