Competition problem — Algebra — Maths Olympiad Prep

Library / /22 of 42

Algebra Difficulty 5.4 AIME, harder Prove it Ireland

Let $f : \mathbb{R} \to \mathbb{R}$ be the fractional part function, defined by $f(x) = x - \lfloor x \rfloor$ , where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ . For instance, $f(5.4) = 0.4$ and $f(-5.4) = 0.6$ .
Let $g : \mathbb{R} \to \mathbb{R}$ be defined by $g(x) = f(1 - 2022x)$ . Find 2023 distinct rational numbers $x$ satisfying $g(x) = x$ . Each such solution $x$ should be explicitly written in the form $m/n$ for some integers $m, n$ .

Solution

The fractional part $f(x)$ of a number $x$ is characterised by
$f(x) \in [0, 1) \text{ and } x - f(x) \in \mathbb{Z}.$
The equation $g(x) = x$ , i.e. $f(1 - 2022x) = x$ , is therefore equivalent to
$x \in [0, 1) \text{ and } (1 - 2022x) - x = 1 - 2023x \in \mathbb{Z}.$
If $x = \frac{m}{n}$ is a rational number, this is equivalent to
$0 \le m < n \text{ and } \frac{2023m}{n} \in \mathbb{Z}.$
If we choose $n = 2023$ , then the 2023 integers $m$ that satisfy $0 \le m < 2023$ give an $x = \frac{m}{n}$ that fulfils the desired condition.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.