Question

Cm: 11ⁿ⁺²+12²ⁿ⁺¹ chia hết cho 133

Answer 1

Lời giải:

Xét modulo $3$ cho $n$ thôi . Ở đây mình xét cụ thể TH $n=3k$. TH \(n=3k+1,3k+2\) ta hoàn toàn làm tương tự

TH1: \(n=3k\)

Ta có :

\(11^3\equiv 1\pmod 7\Rightarrow 11^n=11^{3k}\equiv 1\pmod 7\Rightarrow 11^{n+2}\equiv 11^2\equiv 2\pmod 7\)

\(12^6\equiv 1\pmod 7\Rightarrow 12^{2n}=12^{6k}\equiv 1\pmod 7\Rightarrow 12^{2n+1}\equiv 12\pmod 7\)

\(\Rightarrow 11^{n+2}+12^{2n+1}\equiv 14\equiv 0\pmod 7\) $(1)$

Lại có:

\(11^3\equiv 1\pmod {19}\Rightarrow 11^n=11^{3k}\equiv 1\pmod {19}\Rightarrow 11^{n+2}\equiv 7\pmod {19}\)

\(12^6\equiv 1\pmod {19}\Rightarrow 12^{2n}=12^{6k}\equiv 1\pmod {19}\Rightarrow 12^{2n+1}\equiv 12\pmod {19}\)

\(\Rightarrow 11^{n+2}+12^{2n+1}\equiv 19\equiv 0\pmod {19}\) $(2)$

Từ \((1),(2)\) kết hợp với \((7,19)=1\) suy ra \(11^{n+2}+12^{2n+1}\vdots (7.19=133)\) (đpcm)

Answer 2

11ⁿ⁺²+12²ⁿ⁺¹=121*11ⁿ+12*144ⁿ

=(133-12)*11ⁿ+12*144ⁿ=133*11ⁿ+(144ⁿ-11ⁿ)*12

ta có 133*11ⁿ\(⋮\)133,(144ⁿ-11ⁿ)*12\(⋮\)(144-11)

vậy 11ⁿ⁺²+12²ⁿ⁺¹\(⋮\)133(đpcm)

Answer 3

Xét modulo $3$ cho $n$ thôi . Ở đây mình xét cụ thể TH $n=3k$. TH $n=3k+1,3k+2$ ta hoàn toàn làm tương tự

TH1: $n=3k$

Ta có :

${11}^3\equiv 1\left(\mathrm{mod}7\right)\Rightarrow {11}^n={11}^{3k}\equiv 1\left(\mathrm{mod}7\right)\Rightarrow {11}^{n+2}\equiv {11}^2\equiv 2\left(\mathrm{mod}7\right)$

${12}^6\equiv 1\left(\mathrm{mod}7\right)\Rightarrow {12}^{2n}={12}^{6k}\equiv 1\left(\mathrm{mod}7\right)\Rightarrow {12}^{2n+1}\equiv 12\left(\mathrm{mod}7\right)$

$\Rightarrow {11}^{n+2}+{12}^{2n+1}\equiv 14\equiv 0\left(\mathrm{mod}7\right)$ $\left(1\right)$

Lại có:

${11}^3\equiv 1\left(\mathrm{mod}19\right)\Rightarrow {11}^n={11}^{3k}\equiv 1\left(\mathrm{mod}19\right)\Rightarrow {11}^{n+2}\equiv 7\left(\mathrm{mod}19\right)$

${12}^6\equiv 1\left(\mathrm{mod}19\right)\Rightarrow {12}^{2n}={12}^{6k}\equiv 1\left(\mathrm{mod}19\right)\Rightarrow {12}^{2n+1}\equiv 12\left(\mathrm{mod}19\right)$

$\Rightarrow {11}^{n+2}+{12}^{2n+1}\equiv 19\equiv 0\left(\mathrm{mod}19\right)$ $\left(2\right)$

Từ $\left(1\right),\left(2\right)$ kết hợp với $(7,19)=1$ suy ra ${11}^{n+2}+{12}^{2n+1}⋮(7.19=133)$ (đpcm)

Answer 4

Ta có :\(11^{n+2}+12^{2n+1}=11^n.121+144^n.12\)

\(\equiv\)\(11^n.121+11^n.12\)(mod133)

\(\equiv0\left(mod133\right)\)

Answer 5

\(12^2=144\equiv11\left(mod133\right)\Rightarrow12^{2n}=144^n\equiv11^n\left(mod133\right)\Rightarrow12^{2n+1}=144^n.12\equiv11^n.12\left(mod133\right)\)

\(11^2=121\equiv-12\left(mod133\right)\Rightarrow11^n.11^2=11^{n+2}\equiv\left(-12\right).11^n\left(mod133\right)\)

\(\Rightarrow11^{n+2}+12^{2n+1}\equiv\left(-12\right).11^n+12.11^n\equiv0\left(mod133\right)\Rightarrow11^{n+2}+12^{2n+1}⋮133\)