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 $m + n$ .

note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques

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 $x=\frac15$ $z=\frac{5}{24}.$ 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 3

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 $5$ .

Solution 4

(Hybrid between 1/2)

Because $xyz = 1, \hspace{0.15cm} \frac{1}{x} = yz, \hspace{0.15cm} \frac{1}{y} = xz,$ and $\hspace{0.05cm}\frac{1}{z} = xy$ . Substituting and factoring, we get $x(y+1) = 5$ , $\hspace{0.15cm}y(z+1) = 29$ , and $\hspace{0.05cm}z(x+1) = k$ . Multiplying them all together, we get, $xyz(x+1)(y+1)(z+1) = 145k$ , but $xyz$ is $1$ , and by the Identity property of multiplication, we can take it out. So, in the end, we get $(x+1)(y+1)(z+1) = 145k$ . And, we can expand this to get $xyz+xy+yz+xz+x+y+z+1 = 145k$ , and if we make a substitution for $xyz$ , and rearrange the terms, we get $xy+yz+xz+x+y+z = 145k-2$ This will be important.

Now, lets add the 3 equations $x(y+1) = 5, \hspace{0.15cm}y(z+1) = 29$ , and $\hspace{0.05cm}z(x+1) = k$ . We use the expand the Left hand sides, then, we add the equations to get $xy+yz+xz+x+y+z = k+34$ Notice that the LHS of this equation matches the LHS equation that I said was important. So, the RHS of both equations are equal, and thus $145k-2 = k+34$ We move all constant terms to the right, and all linear terms to the left, to get $144k = 36$ , so $k = \frac{1}{4}$ which gives an answer of $1+4 = \boxed{005}$

-AlexLikeMath

Solution 5

Get rid of the denominators in the second and third equations to get $xz-5z=-1$ and $xy-29x=-1$ . Then, since $xyz=1$ , we have $\tfrac 1y-5z=-1$ and $\tfrac 1z-29x=-1$ . Then, since we know that $\tfrac 1z+x=5$ , we can subtract these two equations to get that $30x=6\implies x=5$ . The result follows that $z=\tfrac 5{24}$ and $y=24$ , so $z+\tfrac 1y=\tfrac 1{24}+\tfrac 5{24}=\tfrac 14$ , and the requested answer is $1+4=\boxed{005}.$

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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