https://docs.oracle.com/javase/8/docs/api/java/util/Set.html
"A collection that contains no duplicate elements. More formally, sets contain no pair of elements e1 and e2 such that e1.equals(e2), and at most one null element."

Good peformance, and in addition very readable.

Good solution. Even more you could have saved some cpu time had you used TreeSet instead of HashSet. In this case you don't need the line -> Arrays.sort(resultArray). And yes I'm aware that TreeSet takes more time than HashSet but still thats better for this case.

Nothing. Regular string concatenation faster that String#format. You can find in google performance of regular concatenation and String#format.

I do not understand how it works, can anyone explain?

Looks like you are using two for inner loops, which will end up with O(N^2) as brute-force solution. You need to improve the performance to at least O(N).

This is a cool result. I think it needs an explaination. Maybe pointer to the proof.

This is great, but you for each number you check if it is divisible. My name is C4rbyn3m4n look at my code. It is only 2 lines of code and runs more efficiently. If you like it please upvote it. Thanks and have fun programming!

If you use a TreeSet, why do you check if the String is allready in the TreeSet before adding it?

What is the advangage of using String.format over regular concatenation?

Why concat a string when you could simply set it? This is (a little bit) slower, because it creates a StringBuilder on the fly to add "" and the other String.

Thanks for the feedback, I updated the details to make it clear that there will always be only one odd integer.

Better output messages: was true or false expect?