We have seen that the repeating block of $\frac{1}{7}$ is a cyclic number.

Try to find more numbers ($n$) whose reciprocals $\frac{1}{n}$ produce decimals with repeating blocks that are cyclic.
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Reciprocals of so-called 'full-reptend primes' produce cyclic numbers. These are primes $p$ for which the repeating block of $\frac{1}{p}$ has length exactly $p - 1$ .

Examples (besides 7):

$\frac{1}{17}=0.\overline{0588235294117647}\rightarrow16\text{-digit cyclic block}$

$\frac{1}{19} = 0.\overline{0526315789473684211} \rightarrow 18\text{-digit}$ cyclic block

$\frac{1}{23} = 0.\overline{0434782608695652173913} \rightarrow$ 22-digit cyclic block

$\frac{1}{29}, \frac{1}{47}, \frac{1}{59}, \frac{1}{61}, \frac{1}{97}$ are further examples.

For such primes $p$ , the blocks for $\frac{2}{p}, \frac{3}{p}, \dots, \frac{p-1}{p}$are cyclic permutations of the block for $\frac{1}{p}$ (as we saw with $\frac{1}{7}$ ).