Harvard-MIT Math Tournament

Solution:

We know $f(a) = n^{2} - f\left(2^{n} - a\right)$ for any $a$, $n$ with $2^{n} > a$; repeated application gives
$$\begin{gathered} f(2002) = 11^{2} - f(46) = 11^{2} - \left(6^{2} - f(18)\right) = 11^{2} - \left(6^{2} - \left(5^{2} - f(14)\right)\right) \\ = 11^{2} - \left(6^{2} - \left(5^{2} - \left(4^{2} - f(2)\right)\right)\right) . \end{gathered}$$
But $f(2) = 2^{2} - f(2)$, giving $f(2) = 2$, so the above simplifies to $11^{2} - \left(6^{2} - \left(5^{2} - \left(4^{2} - 2\right)\right)\right) = 96$.