Period of Reciprocal of 79 is One Sixth of Maximal

Theorem

The decimal expansion of the reciprocal of $79$ has $\dfrac 1 6$ the maximum period, that is: $13$:

$\dfrac 1 {79} = 0 \cdotp \dot 01265 \, 82278 \, 48 \dot 1$

Proof

From Reciprocal of $79$:

$\dfrac 1 {79} = 0 \cdotp \dot 01265 \, 82278 \, 48 \dot 1$

Counting the digits, it is seen that this has a period of recurrence of $13$.

From Maximum Period of Reciprocal of Prime, the maximum period of recurrence of $\dfrac 1 p$ is $p - 1$.

We have that:

$13 = \dfrac {79 - 1} 6$

$\blacksquare$