AIME 2000 I · 第 7 题

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

$$\begin{aligned} (5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\ &=xyz + \frac{xy}{y} + \frac{xz}{x} + \frac{yz}{z} + \frac{x}{xy} + \frac{y}{yz} + \frac{z}{xz} + \frac{1}{xyz}\\ &=1 + x + z + y + \frac{1}{y} + \frac{1}{z} + \frac{1}{x} + \frac{1}{1}\\ &=2 + \left(x + \frac {1}{z}\right) + \left(y + \frac {1}{x}\right) + \left(z + \frac {1}{y}\right)\\ &=2 + 5 + 29 + r\\ &=36 + r \end{aligned}$$ 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 $1+4=\boxed{005}.$

Solution 6

Rewrite the equations in terms of x.

$x+\frac{1}{z}=5$ becomes $z=\frac{1}{x+5}$.

$y+\frac{1}{x}=29$ becomes $y=29-\frac{1}{x}$

Now express $xyz=1$ in terms of x.

$\frac{1}{5-x}\cdot(29-\frac{1}{x})\cdot x=1$.

This evaluates to $29x-1=5-x$, giving us $x=\frac{1}{5}$. We can now plug x into the other equations to get $y=24$ and $z=\frac{5}{24}$.

Therefore, $z+\frac{1}{y}=\frac{5}{24}+\frac{1}{24}=\frac{6}{24}=\frac{1}{4}$.

$1+4=\boxed{5}$, and we are done. ~MC413551

Solution 7

First, we have been given the value of $xyz$, so we should probably figure out a way to use that. Before that, we replace $\dfrac{m}{n}$, because why deal with $2$ variables when you don't have to. We multiply these $3$ equations to get:

$$(x+\dfrac{1}{z})(y+\dfrac{1}{x})(z+\dfrac{1}{y}) \Longrightarrow xyz+x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} = 145a$$ . Plugging the value in for $xyz$, we get:

$$2+x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$$ . Now, we need a way to get these other random terms out of there. We add the three equations to get

$$x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} = 34+a$$ . Plugging this back into the equation we found we get:

$$36+a = 145a \Longrightarrow a = \dfrac{1}{4} \Longrightarrow \boxed{005}$$ -jb2015007

Solution 8

Let $x=\frac{a}{b},y=\frac{b}{c},$ and $z=\frac{c}{a}.$ From $x+\frac{1}{z}=5,$ we get $\frac{a}{b}+\frac{a}{c}=5\Rightarrow ac+ab=5bc.$ Similarly, from $y+\frac{1}{x}=29,ab+bc=29ac.$ We wish to find $\frac{c}{a}+\frac{c}{b}=\frac{ac+bc}{ab}.$ Set $ab=1.$ Then $ac+1=5bc$ and $bc+1=29ac.$ Solving these equations, we get $bc=\frac{5}{24}$ and $ac=\frac{1}{24}$ and thus we get $\frac{5}{24}+\frac{1}{24}=\frac{1}{4}\Rightarrow m=1,n=4$ and thus our answer is $\boxed{5}.$ ~rice_farmer