Question 2

Step 1 · Perform Long Division for $\dfrac{1}{13}$

Perform long division of $1$ by $13$:

The remainder $1$ repeats after $6$ division steps.

$\dfrac{1}{13} = 0.076923076923\dots = 0.\overline{076923}$

The repeating block of digits is $076923$.

Step 2 · Evaluate $\dfrac{2}{13}$

Computing the decimal expansion for $\dfrac{2}{13}$ by long division:

$\dfrac{2}{13} = 0.153846153846\dots = 0.\overline{153846}$

The repeating block is $153846$.

This is a completely different block from $076923$, meaning $\dfrac{2}{13}$ is not a cyclic permutation of $\dfrac{1}{13}$.

Step 3 · Compute $\dfrac{3}{13}$, $\dfrac{4}{13}$, $\dfrac{5}{13}$, and $\dfrac{6}{13}$

Finding the decimal expansions for the remaining fractions:

$\dfrac{3}{13} = 0.230769230769\dots = 0.\overline{230769}$

$\dfrac{4}{13} = 0.307692307692\dots = 0.\overline{307692}$

$\dfrac{5}{13} = 0.384615384615\dots = 0.\overline{384615}$

$\dfrac{6}{13} = 0.461538461538\dots = 0.\overline{461538}$

Step 4 · Observe Cyclic Patterns

Comparing all the repeating blocks:

• $\dfrac{1}{13} = 0.\overline{076923}$
• $\dfrac{2}{13} = 0.\overline{153846}$
• $\dfrac{3}{13} = 0.\overline{230769}$
• $\dfrac{4}{13} = 0.\overline{307692}$
• $\dfrac{5}{13} = 0.\overline{384615}$
• $\dfrac{6}{13} = 0.\overline{461538}$

The decimal expansions form two distinct cyclic families:

1. Family 1 (Digits $\{0, 7, 6, 9, 2, 3\}$):

   • $\dfrac{1}{13} = 0.\overline{\mathbf{0}76923}$
   • $\dfrac{3}{13} = 0.\overline{\mathbf{2}30769}$
   • $\dfrac{4}{13} = 0.\overline{\mathbf{3}07692}$

2. Family 2 (Digits $\{1, 5, 3, 8, 4, 6\}$):

   • $\dfrac{2}{13} = 0.\overline{\mathbf{1}53846}$
   • $\dfrac{5}{13} = 0.\overline{\mathbf{3}84615}$
   • $\dfrac{6}{13} = 0.\overline{\mathbf{4}61538}$