# Solution

Find $23^{534} \pmod{29}$ using Fermat's Little Theorem.

Note, $29$ is prime and $29 \not\mid 23$, so Fermat's Little Theorem guarantees that $29^{28} \equiv 1\pmod{29}$, so...

$$23^{534} = (23^{28})^{19} \cdot 23^2 \equiv 1 \cdot 23^2 \equiv 529 \equiv 7 \pmod{29}$$