Question

sa se determine restul impartirii nr. 2 la puterea 2017 la 3.

Answer 1

[tex]
\\ 2^{2n}=(2^{2})^{n}=4^{n}=(3+1)^{n}=M_{3}+1^{n}=M_{3}+1\\
\\ \Rightarrow restul~impartirii~lui~2^{2n}~la~3~este~1\\
\\ 2^{2n+1}=2^{2n}\cdot 2^{1}=(M_{3}+1)\cdot 2=M_{3}+2\\
\\ \Rightarrow  restul~impartirii~lui~2^{2n+1}~la~3~este~2\\
\\ \\ 2^{2017}=2^{2 \cdot 1008+1}=M_{3}+2 \Rightarrow restul~impartirii~lui~2^{2017}~la~3~este~2\\
\\ \\[/tex]