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 #56.

“Considering natural numbers of the form, a^{b}, where a, b < 100, what is the maximum digital sum?”

I’ll be reusing the code originally started in “Power digit sum” to form the base of this puzzle. The next part is to take the power of a 100 of each digit and check which consecutive iteration of each vector contains the most digits (so: multiply by the same vector each time).

```
fn max_power_digit_sum(digit: u128) -> u64 {
let mul_digits = digit.to_digits();
let mut digits = digit.to_digits();
let mut max = 0;
for _ in 2..=100 {
digits.mul_vec(&mul_digits);
let sum = digits.iter().fold(0, |acc, &n| acc + n as u64);
if sum > max {
max = sum;
}
}
max
}
#[test]
fn test_max_power_digit_sum() {
assert_eq!(max_power_digit_sum(10), 1);
assert_eq!(max_power_digit_sum(99), 972);
}
```

To resolve problem 56 do the following:

```
fn problem_56() -> u64 {
let mut max = 0;
let mut a = 99;
while a > 2 {
let pds = max_power_digit_sum(a);
if pds > max {
max = pds
}
a -= 1;
}
max
}
#[test]
fn test_problem_56() {
assert_eq!(problem_56(), 972);
}
```

The code is not fast, but it results in the right answer. It resolves in 9.16s.

**
The full solution is available on
GitHub.
**