Well adding 9 is like incrementing the tens place by 1 and decrementing the ones place by 1
17 + 9 =
119 + 9 =
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
55 + 27 =
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:
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.