# 101 is Smallest Number whose Period of Reciprocal is 4

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## Theorem

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

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

This sequence is A021105 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).

## Proof

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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.

$\blacksquare$