## My solution to the Problem Of the Week-9 [Todd and Vishal’s blog]: Period of a decimal expansion

pdf: included in Caderno (see “caderno” page)

I submitted a solution to the following Problem that  was accepted.

Find the length of the period of the repeating decimal representation of $\dfrac{1}{65537}.$

My Solution:

The repeating decimal representation of the number $1/65537$ is

$\dfrac{1}{65537}=0.\overset{65536\text{ digits}}{\overline{000\,015\,258\,556\ldots cba}}\quad$.

Let $p$ be a prime number. The period of the repeating decimal of $1/p$ is equal to the order  of $10$ $(\mod p\; )$ and is either $p-1$ or a divisor of $p-1.$ Since $65537$ is a prime number, the period of the repeating decimal of $1/65537$ is therefore either $65536$ or a divisor of $65536=2^{16}$. These divisors are

$k=2^{0},2^{1},2^{2},2^{3},\ldots ,2^{16}$.

By the definition of the order of $10$ $(\mod 65537\; )$, I have to find the smallest of these $k=2^{m}$ such that

$10^{k}\equiv 1\; (\mod 65537)$,

which means $(10^{(2^{m})}-1)/65537$ should be an integer.

Since

$10-1<10^{2}-1<10^{3}-1<10^{4}-1<65537$,

the remaining cases are $m=3,4,\ldots ,16$. From these I have checked in PARI that only

$\dfrac{10^{65536}-1}{65537}=669179\ldots 526527$

is an integer (*). For instance

$\dfrac{10^{16}-1}{65537}=\dfrac{999999999999999}{65537}\notin\mathbb{Z}$.

Conclusion. The length of the period of the repeating decimal representation of $\dfrac{1}{65537}$ is $65536.$

(*) Edited a little bit to improve the English text. $\qquad \blacktriangleleft$

A much more sophisticated and elegant solution by Philipp Lampe was posted by one of the authors of this excellent advanced mathematical blog.

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## Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Caderno, Matemática, Math, Number Theory, Problem, Problemas, Teoria dos Números com as etiquetas , , . ligação permanente.

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