Ta có :

\(M=3^3+3^4+.....+3^{15}+3^{16}\)

\(\Rightarrow M=3^3\left(1+3\right)+......+3^{15}\left(1+3\right)\)

\(\Rightarrow M=3^3.4+......+3^{15}.4\)

=> M chia hết cho 4 .

\(M=\left(3^3+3^5\right)+....+\left(3^{14}+3^{16}\right)\)

\(\Rightarrow M=3^3\left(1+9\right)+.....+3^{14}\left(1+9\right)\)

\(\Rightarrow M=3^3.10+.....+3^{14}.10\)

=> M chia hết cho 10

khó quá

\(M=3^3+3^4+...+3^{15}+3^{16}\)

\(=3^3\times\left(3+1\right)+...+3^{15}\times\left(3+1\right)\)

\(=3^3\times4+...+3^{15}\times4\)

\(=4\times\left(3^3+...+3^{15}\right)⋮4\)

Đặt \(A=3+3^2+3^3+...+3^{15}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{13}+3^{14}+3^{15}\right)\)

\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{13}\left(1+3+3^2\right)\)

\(=3.13+3^4.13+...+3^{13}.13\)

Vì \(13⋮13\)nên \(3.13+3^4.13+...+3^{13}.13⋮13\)

hay \(A⋮13\)

Vậy \(A⋮13.\)

A=3+3^2+3^3+......+3^13+3^14+3^15

=(3+3^2+3^3)+......+(3^13+3^14+3^15)

=3(1+3+3^2)+.......+3^13(1+3+3^2)

=(3+....+3^13)+(1+3+3^2)

=13(3+.....+3^13) chia hết cho 13

M=3³.(1+3)+3⁵.(1+3)+........+3¹⁵.(1+3)

M=4.(3³+3⁵+..............+3¹⁵)

M có thừa số 4 suy ra M chia hết cho 4 

chac co

a) \(4^{13}+4^{14}+4^{15}+4^{16}=4^{13}\left(1+4\right)+4^{14}\left(1+4\right)=4^{13}.5+4^{14}.5=5\left(4^{13}+4^{14}\right)⋮5\Rightarrow dpcm\)

c) \(2^{10}+2^{11}+2^{12}+2^{13}+2^{14}+2^{15}\)

\(=2^{10}\left(1+2+2^2\right)+2^{13}\left(1+2+2^2\right)\)

\(=2^{10}.7+2^{13}.7=7\left(2^{10}+2^{13}\right)⋮7\Rightarrow dpcm\)

Câu c bạn xem lại đê