Recursive digit-sum Let the recursive digit-sum(R.D.) be defined as: continue taking the sum of digits until it becomes <10.
For example, the digit-sum of 5987 = 29, the digit-sum of 29 =11
So, R.D. of 5987 is 2.
Prove that the value of R.D. recurs after each 9 numbers i.e., R.D. of any natural numbers of the form (9.a+b) where 0≤b<9 are same.
 A: R.D of a number $n$ is nothing but the value  $n\pmod9$ which repeats after every $9th$ number. Thus, R.D sum of $n$ and $9k+n$ is same.
A: Recursive digit-sum of an integer gives you the remainder when the number is divided by $9$ (with the exception of $9$ which indicates that the number is indeed divisible by $9$) In other words, $ \overline{a_0a_1a_2...a_n} \equiv 10^na_0+10^{(n−1)}a_{1}+...+10a_{n-1}+a_n  \equiv a_1+a_2+ ... +a_n (mod \ \ 9)$ 
Hence all the numbers of the form $9k+r_0$ where $k \in \mathbb{Z}$ is arbitrary and  $r_0 \in \{1, 2, ... , 9 \}$ is fixed, will share the same recursive digit-sum $r_0$ .... $(1)$
Given two numbers which differ by $9$, we know that they will have the same remainders when divided by $9$. Hence the claim follows from $(1)$.
A: You'll see that your Recursive Digit Sum is similar to the Digit Root.

Definition 1: The Digit Root of a number $n$ is defined by:
$\operatorname{dr}(n) = \begin{cases}
n \pmod9\ & \operatorname{if}\ n \not\equiv 0 \pmod9  \\
                     9\ &\operatorname{if}\ n \equiv 0 \pmod9 
       \end{cases}$
And more generally $\operatorname{dr}(n) = 1 + \Big(n-1 \pmod 9\Big)$
For more information, see Digit Root - MathWorld

Now, prove that  $\operatorname{dr}(n) =  \operatorname{dr}(n + 9)$
$\underline{\textbf{Proof}}$: Because $\operatorname{dr}(n) = 1 + \Big(n-1 \pmod 9\Big)$ and $n + 9 \pmod 9 \iff n \pmod 9 $. So $\operatorname{dr}(n + 9) = 1 + \Big(n + 9 - 1 \pmod 9\Big) = 1 + \Big(n-1 \pmod 9\Big)\ \square$.
And we're done !
