<s>[INST] Strange sum that always end up with 9 If we have any number, example 4896, and sum all digits
sum = 4+8+9+6 = 27
and than substract this number from the original number, we always get a number that is divisible by 9:
4896-27=4869 -> 4869/9 = 541.
Moreover now everytime i sum the digits from new number (4869), i will get number divisible by 9:
(4+8+6+9=27[divisible])
71 - (7+1) = 63
485 - (4+8+5) = 468/9=52 , 4+6+8 = 18(divisible)

Similarly interesting is with 2 digit numbers where with the same process resulting number is always 9
45 -> 45-9 = 36 (3+6=9)
87 -> 87-15 = 72 (7+2=9)
Its quite beyond my comprehension. Can anyone explain this phenomena?
 [/INST] I will give a formal proof for numbers with $4$ digits. You can think about how to generalize it (Edit: or refer to @fleablood's answer who gave a formal proof for arbitrarily many digits).
Let $n=abcd$ be a number where $0\leq a,b,c,d\leq 9$ are it's digits. For example if $n=4821$ then $a=4,b=8,c=2,d=1$.
Another way to write $n$ is as $n=a\cdot 1000 + b\cdot 100 +c \cdot 10 +d$.
On the other hand the sum of digits is $a+b+c+d$.
so $$n - \text{The sum of digits} = a\cdot 1000 + b\cdot 100 + c\cdot 10 + d -a-b-c-d$$
The right hand side turns out to be $a\cdot 999 + b\cdot 99 + c\cdot 9$ which is divisible by $9$.
The second thing that interested you is that the sum of digits of the new number is always divisible by $9$. For this you should refer to @Ross Milikan answer.
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