\(\left(27^{21}-9^{31}-3^{60}\right)\)

\(=\left[\left(3^3\right)^{21}-\left(3^2\right)^{31}-3^{60}\right]\)

\(=\left(3^{63}-3^{62}-3^{60}\right)\)

\(=3^{60}\left(3^3-3^2-3\right)\)

\(=3^{60}.17\)

\(\Rightarrow\left(27^{21}-9^{31}-3^{60}\right)⋮17\)

\(\RightarrowĐPCM\)

\(\left(27^{21}-9^{31}-3^{60}\right)\)

\(=\left(3^3\right)^{21}-\left(3^2\right)^{31}-3^{60}\)

\(=\left(3^{63}-3^{62}-3^{60}\right)\)

\(=3^{60}\left(3^3-3^3-3\right)\)

\(=3^{60}.17\)

\(\Rightarrow\left(27^{21}-9^{31}-3^{60}\right)⋮17\)

Vậy (27²¹ - 9³¹ - 3⁶⁰ ) \(⋮\)17

\(A=17^{18}-17^{16}\\ =17^{16}\cdot\left(17^2-1\right)\\ =17^{16}\cdot\left(289-1\right)\\ =17^{16}\cdot288\\ =17^{16}\cdot18\cdot16⋮18\)

Vậy \(A⋮18\)

\(B=1+3+3^2+...+3^{11}\)

Ta có: \(52=4\cdot13\)

\(B=1+3+3^2+...+3^{11}\\ =\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{10}+3^{11}\right)\\ =1\cdot\left(1+3\right)+3^2\cdot\left(1+3\right)+...+3^{10}\cdot\left(1+3\right)\\ =\left(1+3\right)\cdot\left(1+3^2+...+3^{10}\right)\\ =4\cdot\left(1+3^2+...+3^{10}\right)⋮4\)

Vậy \(B⋮4\)

\(B=1+3+3^2+...+3^{11}\\ =\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^9+3^{10}+3^{11}\right)\\ =1\cdot\left(1+3+3^2\right)+3^3\cdot\left(1+3+3^2\right)+...+3^9\cdot\left(1+3+3^2\right)\\ =\left(1+3+3^2\right)\cdot\left(1+3^3+...+3^9\right)\\ =13\cdot\left(1+3^3+...+3^9\right)⋮13\)

Vậy \(B⋮13\)

Vì \(4\) và \(13\) là hai số nguyên tố cùng nhau nên tao có \(B⋮4\cdot13\Leftrightarrow B⋮52\)

Vậy \(B⋮52\)

\(C=3+3^3+3^5+...3^{31}\)

\(C=3+3^3+3^5+...+3^{31}\\ =\left(3+3^3\right)+\left(3^5+3^7\right)+...+\left(3^{29}+3^{31}\right)\\ =1\cdot\left(3+3^3\right)+3^4\cdot\left(3+3^3\right)+...+3^{28}\cdot\left(3+3^3\right)\\ =\left(3+3^3\right)\cdot\left(1+3^4+...+3^{28}\right)\\ =30\cdot\left(1+3^4+...+3^{28}\right)⋮15\left(\text{vì }30⋮15\right)\)

Vậy \(C⋮15\)

\(D=2+2^2+2^3+...+2^{60}\)

Tao có: \(21=3\cdot7;15=3\cdot5\)

\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\\ =2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\\ =\left(1+2\right)\cdot\left(2+2^3+...+2^{59}\right)\\ =3\cdot\left(2+2^3+...+2^{59}\right)⋮3\)

Vậy \(D⋮3\)

\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^3\right)+\left(2^5+2^7\right)+...+\left(2^{57}+2^{59}\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\\ =2\cdot\left(1+2^2\right)+2^5\cdot\left(1+2^2\right)+...+2^{57}\cdot\left(1+2^2\right)+2^2\cdot\left(1+2^2\right)+...+2^{58}\cdot\left(1+2^2\right)\\ =\left(1+2^2\right)\cdot\left(2+2^5+...+2^{57}+2^2+...+2^{59}\right)\\ =5\cdot\left(2+2^5+...+2^{57}+2^2+...+2^{59}\right)⋮5\)

Vậy \(D⋮5\)

\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\\ =2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+...+2^{58}\cdot\left(1+2+2^2\right)\\ =\left(1+2+2^2\right)\cdot\left(2+2^4+...+2^{58}\right)\\ =7\cdot\left(2+2^4+...+2^{58}\right)⋮7\)

Ta có:

\(D⋮3;D⋮5\Rightarrow D⋮3\cdot5\Leftrightarrow D⋮15\)

\(D⋮3;D⋮7\Rightarrow D⋮3\cdot7\Leftrightarrow D⋮21\)

Vậy \(D⋮15;D⋮21\)

Mình chỉ làm mẫu 1 câu thui nha:

\(A=17^{18}-17^{16}\)

\(A=17^{16}.17^2-17^{16}.1\)

\(A=17^{16}\left(17^2-1\right)\)

\(A=17^{16}.288\)

\(A=17^{16}.16.18\)

\(A⋮18\left(đpcm\right)\)

3^31+9^15+27^11=3^31+3^30+3^33=3^30×(3+1+27)=3^30×31chia hết cho31

Vì n>2 ; n không chia hết cho 3 . Mà 3 là snt =>(n;3)=1.=>n^2 không chia hết cho 3.                                                                              Vì n>2=>n^2-1 lớn hơn hoặc bằng 3.                                                                                                                                                    Sau đó bạn xét số dư n cho 3 rồi c tỏ n^2-1 chia hết cho 3 hoặc n^2+1 chia hết cho 3 nha .

Vì n>2 không chia hết cho 3

=> n² : 3 dư 1 

Nếu (n² - 1) \(⋮\) 3 thì (n² + 1) chia 3 dư 1

Nếu (n² - 1) chia 3 dư 1 thì (n² + 1) \(⋮\)3

=> (n² - 1) và (n² + 1) không thể đồng thời là số nguyên tố được.

S = 1 + 3 + 3^2 + 3^3 + ... + 3^9

S = 1 x 1 + 3 x 1 + 3^2 x 1 + 3^2 x 3 + ... + 3^8 x 1 + 3^8 x 3

S = 1 x (1 + 3) + 3 x (1 + 3) + ... + 3^8 x (1 + 3)

S = 1 x 4 + 3 x 4 + ... + 3^8 x 4

S = 4 x (1 + 3 + ... + 3^8)\(⋮\)4

ta có (1+3)+(3^2+3^3)+...+(3^8+3^9)

=(1+3)+3^2x(1+3)+...+3^8x(1+3)

=4+3^2x4+...+3^8x4

=4x(3^2+...+3^8)

ta thấy 4 chia hết cho 4 nên S chia hết cho 4

kết luận S chia hết cho 4