How to find remainder modulo $n$, when $n$ is a large number I am doing RSA questions and I really could use help! Can someone show me a simple way to find $25^9 \pmod{33}$?

For variety, let's invoke the Chinese remainder theorem
$$ 25^9 \equiv 1^9 \equiv 1 \pmod 3 $$
$$ 25^9 \equiv 3^9 \pmod{11}$$
That latter one isn't so bad to compute by repeated squaring:
$$ 3^9 \equiv 3 \cdot 3^8 \equiv 3 \cdot 9^4 \equiv 3 \cdot 81^2 \equiv 3 \cdot 4^2 \equiv 3 \cdot 5 \equiv 4 \pmod{11}$$
but I'm even lazier. Since $3^{10} \equiv 1 \pmod{11}$ (because $11$ is prime):
$$ 3^9 \equiv 3^{-1} \equiv 1/3 \equiv 12/3 \equiv 4 \pmod{11} $$
The extended Euclidean algorithm (or inspection) lets us compute
$$ 4 \cdot 3 + (-1) \cdot 11 = 1 $$
and therefore
$$ 25^9 \equiv 1 \cdot (-1 \cdot 11) + 4 \cdot (4 \cdot 3)
\equiv -11 + 48 \equiv -11 + 15 \equiv 4 \pmod{33} $$
