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# What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $$\frac{1}{17}$$ ? Perform the division to check your answer.

We know that, the maximum number of digits in the repeating block of digits in the decimal expansion of $$\frac{1}{17}$$ is 17 – 1 = 16. $$\require{enclose} \begin{array}{rll} 0.0588235294117647 && \hbox{(Explanations)} \\[-3pt] 17 \enclose{longdiv}{100}\kern-.2ex \\[-3pt] \underline{85\phantom{00}} && \hbox{(5 \times 17 = 85)} \\[-3pt] \phantom{0}150 && \hbox{(100 - 85 = 15)} \\[-3pt] \underline{136\phantom{00}} && \hbox{(8 \times 17 = 136)} \\[-3pt] \phantom{0}140 && \hbox{(150 - 136 = 14)} \\[-3pt] \underline{136\phantom{00}} && \hbox{(8 \times 17 = 136)} \\[-3pt] \phantom{0}40 && \hbox{(140 - 136 = 4)} \\[-3pt] \underline{34\phantom{00}} && \hbox{(2 \times 17 = 34)} \\[-3pt] \phantom{0}60 && \hbox{(40 - 34 = 6)} \\[-3pt] \underline{51\phantom{00}} && \hbox{(3 \times 17 = 51)} \\[-3pt] \phantom{0}90 && \hbox{(60 - 51 = 9)} \\[-3pt] \underline{85\phantom{00}} && \hbox{(5 \times 17 = 85)} \\[-3pt] \phantom{0}50 && \hbox{(90 - 85 = 5)} \\[-3pt] \underline{34\phantom{00}} && \hbox{(2 \times 17 = 34)} \\[-3pt] \phantom{0}160 && \hbox{(50 - 34 = 16)} \\[-3pt] \underline{153\phantom{00}} && \hbox{(9 \times 17 = 153)} \\[-3pt] \phantom{0}70 && \hbox{(160 - 153 = 7)} \\[-3pt] \underline{68\phantom{00}} && \hbox{(4 \times 17 = 68)} \\[-3pt] \phantom{0}20 && \hbox{(70 - 68 = 2)} \\[-3pt] \underline{17\phantom{00}} && \hbox{(1 \times 17 = 17)} \\[-3pt] \phantom{0}30 && \hbox{(20 - 17 = 3)} \\[-3pt] \underline{28\phantom{00}} && \hbox{(2 \times 17 = 28)} \\[-3pt] \phantom{0}130 && \hbox{(30 - 17 = 13)} \\[-3pt] \underline{119\phantom{00}} && \hbox{(7 \times 17 = 119)} \\[-3pt] \phantom{0}110 && \hbox{(130 - 119 = 11)} \\[-3pt] \underline{102\phantom{00}} && \hbox{(6 \times 17 = 136)} \\[-3pt] \phantom{0}80 && \hbox{(110 - 102 = 8)} \\[-3pt] \underline{68\phantom{00}} && \hbox{(4 \times 17 = 68)} \\[-3pt] \phantom{0}40 && \hbox{(80 - 68 = 120)} \\[-3pt] \underline{\phantom{0}119} && \hbox{(7 \times 400 = 119)} \\[-3pt] \phantom{00}1 \end{array}$$ We have, Thus, $$\frac{1}{17}$$= $$= 0\overline{.0588235294117647}$$