Period of Reciprocal of 31 is of Odd Length

Theorem
The decimal expansion of the reciprocal of $31$ has an odd period, that is: $15$:
 * $\dfrac 1 {31} = 0 \cdotp \dot 03225 \, 80645 \, 1612 \dot 9$

It is the smallest prime number to have an odd period greater than $1$.

Proof
Performing the calculation using long division:

0.03225806451612903... --- 31)1.00000000000000000000     93     155     --     ---      70      50      62      31      --      --       80     190       62     186       --     ---       180      40       155      31       ---      --        250      90        248      62        ---      --          200    280          186    279          ---    ---           140     100           124      93           ---     ---            160    ...            155            ---

Therefore $\dfrac 1 {31} = 0 \cdotp \dot 03225 \, 80645 \, 1612 \dot 9$, with an odd period of $15$.

The prime numbers less than $31$ are $2, 3, 5, 7, 11, 13, 17, 19, 23, 29$.

We have:
 * $\dfrac 1 2 = 0 \cdotp 5$: not recurring.


 * $\dfrac 1 3 = 0 \cdotp \dot 3$: recurring with period $1$.


 * $\dfrac 1 5 = 0 \cdotp 2$: not recurring.


 * $\dfrac 1 7 = 0 \cdotp \dot 14285 \dot 7$: recurring with period $6$.


 * $\dfrac 1 {11} = 0 \cdotp \dot 0 \dot 9$: recurring with period $2$.


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


 * $\dfrac 1 {17} = 0 \cdotp \dot 058823529411764 \dot 7$: recurring with period $16$.


 * $\dfrac 1 {19} = 0 \cdotp \dot 05263157894736842 \dot 1$: recurring with period $18$.


 * $\dfrac 1 {23} = 0 \cdotp \dot 043478260869565217391 \dot 3$: recurring with period $22$.


 * $\dfrac 1 {29} = 0 \cdotp \dot 034482758620689655172413793 \dot 1$: recurring with period $28$.

All of the above either terminates, has period $1$ or has even periods.