Question

What is the remainder when $5^{99}$ is divided by $13.$

Answer 1

$5^{99}={5.5}^{98}=5.{\left(25\right)}^{49}=5{(26-1)}^{49}$
$=5[{}^{49}{\text{C}}_0{\left(26\right)}^{49}{-}^{49}{\text{C}}_1{\left(26\right)}^{48}+...{+}^{49}{\text{C}}_{48}{\left(26\right)}^1{-}^{49}{\text{C}}_{49}{\left(26\right)}^0]$
$=5[{}^{49}{\text{C}}_0{\left(26\right)}^{49}{-}^{49}{\text{C}}_1{\left(26\right)}^{48}+...{+}^{49}{\text{C}}_{48}{\left(26\right)}^1-1]$
$=13\left(k\right)-5$ (where k is a positive integer)
$=13\left(k_1\right)+8$
Hence, remainder is $8.$