As guriyeller correctly observes, for all numbers greater than 1, multiplying will give the maximum result. So I special-cased where numbers are equal to 1, because then it's better to add them to another multiplier. For a and c that can only be to b, for b it depends on whether a<c or not; (a+1)c==ac+c and a(c+1)==ac+a. I made use of invariants throughout the various steps to arrive at a compact final expression, carefully avoiding overwriting values. All paths that add to variables make sure the correct variables are still 1 to not affect the result.

i love your solution, you must be an Obfuscated C code contestant ;-)

you inspired my insanity ;)

@roswold. But that's not shorter unless you are counting whitespace. You gained the,

Frack me, it looks awful.

I... I think I need a proof.

Good question! Good question indeed... ))

@pvtroswold

Looks Yummy , Revised and APPROVED

This is my solution and Troj2 is a copycat (copied solutions in other kata as well).

You can figure out how this solution works by reading this comment (you need to solve a simple 8 kyu kata first).

NASM 2.11 on Linux, assembled with nasm -f elf64.

https://github.com/Codewars/codewars.com/wiki/Language-NASM

Maybe someone can help you if you pasted your solution marking with spoiler flag.