$3^{123} \mod 100$

You are indeed correct. There is, however, one minor improvement. Using the Carmichael function, you can argue that a smaller power of $3$, namely $3^{\lambda(100)}=3^{20}\equiv 1\bmod 100$. The Carmichael function of divides half the Euler totient function when the argument is even and the Euler totient is a multiple of $4$, which is true for $\lambda(100)$; thus $3^{20}$ can replace $3^{40}$ in the argument.

At a more elementary level, you can render $3^4=80+1$ and raise both sides to the fifth power, thus $3^{20}\equiv1\bmod 100$ as the Binomial Theorem for $(80+1)^5$ gives multiples of $100$ plus $1$.