What is the remainder when $5^{30}$ is divided by 7?
I know that $5^6 \equiv 1 \pmod{7}$ by Fermat's little theorem, since 7 is prime and 5 is not a multiple of 7. So, I can write $5^{30} = (5^6)^5 \times 5^0$. Then, I can use the property that if $a \equiv b \pmod{n}$ and $c \equiv d \pmod{n}$, then $ac \equiv bd \pmod{n}$. This means that $(5^6)^5 \times 5^0 \equiv (1)^5 \times 1 \pmod{7}$. So, the remainder is 1. 