Question

# Use modulo reduction of the base, if applicate, and the power decomposition calculations on congruences, to calculate the least non-negative residue (or equivalently the congruence class) of the folowing: 8052^403 mod 5

Congruence
Use modulo reduction of the base, if applicate, and the power decomposition calculations on congruences, to calculate the least non-negative residue (or equivalently the congruence class) of the folowing:
$$\displaystyle{8052}^{{403}}\text{mod}{5}$$

2020-11-09
Given that $$\displaystyle{8052}^{{403}}$$
This can be rewritten as,
$$\displaystyle{8052}^{{403}}\equiv{\left({8050}+{2}\right)}^{{403}}$$
$$\displaystyle\equiv{2}^{{403}}\text{mod}{5}$$
So, the equation can be rewritten as,
$$\displaystyle{8052}^{{403}}\equiv{2}^{{402}}{\left({2}\right)}\text{mod}{5}$$
$$\displaystyle\equiv{4}^{{201}}{\left({2}\right)}\text{mod}{5}$$
$$\displaystyle={\left({5}-{1}\right)}^{{201}}{\left({2}\right)}\text{mod}{5}$$
$$\displaystyle\equiv-{1}^{{201}}{\left({2}\right)}\text{mod}{5}$$
$$\displaystyle\equiv-{2}\text{mod}{5}$$
$$\displaystyle\equiv{\left({5}-{2}\right)}\text{mod}{5}$$
$$\displaystyle\equiv{3}\text{mod}{5}$$