Question

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 $m + n.$

Answer

Answer: 5

Concepts:system of equations algebraic manipulation symmetry (algebra)

Difficulty rating: 2330

Solution:

Let $t = z + \frac{1}{y}.$ Expanding the product of all three expressions, $\begin{aligned} &\left(x + \frac{1}{z}\right)\!\left(y + \frac{1}{x}\right)\!\left(z + \frac{1}{y}\right) \\ &= xyz + \frac{1}{xyz} + \left(x + \frac{1}{z}\right) \\ &\quad {}+ \left(y + \frac{1}{x}\right) + \left(z + \frac{1}{y}\right). \end{aligned}$ Since $xyz = 1,$ the left side is $5 \cdot 29 \cdot t = 145t$ and the right side is $2 + 5 + 29 + t = 36 + t.$

So $145t = 36 + t,$ giving $t = \frac{36}{144} = \frac{1}{4}.$ Thus $m + n = 1 + 4 = 5.$

2000 AIME I Problem 7

Answer: 5

Difficulty rating: 2330

Solution:

Problem 7 in Other Years