When 5⁹⁹ is divided by 13, the remainder is

We have,

$\begin{array}{rl} & 5^{99}=5^3\cdot 5^{96}=\left(125\right)(625{)}^{24}\\ & =[13\times 9+8](1+48\times 13{)}^{24}\\ & =(13\times 9+8)[1{+}^{24}{\mathrm{C}}_1\times (48\times 13){+}^{24}{\mathrm{C}}_2(48\times 13{)}^2+\dots +(48\times 13{)}^{24}]\end{array}$

= 8 + terms containing powers of 13. Hence remainder = 8.