Find the period of the repetend of the fraction 391428 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.b1b2⋯bsa1a2⋯aka1a2⋯aka1a2⋯ak⋯. Here the repeated sequence a1a2⋯ak 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 13476. Note that 476=119⋅4 and hence the binary expansion of 13476 should be exactly the same as 13119 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⋅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.