# 101 is Smallest Number whose Period of Reciprocal is 4

## Theorem

$101$ is the first positive integer the decimal expansion of whose reciprocal has a period of $4$:

$\dfrac 1 {101} = 0 \cdotp \dot 009 \dot 9$

## Proof

Let the positive integer reciprocal be $\dfrac1 k$ for some $k \in \Z_{\ge 0}$.

For it to have a period of recurrence of $4$ in base $10$, it must be able to be expressed as $\dfrac a {10^4-1}$ for some $a \in \Z_{\ge 0}$.

 $\ds 10^4-1$ $=$ $\ds \paren { 10^2 - 1 } \paren { 10^2 + 1 }$ Difference of Two Squares $\ds$ $=$ $\ds 99 \cdot 101$

Since this rational number must have a period of recurrence minimum of $4$, it cannot also have a period of $2$ or $1$. It will be sufficient for this proof to only consider periods of $2$. This rational number must not be able to be expressed as $\dfrac b {10^2-1}$ for any $b \in \Z_{\ge 0}$.

 $\ds \dfrac1 k$ $=$ $\ds \dfrac a {99 \cdot 101}$ $\ds \leadsto \ \$ $\ds a k$ $=$ $\ds 99 \cdot 101$

$\dfrac1 k \neq \dfrac b {10^2-1}$

$\leadsto \dfrac a {99 \cdot 101} \neq \dfrac b {99}$

$\leadsto a \neq 101 b$

Since $101$ is prime, $a$ is not a multiple of $101$ but $ak$ is, then $k$ must be a multiple of $101$. $k = 101$ is the first instance which satisfies these conditions and produces a reciprocal with a period of recurrence of 4.

$\blacksquare$