Question

[Thái Gia Bảo]

chứng tỏ:B= 3 + 3² + 3³ + 3⁴+ ... +3¹⁹⁹¹chia hết cho 13

              B= 3 + 3² + 3³ + 3⁴ + ... +3¹⁹⁹¹chia hết cho 41

Answer 1

[Akai Haruma]

Lời giải:

$B=3+3^2+(3^3+3^4+3^5)+(3^6+3^7+3^8)+....+(3^{1989}+3^{1990}+3^{1991})$

$=12+3^3(1+3+3^2)+3^6(1+3+3^2)+...+3^{1989}(1+3+3^2)$

$=12+(1+3+3^2)(3^3+3^6+...+3^{1989})$

$=12+13(3^3+3^6+...+3^{1989})$

$\Rightarrow B$ chia $13$ dư $12$.

Answer 2

[Akai Haruma]

2/

$B=3+3^2+3^3+...+3^{1991}$

$3B=3^2+3^3+3^4+...+3^{1992}$
$\Rightarrow 3B-B=3^{1992}-3$

$\Rightarrow 2B=3^{1992}-3$

Có:

$3^4\equiv -1\pmod {41}$

$\Rightarrow 3^{1992}=(3^4)^{498}\equiv (-1)^{498}\equiv 1\pmod {41}$

$\Rightarrow 3^{1992}-3\equiv 1-3\equiv -2\pmod {41}$

$\Rightarrow 2B\equiv -2\pmod {41}$

$\Rightarrow 2B\not\vdots 41$

$\Rightarrow B\not\vdots 41$.