Question

cmr q(x)=\(10^{6x+2}+10^{3x+1}+1⋮91\forall x\in N\), x lẻ

Answer 1

Lời giải:

Đặt \(x=2t+1\). Khi đó, \(q(x)=10^{6x+2}+10^{6t+4}+1\)

Ta thấy: \(10^6\equiv 1\pmod {91}\). Do đó:

\(\left\{\begin{matrix} 10^{6k}\equiv 1\pmod {91}\\ 10^{6t}\equiv 1\pmod {91}\end{matrix}\right.\)

\(\Rightarrow q(x)\equiv 10^2+10^4+1\equiv 10101\equiv 0\pmod {91}\)

Do đó, \(q(x)\vdots 91\) với \(x\in\mathbb{N}\) lẻ.