Exponents and mod (Euler's theorem) I know how to compute $7^{402} \pmod{10}$ using Euler's theorem since $7$ and $10$ are relatively prime.
But is there an easy way without using a calculator to compute $12^{720} \pmod{10}$. I don't think Euler's Theorem can be applied because $12$ and $10$ are not relatively prime...
Also, for $5^{1806} \pmod{63}$, finding $\varphi(63)$ is kinda difficult. Is there an easy way to solve that?

First observe $(12)^{720} \equiv (2)^{720} \pmod {10}$ (I think this is obvious).
Anyways we can easily prove it using binomial theorem on $(2+10)^{270}$
Now, try to find $x$ such that $2^{719} \equiv x \pmod 5$. This is easy by Euler's theorem.
$2^{719} \equiv 3 \pmod 5$.
So, $2^{720} \equiv 6 \pmod {10}$.
For your second question,
$5^{1806} \equiv 125^{602} \equiv (63 \times 2 -1)^{602} \equiv (-1)^{602} \equiv 1 \pmod {63}$.
