Project Euler #52: Permuted multiples

This article is part of a series where I'll be diving head first into the Project Euler puzzles. I want to document the challenge of solving such a puzzle and how I got to the answer. I want to prefix this by stating that I can't cheat for any of these challenges; with that I mean I can't look up any other implementations online. After the implementation, I will validate the answer by using this document or a similar sheet.

In this article I'll be solving: Project Euler #52.


“It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.

Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.”


I’ll reuse the unique_digits method from the previous Euler puzzle and slightly alter it to be more in line with int_to_vec() of “Power digit sum”. The next thing I did was to sort the digits that come out in the end.

To resolve problem 52:

fn problem_52() -> u64 {
    let mut x = 1;

    loop {
        let dx = to_digits(x);

        if (2..=6).all(|m| to_digits(x * m) == dx) {
            break x

        x += 1

fn test_problem_52() {
    assert_eq!(problem_52(), 142857);

Solved! This one was too easy.

The full solution is available on GitHub.