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, \(\frac{1}{17} \)= \(= 0\overline{.0588235294117647}\)
i.e.a block of 16 digits is repeated.