Đây

Ta có: \(3^{2n}+3^n+1\)

Vì n không chia hết cho 3 nên: n có dạng là \(3k+1\)

Thế vào: Ta có: \(3^{6k+2}+3^{3k+1}+1\)

\(=729^k\cdot9+27^k\cdot3+1\)

Mặt khác: \(729\equiv27\equiv1\)(mod 13)

Do đó: \(729^k\cdot9+27^k\cdot3+1\equiv1\cdot9+1\cdot3+1=13\)(mod 13)

Vậy .............

P/s: Xét luôn trường hợp \(n=3k+2\)với cách làm tương tự trên

1)

a)2⁵¹-1

=(2³)¹⁷-1\(⋮\)2³-1=7

Vậy 2⁵¹-1\(⋮\)7

b)2⁷⁰+3⁷⁰

=(2²)³⁵+(3²)³⁵\(⋮\)2²+3²=13

Vậy 2⁷⁰+3⁷⁰\(⋮\)13

c)17¹⁹+19¹⁷

=(BS18-1)¹⁹+(BS18+1)¹⁷

=BS18-1+BS18+1

=BS18\(⋮\)18

d)36⁶³-1\(⋮\)35\(⋮\)7

Vậy 36⁶³-1\(⋮\)7

36⁶³-1

=36⁶³+1-2

=BS37-2\(⋮̸\)37

Vậy 36⁶³-1\(⋮̸\)37

e)2⁴ⁿ-1

=(2⁴)ⁿ-1\(⋮\)2⁴-1=15

Vậy 2⁴ⁿ-1\(⋮\)15

Đề bài là tìm n chứ:

a) Ta có:

\(n+5⋮n+2\)

\(\Rightarrow\left(n+2\right)+3⋮n+2\)

\(\Rightarrow3⋮n+2\)

\(\Rightarrow n+2\in U\left(3\right)=\left\{-1;1;-3;3\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}n+2=-1\Rightarrow n=-3\\n+2=1\Rightarrow n=-1\\n+2=-3\Rightarrow n=-5\\n+2=3\Rightarrow n=1\end{matrix}\right.\)

Vậy \(n\in\left\{-3;-1;-5;1\right\}\)

b) Ta có:

\(2n+1⋮n-5\)

\(\Rightarrow\left(2n-10\right)+11⋮n-5\)

\(\Rightarrow2\left(n-5\right)+11⋮n-5\)

\(\Rightarrow11⋮n-5\)

\(\Rightarrow n-5\in U\left(11\right)=\left\{-1;1;-11;11\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}n-5=-1\Rightarrow n=4\\n-5=1\Rightarrow n=6\\n-5=-11\Rightarrow n=-6\\n-5=11\Rightarrow n=16\end{matrix}\right.\)

Vậy \(n\in\left\{4;6;-6;16\right\}\)

c) Ta có:

\(n^2+3n-13⋮n+3\)

\(\Rightarrow n\left(n+3\right)-13⋮n+3\)

\(\Rightarrow-13⋮n+3\)

\(\Rightarrow n+3\in U\left(13\right)=\left\{-1;1;-13;13\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}n+3=-1\Rightarrow n=-4\\n+3=1\Rightarrow n=-2\\n+3=-13\Rightarrow n=-16\\n+3=13\Rightarrow n=10\end{matrix}\right.\)

Vậy \(n\in\left\{-4;-2;-16;10\right\}\)