The remainder when $5^{99}$ is divided by 13 is

Given number is, $5^{99}$

It is divided by the number 13.

It can be given as,

$5^{99}=5(26-1{)}^{49}=5(26m-1)$ where $m$ is some integer.

$\begin{array}{l}{5}^{99}=5(5{)}^{98}=5(25{)}^{49}\\ =5(26-1{)}^{49}\\ =5\left[{ }^{49}{C}_0\left(26{)}^{49}{-}^{49}{C}_1\right(26{)}^{48}{+}^{49}{C}_2(26{)}^{47}+\dots -1\right]\\ =5[26k-1],k\in Z\\ \text{Here,}\\ \frac{5^{99}}{13}=\frac{5[26k-1]}{13}\\ =10k-\frac{5}{13}=10k-1+\frac{8}{13}\end{array}$

So, we get the 8 as reminder.