Question

[Nguyễn Thanh Tâm]

 GIÚP EM NHANHVỚI Ạ,EM CẢM ƠN
(SỬ DỤNG ĐỒNG DƯ NHÉ Ạ)

Answer 1

[Akai Haruma]

Bài 8:

$A=2^{2n+1}+3^{2n+1}=4^n.2+9^n.3$

$\equiv (-1)^n.2+(-1)^n.3\pmod 5$

Nếu $n$ chẵn:

$A\equiv (-1)^n.2+(-1)^n.3\equiv 2+3\equiv 5\equiv 0\pmod 5$

$\Rightarrow A\vdots 5$

Nếu $n$ lẻ:

$A\equiv (-1)^n.2+(-1)^n.3\equiv -2+(-3)\equiv -5\equiv 0\pmod 5$

$\Rightarrow A\vdots 5$

Vậy $A$ chia hết cho $5$

Answer 2

[Akai Haruma]

Bài 9:

Có: $2^5=32\equiv 1\pmod {31}$

$\Rightarrow 2^{2002}=(2^5)^{400}.2^2\equiv 1^{400}.2^2\equiv 4\pmod {31}$

$\Rightarrow 2^{2002}-4\equiv 0\pmod {31}$

$\Rightarrow 2^{2002}-4\vdots 31$