We have seen that the repeating block of $\dfrac{1}{7}$ is a cyclic number. Try to find more numbers (n) whose reciprocals $\left(\dfrac{1}{n}\right)$ produce decimals with repeating blocks that are cyclic.

Numbers n whose reciprocals $\dfrac{1}{n}$ produce fully cyclic repeating blocks are called full reptend primes. These are prime numbers p for which the period of $\dfrac{1}{p}$ has exactly p - 1 digits.

Examples of such numbers :

1. n = 7 : $\dfrac{1}{7} = 0.\overline{142857}$ (6-digit cyclic block).

2. n = 17 : $\dfrac{1}{17} = 0.\overline{0588235294117647}$ (16-digit cyclic block).

3. n = 19 : $\dfrac{1}{19} = 0.\overline{052631578947368421}$ (18-digit cyclic block).

4. n = 23 : $\dfrac{1}{23} = 0.\overline{0434782608695652173913}$ (22-digit cyclic block).

5. n = 29 : $\dfrac{1}{29}$ has a 28-digit cyclic block.

6. Other examples : 47, 59, 61, 97, 109, etc.

The cyclic property : If n is a full reptend prime, then multiplying the repeating block of $\dfrac{1}{n}$ by integers 1, 2, …, n - 1 gives cyclic permutations of the same digits.

For example, for n = 7 :

⇒ 142857 × 1 = 142857

⇒ 142857 × 2 = 285714

⇒ 142857 × 3 = 428571 (and so on, all permutations of the same digits).

Hence, prime numbers like 7, 17, 19, 23, 29, 47, 59, 61, 97, etc. (full reptend primes) produce reciprocals with cyclic repeating blocks.