Question

C/m: a) 2^2^2n+1+3 chia hết cho 7.

b) 2^2^6n+1+4 cha hết cho 11.

Giúp mình nhé!

Answer 1

Đặt A= \(2^{2n+1}\)

Ta có:\(2^{2n+1}\)\(⋮\)2

\(2^{2n+1}\)= \(4^n\).2\(\equiv\)2(mod 3)

\(\Rightarrow\)\(\hept{\begin{cases}A⋮2\\A-2⋮3\end{cases}}\)

\(\Rightarrow\)A-2\(⋮\)6

\(\Rightarrow\)A=6k+2

Thay vào:\(2^{2^{2n+1}}\)=\(2^{6k+2}\)\(\equiv\)4(mod 7)

\(2^{2^{2n+1}}\)+3\(\equiv\)4+3(mod7)

                    \(\equiv\)0(mod 7)\(\Rightarrow\)\(2^{2^{2n+1}}\)+3\(⋮\)7