Honors Math 7.5B Final Exam \( \begin{array}{l}\text { 17. [8 points] In which positive integer base } b \text { is the base- } b \text { radix expansion of the fraction } \frac{1}{8} \text { equal to } \\ 0 . \overline{18}_{b} \text { ? }\end{array} \)

To find the positive integer base \( b \) for which the base-\( b \) representation \( 0.\overline{18}_b \) is equal to \( \frac{1}{8} \), we first recognize that \( 0.\overline{18}_b \) is a repeating decimal. We can express it as: \[ x = 0.181818\ldots_b \] Multiplying both sides by \( b^2 \) (since the repeating part has a length of 2) gives us: \[ b^2 x = 18.b181818\ldots_b = 18_b + x \] Now, we need to convert \( 18_b \) into decimal form. In base \( b \), \( 18_b \) equals: \[ 1 \cdot b^1 + 8 \cdot b^0 = b + 8 \] Substituting this back in, we get: \[ b^2 x = (b + 8) + x \] \[ b^2 x - x = b + 8 \] \[ (b^2 - 1)x = b + 8 \] \[ x = \frac{b + 8}{b^2 - 1} \] We set this equal to \( \frac{1}{8} \): \[ \frac{b + 8}{b^2 - 1} = \frac{1}{8} \] Cross multiplying gives: \[ 8(b + 8) = b^2 - 1 \] \[ 8b + 64 = b^2 - 1 \] \[ b^2 - 8b - 65 = 0 \] Now we solve this quadratic equation using the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \), where \( A = 1, B = -8, C = -65 \): \[ b = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot (-65)}}{2 \cdot 1} = \frac{8 \pm \sqrt{64 + 260}}{2} = \frac{8 \pm \sqrt{324}}{2} = \frac{8 \pm 18}{2} \] This gives us: \[ b = \frac{26}{2} = 13 \quad \text{or} \quad b = \frac{-10}{2} = -5 \] Since we are looking for a positive integer base, the solution is: \[ b = 13 \] Thus, the positive integer base \( b \) in which the base-\( b \) radix expansion of the fraction \( \frac{1}{8} \) equals \( 0.\overline{18}_{b} \) is: \[ \boxed{13} \]