Respostas

  • Usuário do Brainly
2013-05-30T23:27:44-03:00

Pelo Algoritmo do MDC de Euclides:

 

\text{mdc}(\text{a}, \text{b})=\text{mdc}(\text{a}, \text{b}-\text{a}\times\text{c})=\text{mdc}(\text{a}, \text{r}),

 

onde, \text{r}=\text{b}-\text{a}\times\text{c}

 

Desta maneira, podemos afirmar que:

 

\text{mdc}(2^{100}-1, 2^{20}-1)=\text{mdc}(2^{20}-1, 2^{100}-1-(2^{20}-1)\times2^{80}

 

=\text{mdc}(2^{20}-1, 2^{80}-1)=\text{mdc}(2^{20}-1, 2^{80}-1-(2^{20}-1)\times2^{60}

 

=\text{mdc}(2^{20}-1. 2^{60}-1)=\text{mdc}(2^{20}-1, 2^{80}}-1-(2^{20}-1)\times2^{40}

 

=\text{mdc}(2^{20}-1, 2^{40}-1)=\text{mdc}(2^{20}-1, 2^{40}-1-(2^{20}-1)\times2^{20}

 

=\text{mdc}(2^{20}-1, 2^{20}-1)=\text{mdc}(2^{20}-1, 2^{20}-1-(2^{20}-1)

 

=\text{mdc}(2^{20}-1,0)=2^{20}-1

 

Logo, chegamos à conclusão de que:

 

\text{mdc}(2^{100}-1, 2^{20}-1)=2^{20}-1

 

Alternativa E

2013-05-30T23:35:36-03:00

Olá !! veja a solução no anexo!!!

 

espero que goste !!