Respostas

A melhor resposta!
2013-05-30T23:36:51-03:00

Olá!!! vaja a solução no anexo!!espero que goste!!!

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  • Usuário do Brainly
2013-05-31T11:13:29-03:00

Pelo Algoritmo do MDC de Euclides, temos 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^{60}-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})

 

=\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