Period of Reciprocal of 53 is One Quarter of Maximal

Theorem

The decimal expansion of the reciprocal of $53$ has $\dfrac 1 4$ the maximum period, that is: $13$:

$\dfrac 1 {53} = 0 \cdotp \dot 01886 \, 79245 \, 28 \dot 3$

Proof

From Reciprocal of $53$:

$\dfrac 1 {53} = 0 \cdotp \dot 01886 \, 79245 \, 28 \dot 3$

Counting the digits, it is seen that this has a period of recurrence of $13$.

From Maximum Period of Reciprocal of Prime, the maximum period of recurrence of $\dfrac 1 p$ is $p - 1$.

We have that:

$13 = \dfrac {53 - 1} 4$

Hence the result.

$\blacksquare$

Sources

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $53$
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $53$