333,667 is Unique Period Prime with Period 9
Theorem
$333 \, 667$ is a unique period prime whose reciprocal has a period of $9$:
- $\dfrac 1 {333 \, 667} = 0 \cdotp \dot 00000 \, 299 \dot 7$
Proof
By long division:
0.000002997000002... --------------------- 333667)1.000000000000000000 667334 -------- 3326660 3003003 ------- 3236570 3003003 ------- 2335670 2335669 ------- 1000000 667334 ------- ......
It remains to be shown that $333 \, 667$ is the only prime number whose reciprocal has a period of $9$.
We have that:
- $333 \, 667 \nmid 10$
From Period of Reciprocal of Prime, the period of such a prime is the order of $10$ modulo $p$.
That is, the smallest integer $d$ such that:
- $10^d \equiv 1 \pmod p$
From the above long division we know that the period of $\dfrac 1 {333 \, 667}$ is $9$, so $10^9 \equiv 1 \pmod {333 \, 667}$.
The only other possible primes $p$ whose reciprocals might have a period of $9$ must also satisfy:
- $10^9 \equiv 1 \pmod p$
that is:
- $p \divides 10^9 - 1$
Consider:
\(\ds 10^9 - 1\) | \(=\) | \(\ds 999 \, 999 \, 999\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3^4 \times 37 \times 333 \, 667\) | prime factorization |
Therefore the only other possible primes whose reciprocals might have a period of $9$ are $3$ and $37$.
From Reciprocal of $37$:
- $\dfrac 1 {37} = 0 \cdotp \dot 02 \dot 7$
and trivially:
- $\dfrac 1 3 = 0 \cdotp \dot 3$
which has a period of $1$.
As required, the only prime number whose reciprocal has a period of $9$ is $333 \, 667$.
$\blacksquare$
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $333,667$