A well presented solution from Richard of The Royal Hospital School reflected those of a number of other solvers including Kevin of Langley Grammar, Jeff from New Zealand and Andrei of Tudor Vianu School. Well done to all of you.


We are given that:
F0(x)
=
1 / (1 - x)
Fn(x)
=
F0 (Fn-1(x))
This implies that:
Fn(x)
=
1 / (1 - Fn-1(x))
and can be extended to:
Fn(x)
=
1 / (1 - (1 / (1 - Fn-2(x))))
Creating functions of x when n = 1, 2 and 3 gives:
F1(x)
=
1 / (1 - F0(x))
=
1 / (1 - (1 / (1 - x)))
=
1 / ((1 - x - 1) / (1 - x))
=
1(1 - x) / -x
=
(1 - x) / -x
F2(x)
=
1 / (1 - F1(x))
=
1 / (1 - ((1 - x) / -x))
=
1 / ((-x - 1 + x) / -x)
=
-x / -1
=
x
F3(x)
=
1 / (1 - F2(x))
=
1 / (1 - x)
=
F0(x)
Therefore the function of x will repeat itself every three times.
F0(x)
=
F3(x)
F1(x)
=
F4(x)
Etc. etc. So, to find F2000(x), we must find the remainder given when 2000 is divided by three.


Mod3 2000 = 2
Thus, F2000(x) = F2(x), and F2(x) = x

Therefore:
F22000 = 2000