Recurring Parts of Multiples of One Thirteenth

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Theorem

The multiples of $\dfrac 1 {13}$ from $\dfrac 1 {13}$ to $\dfrac {12} {13}$ can be divided into two sets of equal size:

one where the digits of the recurring part consists of a cyclic permutation of $076923$
one where the digits of the recurring part consists of a cyclic permutation of $153846$.
ReciprocalOf13Cycles.png


Proof

\(\ds \dfrac 1 {13}\) \(=\) \(\ds 0 \cdotp 076923 \, 076923 \ldots\)
\(\ds \dfrac 2 {13}\) \(=\) \(\ds 0 \cdotp 153846 \, 153846 \ldots\)
\(\ds \dfrac 3 {13}\) \(=\) \(\ds 0 \cdotp 230796 \, 230796 \ldots\)
\(\ds \dfrac 4 {13}\) \(=\) \(\ds 0 \cdotp 307692 \, 307692 \ldots\)
\(\ds \dfrac 5 {13}\) \(=\) \(\ds 0 \cdotp 384615 \, 384615 \ldots\)
\(\ds \dfrac 6 {13}\) \(=\) \(\ds 0 \cdotp 461538 \, 461538 \ldots\)
\(\ds \dfrac 7 {13}\) \(=\) \(\ds 0 \cdotp 538461 \, 538461 \ldots\)
\(\ds \dfrac 8 {13}\) \(=\) \(\ds 0 \cdotp 615384 \, 615384 \ldots\)
\(\ds \dfrac 9 {13}\) \(=\) \(\ds 0 \cdotp 692307 \, 692307 \ldots\)
\(\ds \dfrac {10} {13}\) \(=\) \(\ds 0 \cdotp 769230 \, 769230 \ldots\)
\(\ds \dfrac {11} {13}\) \(=\) \(\ds 0 \cdotp 846153 \, 846153 \ldots\)
\(\ds \dfrac {12} {13}\) \(=\) \(\ds 0 \cdotp 923076 \, 923076 \ldots\)

$\blacksquare$


Also see


Sources

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