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
ANSWERED
asked 2020-11-08
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:
\(\displaystyle{8052}^{{403}}\text{mod}{5}\)

Answers (1)

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,
\(\displaystyle{8052}^{{403}}\equiv{2}^{{402}}{\left({2}\right)}\text{mod}{5}\)
\(\displaystyle\equiv{4}^{{201}}{\left({2}\right)}\text{mod}{5}\)
\(\displaystyle={\left({5}-{1}\right)}^{{201}}{\left({2}\right)}\text{mod}{5}\)
\(\displaystyle\equiv-{1}^{{201}}{\left({2}\right)}\text{mod}{5}\)
\(\displaystyle\equiv-{2}\text{mod}{5}\)
\(\displaystyle\equiv{\left({5}-{2}\right)}\text{mod}{5}\)
\(\displaystyle\equiv{3}\text{mod}{5}\)
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