Find the period of the repetend of the fraction $\frac{39}{1428}$ by using binary numbers, i.e. its binary decimal representation. (Note: When a proper fraction is expressed as a decimal number (of any base), either the decimal number terminates after finite steps, or it is of the form $0.b_1b_2\cdots b_sa_1a_2\cdots a_ka_1a_2\cdots a_ka_1a_2 \cdots a_k \cdots$. Here the repeated sequence $a_1a_2\cdots a_k$ is called the repetend of the fraction, and the smallest length of the repetend, $k$, is called the period of the decimal number.)
Problem
Source: 2023 Hong Kong TST 3 (CHKMO)
Tags: number theory
MC314159
14.12.2022 14:47
First, reduce this fraction into $\frac{13}{476}$. Note that $476 = 119\cdot 4$ and hence the binary expansion of $\frac{13}{476}$ should be exactly the same as $\frac{13}{119}$ except it is shifted two digits rightwards. Then consider the form given in the question, expand it and we have the sum of two terms, one of which is the sum from the non-repeating part and another is from the repeating part. $119$ should divide the denominator, or the RHS would not be an integer when two sides are multiplied by $119$. Now $119 = 7 \cdot 17$, and it suffices to find $k$ such that $2^k \equiv 1 \pmod{117}$. Now use Fermat-Euler Theorem to find the order of $2$ modulo $117$, which should be exactly $24$.
MC314159
14.12.2022 14:48
btw some dude calculated $\frac{39}{1428}$
analysis90
23.09.2024 18:08
The period is order of $2$ modulo $119$.