Question

Let(n) denotes the sum of the digits of n For example s(197) 1 + 9 + 7 = 17. Let s^2 (n) = s(s(n)) and s3 = s(s(s(n)) and so on of s^ 1996 (1996)​

Answer 1

Given that $S(n)$ is the sum of digits of the number $n$ and $S^{k+1}(n)=S(S^k(n))=S^k(S(n))\quad\!\forall k\in\mathbb{N}.$

But if $n$ is a one digit number, the sum of its digits is equal to the number $n$ itself.

$\longrightarrow S(n)=n$

and so,

$\longrightarrow S^2(n)=S(S(n))$

$\longrightarrow S^2(n)=S(n)$

$\longrightarrow S^2(n)=n$

And,

$\longrightarrow S^3(n)=S(S^2(n))$

$\longrightarrow S^3(n)=S(n)$

$\longrightarrow S^3(n)=n$

Assume,

$\longrightarrow S^k(n)=n\quad\!\forall k\in\mathbb{N}$

Let's check if $S^{k+1}(n)=n$ is true.

$\longrightarrow S^{k+1}(n)=S(S^k(n))$

By our assumption,

$\longrightarrow S^{k+1}(n)=S(n)$

$\longrightarrow S^{k+1}(n)=n$

We see it's true, therefore, $S^k(n)=n\quad\forall k\in\mathbb{N}$ if $n$ is a one digit number.

So,

$\longrightarrow S^{1996}(1996)=S^{1995}(S(1996))$

$\longrightarrow S^{1996}(1996)=S^{1995}(1+9+9+6)$

$\longrightarrow S^{1996}(1996)=S^{1995}(25)$

$\longrightarrow S^{1996}(1996)=S^{1994}(S(25))$

$\longrightarrow S^{1996}(1996)=S^{1994}(2+5)$

$\longrightarrow S^{1996}(1996)=S^{1994}(7)$

Since 7 is a one digit number,

$\longrightarrow\underline{\underline{S^{1996}(1996)=7}}$

Hence 7 is the answer.