19. The sequence of functions $F_{1}(x), F_{2}(x), \ldots$ satisfies the following conditions:
$$
F_{1}(x)=x, \quad F_{n+1}(x)=\frac{1}{1-F_{n}(x)} .
$$

The integer $C$ is a three-digit cube such that $F_{C}(C)=C$.
What is the largest possible value of $C$ ?
19. 343
$$
F_{1}(x)=x ; F_{2}(x)=\frac{1}{1-x} ; F_{3}(x)=\frac{1}{1-\frac{1}{1-x}}=1-\frac{1}{x} ; F_{4}(x)=\frac{1}{1-\left(1-\frac{1}{x}\right)}=x .
$$

Hence we have, $F_{1}(x)=F_{4}(x)=F_{7}(x)=\ldots=F_{3 k-2}(x)$.
Therefore we are required to solve $F_{3 k-2}(3 k-2)=3 k-2$ where $3 k-2$ is a threedigit cube (given). The cubes are 125, 216,343, 512, 729. The only one of the format $3 k-2$ for some positive integer $k$ is 343 .