Lời giải:
Ta xét các TH sau:
TH1: $n$ chia hết cho $3$: $n=3k$ với $k\in\mathbb{N}$
\(10^n+18n-28=10^{3k}+18.3k-28\)
Ta thấy:
\(10^3\equiv 1\pmod {27}\Rightarrow 10^{3k}\equiv 1^k\equiv 1\pmod {27}\)
\(18.3k=27.2k\equiv 0\pmod {27}\)
\(28\equiv 1\pmod {27}\)
\(\Rightarrow 10^n+18n-28\equiv 1+0-1\equiv 0\pmod {27}(1)\)
TH2: $n$ chia 3 dư $1$: $n=3k+1$ với $k\in\mathbb{N}$
\(10^n+18n-28=10^{3k+1}+18(3k+1)-28=10^{3k}.10+54k-10\)
Ta thấy:
\(10^{3k}\equiv 1\pmod {27} \) (cmt) \(\Rightarrow 10^{3k}.10\equiv 10\pmod {27}\)
\(54k\equiv 0\pmod {27}\)
\(10\equiv 10\pmod {27}\)
\(\Rightarrow 10^n+18n-28\equiv 10-0-10\equiv 0\pmod {27}(2)\)
TH3: $n$ chia 3 dư $2$: $n=3k+2$
\(10^n+18n-28=10^{3k}.100+54k+8\equiv 100+0+8\equiv 0\pmod {27}(3)\)
Từ (1);(2);(3) suy ra $10^n+18n-28$ chia hết cho $27$ với mọi số tự nhiên $n$
Lời giải:
Ta xét các TH sau:
TH1: $n$ chia hết cho $3$: $n=3k$ với $k\in\mathbb{N}$
\(10^n+18n-28=10^{3k}+18.3k-28\)
Ta thấy:
\(10^3\equiv 1\pmod {27}\Rightarrow 10^{3k}\equiv 1^k\equiv 1\pmod {27}\)
\(18.3k=27.2k\equiv 0\pmod {27}\)
\(28\equiv 1\pmod {27}\)
\(\Rightarrow 10^n+18n-28\equiv 1+0-1\equiv 0\pmod {27}(1)\)
TH2: $n$ chia 3 dư $1$: $n=3k+1$ với $k\in\mathbb{N}$
\(10^n+18n-28=10^{3k+1}+18(3k+1)-28=10^{3k}.10+54k-10\)
Ta thấy:
\(10^{3k}\equiv 1\pmod {27} \) (cmt) \(\Rightarrow 10^{3k}.10\equiv 10\pmod {27}\)
\(54k\equiv 0\pmod {27}\)
\(10\equiv 10\pmod {27}\)
\(\Rightarrow 10^n+18n-28\equiv 10-0-10\equiv 0\pmod {27}(2)\)
TH3: $n$ chia 3 dư $2$: $n=3k+2$
\(10^n+18n-28=10^{3k}.100+54k+8\equiv 100+0+8\equiv 0\pmod {27}(3)\)
Từ (1);(2);(3) suy ra $10^n+18n-28$ chia hết cho $27$ với mọi số tự nhiên $n$
