Well adding 9 is like incrementing the tens place by 1 and decrementing the ones place by 1

17 + 9 =

26

119 + 9 =

128

etc

so adding a multiple of 9 is like doing that multiple times

adding 27 (9*3) is like incrementing the tens place by 3 and decrementing the ones place by 3

example:

55 + 27 =

82

I'm still trying to see how this relates to this riddle, but I think there's probably a connection

I did sorta find one in the other post:

abc+cba = m9 + 2(cba)

where m9 is any multiple of 9, abc is a number (of any amount of digits, i made it abc arbritrarily) and cba is its inverse

on a different point, what IS inversing?

abc would be a*b*c, so what I mean by abc must be:

100a+10b+c

flipped is:

100c+10b+a

so in a three-digit number, flipping means adding 99c and subtracting 99a

is there some sort of pattern we can get for x amount of digits?

a = a, no change needed, keep a the same

10a+b = 10b+a, you need to add 9b and subtract 9a

100a+10b+c = 100c+10b+a, you need to add 99c and subtract 99a, keep be the same

1000a+100b+10c+d = 1000d+100c+10b+a, you need to add 999d, add 90c, subtract 90b and subtract 999a

10000a+1000b+100c+10d+e = 10000e+1000d+100c+10b+a, you need to add 9999e, add 990d, keep c the same, subtract 990b, subtract 9999a

yeah I see a pattern emerging....

I agree unreality. You're definitely onto something here. I'm going to play with those formulas and see if I can come up with something.