modular exponentiation $14^{20} \pmod{33}$

Question

How do I find 14^20 mod 33 ?

I tried writing 14^20 as 14^(2+4+8+6) but still no simplification.

Should I just check all the power of 14 mod 33 and hope that some of them give nice numbers (1 or 2)

What is a good method ?

Answer 1

Hint: Consider $14^{20}$ mod $3$ and mod $11$. Use Fermat's theorem.

Answer 2

By Euler's Theorem, as $(14,33)=1$, and $\phi(33)=20$, so $14^{20}\equiv1\pmod{33}$.