, 2015

Solution:
For convenience, let $n=101$. Compute the number of functions such that $f^{n}(1)=1$. Since $n$ is a prime, there are 2 cases: the order of $1$ is either $1$ or $n$. The first case gives $n^{n-1}$ functions, and the second case gives $(n-1)!$ functions. By symmetry, the number of ways for $f^{n}(1)=2$ is
$$\frac{1}{n-1} \cdot \left(n^{n}-n^{n-1}-(n-1)!\right) = n^{n-1}-(n-2)!$$
Plugging in $n=101$, we need to find
$$\begin{gathered} 101^{100}-99! \equiv (-2)^{-2} - \frac{101!}{6} \\ = 1/4 - 1/6 = 1/12 = 43 \quad (\bmod 103). \end{gathered}$$