# 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

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