Lời giải:
Đặt $A=10^n+18^n$.
Nếu $n=0$ thì $A$ chia $27$ dư $2$
Nếu $n=1$ thì $A=28$ chia $27$ dư $1$
Nếu $n\geq 2$. Xét các TH sau
TH1: Nếu $n=3k$ ( $k\in\mathbb{N} >1$)
Có \(10^{3}\equiv 1\pmod {27}\Rightarrow 10^n=(10^3)^k\equiv 1\pmod {27}\)
\(18^n=18^{3k}\equiv (-9)^{3k}\equiv 0\pmod{27}\)
\(\Rightarrow A\equiv 1\pmod{27}\), tức $A$ chia $27$ dư $1$
TH2: $n=3k+1$ ( $k\in\mathbb{N} >1$)
\(10^{n}=10^{3k+1}=10^{3k}.10\equiv 1.10\equiv 10\pmod {27}\)
\(18^{n}=18^{3k+1}\equiv (-9)^{3k+1}\equiv 0\pmod{27}\)
\(\Rightarrow A\equiv 10\pmod{27}\)
TH3: $n=3k+2$
\(10^{n}=10^{3k+2}=10^{3k}.100\equiv 100\equiv 19\pmod{27}\)
\(18^n=18^{3k+2}\equiv (-9)^{3k+2}\equiv 0\pmod {27}\)
\(\Rightarrow A\equiv 19\pmod {27}\)