Instructions: DO NOT USE CALCULATOR. Figures are not drawn to scale. If x and y are positive integers and \(\frac{1}{x} + \frac{1}{y} = \frac{1}{9}\), what is the difference between the maximum and minimum value of x?

Explanation: (C) Given \(\frac{1}{x} + \frac{1}{y} = \frac{1}{9}\). Solving for x: \(\frac{1}{x} = \frac{1}{9} - \frac{1}{y} = \frac{y-9}{9y} \implies x = \frac{9y}{y-9}\). For x to be a positive integer, \(y-9\) must be a divisor of \(9y\), and \(y-9\) must be positive, so \(y > 9\). Maximize x: The denominator \(y-9\) must be minimized. The smallest integer value for \(y-9\) is 1, so \(y = 10\). \(x_{max} = \frac{9 \cdot 10}{10-9} = 90\). Minimize x: Rewrite x: \(x = \frac{9y}{y-9} = \frac{9(y-9) + 81}{y-9} = 9 + \frac{81}{y-9}\). For x to be minimum, \(\frac{81}{y-9}\) must be minimum. Since \(y > 9\), \(\frac{81}{y-9}\) is positive. For a positive value of \(y-9\), the smallest value of \(\frac{81}{y-9}\) occurs when \(y-9\) is largest. The largest value of \(y-9\) for which x is an integer is when \(y-9=81\), so \(y=90\). \(x_{min} = 9 + \frac{81}{81} = 10\). Difference = \(x_{max} - x_{min} = 90 - 10 = 80\).