Question

Simplest form please

Answer 1

hmmm what we do is firstly make the recurring part a variable, then we multiply it such that, the recurring digits move over from the decimal point to the left, so we'd multiply it by some power of 10, in this case power of 3, because we have three digits to move, 246, so let's do all that

`0.\overline{246}\hspace{5em}x=0.\overline{246}\hspace{5em} \begin{array}{llll} 1000x&=&246.\overline{246}\\\\ &&246+0.\overline{246}\\\\ &&246+x \end{array} \\\\[-0.35em] ~\dotfill\\\\ 1000x=246+x\implies 999x=246\implies x=\cfrac{246}{999}\implies x=\cfrac{82}{333}`