Bit Tricky(Basic algebra)

Hint :
5⁹⁹
=5.(5²)⁴⁹
=5.(26-1)⁴⁹
Answer will come out to be 8

\hspace{-16}\mathbf{5^{99}=5.(5^2)^{49}=5(1-26)^{49}}$\\\\\\ $\mathbf{=-5.\left\{\binom{49}{0}-\binom{49}{1}.26+\binom{49}{2}.(26)^2-..................-\binom{49}{49}.(26)^{49}\right\}}$\\\\\\ $\mathbf{=-5.\left\{1-M(26)\right\}}$\\\\ Here $\mathbf{M(26)=}$ Multiple of $\mathbf{26}$\\\\\\ So Remainder is $\mathbf{-5=\underbrace{-5-8}+8=8}$\\\\\\ so remainder is $\mathbf{=8}$

aditiya i have not seen your solution.

Nice solution

\hspace{-16}$Using Congruency::\\\\ $\mathbf{5^2=(-1)mod(13)}$\\\\ $\mathbf{(5^2)^{49}=(-1)^{49}mod(13)}$\\\\ $\mathbf{\hspace{+36}=(-1)mod(13)}$\\\\ $\mathbf{5.(5)^{98}=(-5)mod(13)=(8)mod(13)}$\\\\ So remainder is $\mathbf{=8}$

\begin{align*} & \textup{by chinese reamainder theorum}\\ & 5^3 \equiv 8mod(13)\\ & \implies 5^{99} \equiv 8^{33}mod(13)....(1)\\ & 8^2\equiv -1mod(13)\\ & \implies8^{33}\equiv 8mod(13)\\ & \textup{Therefore the remainder is 8.}\\ & \\ & \\ & \\ & \\ & \end{align*}