# 31 is Smallest Prime whose Reciprocal has Odd Period

## Theorem

$31$ is the smallest prime number to have a decimal expansion of the reciprocal with an odd period greater than $1$:

$\dfrac 1 {31} = 0 \cdotp \dot 03225 \, 80645 \, 1612 \dot 9$

## Proof

From Reciprocal of $31$:

$\dfrac 1 {31} = 0 \cdotp \dot 03225 \, 80645 \, 1612 \dot 9$

Counting the digits, it is seen that this has a period of recurrence of $15$, an odd integer.

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

We investigate the reciprocal of each of these:

 $\ds \dfrac 1 2$ $=$ $\ds 0 \cdotp 5$ Reciprocal of $2$: not recurring $\ds \dfrac 1 3$ $=$ $\ds 0 \cdotp \dot 3$ Reciprocal of $3$: period $1$ $\ds \dfrac 1 5$ $=$ $\ds 0 \cdotp 2$ Reciprocal of $5$: not recurring $\ds \dfrac 1 7$ $=$ $\ds 0 \cdotp \dot 14285 \dot 7$ Reciprocal of $7$: period $6$ $\ds \dfrac 1 {11}$ $=$ $\ds 0 \cdotp \dot 0 \dot 9$ Reciprocal of $11$: period $2$ $\ds \dfrac 1 {13}$ $=$ $\ds 0 \cdotp \dot 07692 \dot 3$ Reciprocal of $13$: period $6$ $\ds \dfrac 1 {17}$ $=$ $\ds 0 \cdotp \dot 05882 \, 35294 \, 11764 \dot 7$ Reciprocal of $17$: period $16$ $\ds \dfrac 1 {19}$ $=$ $\ds 0 \cdotp \dot 05263 \, 15789 \, 47368 \, 42 \dot 1$ Reciprocal of $19$: period $18$ $\ds \dfrac 1 {23}$ $=$ $\ds 0 \cdotp \dot 04347 \, 82608 \, 69565 \, 21739 \, 1 \dot 3$ Reciprocal of $23$: period $22$ $\ds \dfrac 1 {29}$ $=$ $\ds 0 \cdotp \dot 03448 \, 27586 \, 20689 \, 65517 \, 24137 \, 93 \dot 1$ Reciprocal of $29$: period $28$

and it is seen that none has an odd period greater than $1$.

Hence the result.

$\blacksquare$