What would be the remainder ???

$\begin{aligned} 3^{1}\mod5 = 3 \\ 3^{2}\mod5 = 4 \\ 3^{3}\mod5 = 2 \\ 3^{4}\mod5 = 1 \\ 3^{5}\mod5 = 3 \\ 3^{6}\mod5 = 4 \\ 3^{7}\mod5 = 2 \\ 3^{8}\mod5 = 1 \\ \cdots \\ 3^{55}\mod5 = 2 \\ \end{aligned}$

You can solve this question by using the concept of cyclicity

We know cyclicity of 3 is 4

So $\frac{55}{4}$ = remainder 3

$\frac{3³}{5}$

$\frac{27}{5}$

So the remainder is 2