Pick any three digit number

Question

A Link to it is here: http://qr.ae/Rgrdrk
In short, you take a 3 digit number, subtract the sum of it's digits. Any given digit in the result is equal to the difference between the other two numbers made into a 2-digit number and the next largest multiple of 9, where the order in which the remaining 2 digits does not matter.
I attempted to find how this works and i can get as far as simplifying to 99a+9b where the original number uses a, b, and c as the 100, 10, and 1's digits respectively.
(100 a + 10 b + c)-(a + b + c) = 99 a + 9 b = 9 (11 a + b)
Might this trick also have to do with how the sum of the digits in all multiples of 9 add up to a multiple of 9?

Answer 1

This property is true for all the $3$-digits multiples of $9$: take $abc$ one of this numbers. You have that $a+b+c=9$, and this implies that $$9(a+1)=9a+a+b+c=10a+b+c$$ is a multiple of $9$. In particular $$c=9(a+1)-(10a+b)$$ and this is the property you are looking for.

EDIT: to see that the order doesn't matter, add $9b$ instead of $9a$.

REMARK: You have to be careful with this trick: if you take for example $111$ and $201$ and subtract the sum of the digits you get $108$ and $198$. If you choose the middle digit in both cases the other two are $1$ and $8$. How can you tell if the third digit is $9$ or $0$?