2011 AMC 12A Problem 21

Each step requires $\sqrt{n^2 - x}$ to lie in the domain of $f_{n-1}.$ Tracking the domains: $f_1: (-\infty, 1].$ $f_2: \sqrt{4 - x} \in (-\infty, 1] \Rightarrow [3, 4].$ $f_3: \sqrt{9 - x} \in [3, 4] \Rightarrow [-7, 0].$ $f_4: \sqrt{16 - x} \in [-7, 0] \Rightarrow \{16\}$ (only the value $0$ is possible). $f_5: \sqrt{25 - x} = 16 \Rightarrow \{-231\}.$ For $f_6$ we would need $\sqrt{36 - x} = -231,$ impossible, so the domain is empty. Hence $N = 5,$ $c = -231,$ and $N + c = -226.$ Thus, the correct answer is A.