Matemática, perguntado por arendtarthur986, 11 meses atrás

Determine o maximo divisor comum dos seguintes números: a)15 e 30 b)16 e 36 c)21 e 35 d)75 e 120 e)10,32 e 56.

Soluções para a tarefa

Respondido por CyberKirito
0

\large\boxed{\begin{array}{l}\rm a)\\\begin{array}{c|c}\sf15,30&\sf2\\\sf15,15&\boxed{\sf3}\\\sf5,5&\boxed{\sf5}\\\sf1,1\end{array}\\\sf  MDC(15,30)=3\cdot5=15\\\rm b)\\\begin{array}{c|c}\sf16,36&\boxed{\sf2}\\\sf8,18&\boxed{\sf2}\\\sf4,9&\sf2\\\sf2,9&\sf2\\\sf1,9&\sf3\\\sf1,3&\sf3\\\sf1,1\end{array}\\\sf MDC(16,36)=2\cdot2=4\\\rm c)\\\begin{array}{c|c}\sf21,35&\sf3\\\sf7,35&\sf5\\\sf7,7&\boxed{\sf7}\\\sf1,1\end{array}\\\sf MDC(21,35)=7\end{array}}

\large\boxed{\begin{array}{l}\rm d)\\\begin{array}{c|c}\sf75,120&\sf2\\\sf75,60&\sf2\\\sf75,30&\sf2\\\sf75,15&\boxed{\sf3}\\\sf25,5&\boxed{\sf5}\\\sf5,1&\sf5\\\sf1,1\end{array}\\\sf MDC(75,120)=3\cdot5=15\\\rm e)\\\begin{array}{c|c}\sf10,32,56&\boxed{\sf2}\\\sf5,16,28&\sf2\\\sf5,8,14&\sf2\\\sf5,4,7&\sf2\\\sf5,2,7\7\sf&\sf2\\\sf5,1,7&\sf5\\\sf1,1,7&\sf7\\\sf1,1,1\end{array}\\\sf MDC(10,32,56)=2\end{array}}

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