Reciprocal of 103

Theorem
$103$ is the smallest prime number the period of whose decimal expansion of its reciprocal is $\dfrac 1 3$ of its maximum length, that is: $34$:
 * $\dfrac 1 {103} = 0 \cdotp \dot 00970 \, 87378 \, 64077 \, 66990 \, 29126 \, 21359 \, 223 \dot 3$

Proof
Performing the calculation using long division:

0.0097087378640776699029126213592233009... 103)1.0000000000000000000000000000000000000...     927     618     927     103    309      ---     ---     ---     ---    ---       730     420      300    370     1000       721     412      206    309      927       ---     ---      ---    ---              900     800     940    610    ....         824     721     927    515         ---     ---     ---    ---          760     790     130    950          721     721     103    927          ---     ---     ---    ---           390     690     270    230           309     618     206    206           ---     ---     ---    ---            810     720     640    240            721     618     618    206            ---     ---     ---    ---             890    1020     220    340             824     927     206    309             ---         ---    ---              660     930     140    310              618     927     103    309

By Maximum Period of Reciprocal of Prime, the maximum period of recurrence of the reciprocal of $p$ when expressed in decimal notation is $p - 1$.

Therefore in order for a prime number to have its period of its reciprocal to equal $\dfrac 1 3$ of its maximum length, we must have:
 * $3 \divides p - 1$

The prime numbers less than $103$ with this property are $7, 13, 19, 31, 37, 43, 61, 67, 73, 79, 97$.

We have:
 * Reciprocal of 7
 * Period of Reciprocal of 19 is of Maximal Length
 * Reciprocal of 61
 * Reciprocal of 97

which shows these primes have maximum period $p - 1$.

We have:
 * $\dfrac 1 {13} = 0 \cdotp \dot 07692 \dot 3$: recurring with period $6 = \dfrac {p - 1} 2$.


 * $\dfrac 1 {31} = 0 \cdotp \dot 03225806451612 \dot 9$: recurring with period $15 = \dfrac {p - 1} 2$.


 * $\dfrac 1 {37} = 0 \cdotp \dot 02 \dot 7$: recurring with period $3 = \dfrac {p - 1} {12}$.


 * $\dfrac 1 {43} = 0 \cdotp \dot 02325581395348837209 \dot 3$: recurring with period $21 = \dfrac {p - 1} 2$.


 * $\dfrac 1 {67} = 0 \cdotp \dot 01492537313432835820895522388059 \dot 7$: recurring with period $33 = \dfrac {p - 1} 2$.


 * $\dfrac 1 {73} = 0 \cdotp \dot 0136986 \dot 3$: recurring with period $8 = \dfrac {p - 1} 9$.


 * $\dfrac 1 {79} = 0 \cdotp \dot 012658227848 \dot 1$: recurring with period $13 = \dfrac {p - 1} 6$.

Thus $103$ is the smallest prime number with this property.