The last digit of $(3^p+2)$ is, where $P=3^{4n}\text{\hspace{0.17em}and\hspace{0.17em}}n\in N$

To find last digit of $(3^P+2)$, where $P=3^{4n}$

As, $3^{4n}$ where $n\in N$

$\Rightarrow {\left(81\right)}^n={(80+1)}^n=80\lambda +1$

$\therefore 3^P+2=3^{80\lambda +1}+2={\left(3^4\right)}^{20\lambda }\cdot 3+2=3\cdot {\left(81\right)}^{20\lambda }+2$

$$\begin{gathered} =3{[80+1]}^{20\lambda }+2 \\ \begin{array}{l}=3[80k+1]+2\\ =240k+3+2=240k+5\end{array} \end{gathered}$$

$\therefore \text{\hspace{0.17em}Last\hspace{0.17em}digit\hspace{0.17em}of\hspace{0.17em}}{3}^P+2$, is 5.