# 103 is Smallest Prime whose Period of Reciprocal is One Third of Maximal

## Theorem

$103$ is the smallest prime number the period of whose decimal expansion of its reciprocal is $\dfrac 1 3$ of its maximum, that is: $34$:

$\dfrac 1 {103} = 0 \cdotp \dot 00970 \, 87378 \, 64077 \, 66990 \, 29126 \, 21359 \, 223 \dot 3$

## Proof

From Reciprocal of $103$:

$\dfrac 1 {103} = 0 \cdotp \dot 00970 \, 87378 \, 64077 \, 66990 \, 29126 \, 21359 \, 223 \dot 3$

and so by counting it can be seen that its period of recurrence is $34$.

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 $7$, $13$, $19$, $31$, $37$, $43$, $61$, $67$, $73$, $79$ and $97$.

We have:

which shows that each of these primes has maximum period $p - 1$.

We have:

Reciprocal of $13$: $\dfrac 1 {13} = 0 \cdotp \dot 07692 \dot 3$: recurring with period $6 = \dfrac {p - 1} 2$.
Reciprocal of $31$: $\dfrac 1 {31} = 0 \cdotp \dot 03225 \, 80645 \, 1612 \dot 9$: recurring with period $15 = \dfrac {p - 1} 2$.
Reciprocal of $37$: $\dfrac 1 {37} = 0 \cdotp \dot 02 \dot 7$: recurring with period $3 = \dfrac {p - 1} {12}$.
Reciprocal of $43$: $\dfrac 1 {43} = 0 \cdotp \dot 02325 \, 58139 \, 53488 \, 37209 \dot 3$: recurring with period $21 = \dfrac {p - 1} 2$.
Reciprocal of $67$: $\dfrac 1 {67} = 0 \cdotp \dot 01492 \, 53731 \, 34328 \, 35820 \, 89552 \, 23880 \, 59 \dot 7$: recurring with period $33 = \dfrac {p - 1} 2$.
Reciprocal of $73$: $\dfrac 1 {73} = 0 \cdotp \dot 01369 \, 86 \dot 3$: recurring with period $8 = \dfrac {p - 1} 9$.
Reciprocal of $79$: $\dfrac 1 {79} = 0 \cdotp \dot 01265 \, 82278 \, 48 \dot 1$: recurring with period $13 = \dfrac {p - 1} 6$.

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

$\blacksquare$