Use binary 111110000 as example...
Add up the powers of 2, then use bit shift/multiplication/addition identities to simplify the sum of all divisors. You'll get the original Perfect number
Divisors: 1, 10, 100, 1000, 10000, 11111, 111110, 1111100, 11111000
(WIP) I'm planning on improving this, then I'll make a YT video or a GHP-blog/post about it