2019 AMC 12A Problem 11

The repeating block gives $$0.\overline{23}_k = \dfrac{2k + 3}{k^2 - 1} = \dfrac{7}{51}.$$ Cross-multiplying, $51(2k + 3) = 7(k^2 - 1),$ so $7k^2 - 102k - 160 = 0.$ The quadratic formula gives $$\begin{aligned} k &= \dfrac{102 + \sqrt{14884}}{14} \\ &= \dfrac{102 + 122}{14} = 16. \end{aligned}$$ Thus, the correct answer is D.