How can I calculate $24^{33} \bmod 80$

Question

How can I calculate $24^{33} \mod 80$. I don't see how to start. I only know that $\phi (80) = 32$. A hint would suffice, thanks!

Answer 1

Hint 1: $80=2^4\cdot 5$ and $24^{33}=2^{99}\cdot 3^{33}$.

Hint 2: Chinese Remainder Theorem.

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Solution: $24^{33}\equiv (-1)^{33}\equiv -1\bmod{5}$ and $24^{33}\equiv 0\pmod{16}$, hence by Chinese Remainder Theorem $24^{33}\equiv 64\bmod{80}$.

Answer 2

Applying MDL $ =\, $ mod Distributive Law $ $ to factor out $\,\color{#c00}{c=16}\,$ below yields

$$ \begin{align} \color{#c00}c\:\!a\ \bmod\ \color{#c00}c\cdot n\ &=\ \ \color{#c00}c\ \ (a\bmod n)\qquad\quad\ {\rm [MDL]}\\[.2em] {\rm so}\ \ \ 16\mid 24^{33}\Rightarrow\, 24^{\large 33}\!\bmod {16\cdot 5}\, &=\, \color{#c00}{16}\:(\ 24^{\large 33}/16\ \bmod 5)\\[.2em] &=\, 16\,({(-1)^{\large 33}}\!/1\, \bmod 5)\\[.2em] &=\, \bbox[5px,border:1px solid #0a0]{\!\!\!16\,(4)} \end{align}$$

Note how much simpler this is than using CRT - it can be done purely mentally. As explained in the prior linked answer, MDL is an equivalent operational form of CRT, which proves much more convenient for calculations due to its operational form.