In order to generate Pythagorean Triples we can use Euclid's Formula

The problem with this approach is that if we change the range, say from [1 - 100] to [1 - 1 000 000], it would take a much longer time to calculate. What if we could skip all this and have a much faster approach?

The solution involves what we have seen in previous problems: prime factorization. Let's look at the example given in the problem.

The solution here may not be the most efficient, but it works. The program loops through all 3-digit number products, converts them to a string, reverses the string, and then checks if the reversed and unreversed strings are equivalent.

If you understand how a Fibonacci sequence works, then the code below is understandable. The solution has been written in both Python and Rust, and either one will work.

We have to find all the multiples of 3 or 5 below 1000 and get the sum. One thing to be careful of are multiples of 15, as they are multiples of both 3 and 5, and we want multiples of 3 or 5.