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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.

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You might put it this way. Two natural numbers which are same modulo 9 have the same RD. In fact a natural number is the same as its RD modulo 9. Hint: Any integer $N$ can be written as $10^na_n+10^{n-1}a_{n-1}+...+10a_1+a_0$ Here, clearly $a_i$ are the digits of $N$. Just show that $N$ is the same as its digit sum, mod 9. Then by induction you'll be done! –  Host-website-on-iPage Jul 10 '12 at 16:16
    
See this. –  J. M. Jul 10 '12 at 16:19
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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.

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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 $(1)$.

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