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.

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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