Project Euler #63: Powerful digit counts

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 #63.


The 5-digit number, 16807=75, is also a fifth power. Similarly, the 9-digit number, 134217728=89, is a ninth power.

How many n-digit positive integers exist which are also an nth power?

The solution

Because this problem is quite easy to solve, I’m jumping straight to the solution. The code looks as follows:

let mut count = 0;
let mut start: u128 = 1;
let mut pow = 1;
let mut prev_count = 0;

loop {
    let n = start.pow(pow);
    let n_len = ((n as f64).log10() + 1.0) as u32;

    if pow == n_len {
        count += 1;
    } else if n_len > pow {
        start = 1;
        pow += 1;

        if prev_count == count {

        prev_count = count;

    start += 1;


The main gist is to start an infinite loop, and have two functions that are in the form of n = xy, and n_len = log10(n). If y equals to n_len, it counts towards the total, if it doesn’t increase x, and if y exceeds n_len we can increase y by 1 and reset x to 1.

We keep the loop going until the total count no longer increases for said power. This will result in a total of 49 integers.

The full solution is available on GitHub.