jmc

The function $f_1(x)=\sqrt{1-x}$ is defined when $x\le 1$. Next, we have $$f_2(x)=f_1(\sqrt{4-x})=\sqrt{1-\sqrt{4-x}}.$$For this to be defined, we must have $4-x\ge 0$ or $x\le 4,$ and the number $\sqrt{4-x}$ must lie in the domain of $f_1,$ so $\sqrt{4-x}\le 1,$ or $x\ge 3.$ Thus, the domain of $f_2$ is $[3,4].$

Similarly, for $f_3(x)=f_2\left(\sqrt{9-x}\right)$ to be defined, we must have $x\le 9,$ and the number $\sqrt{9-x}$ must lie in the interval $[3,4].$ Therefore, $$3\le \sqrt{9-x}\le 4.$$Squaring all parts of this inequality chain gives $9\le 9-x\le 16,$ and so $-7\le x\le 0.$ Thus, the domain of $f_3$ is $[-7,0].$

Similarly, for $f_4(x)=f_3\left(\sqrt{16-x}\right)$ to be defined, we must have $x\le 16,$ and $\sqrt{16-x}$ must lie in the interval $[-7,0].$ But $\sqrt{16-x}$ is always nonnegative, so we must have $\sqrt{16-x}=0,$ or $x=16.$ Thus, the domain of $f_4$ consists of a single point $\{16\}.$

We see, then, that $f_5(x)=f_4\left(\sqrt{25-x}\right)$ is defined if and only if $\sqrt{25-x}=16,$ or $x=25-16^2=-231.$ Therefore, the domain of $f_5$ is $\{-231\}.$

The domain of $f_6(x)$ is empty, because $\sqrt{36-x}$ can never equal a negative number like $-231.$ Thus, $N=5$ and $c=\boxed{-231}.$