How to get smallest positive integer N?

Let $T(N) = N + S(N) + S(S(N))$ . Since neither $S(N)$ nor $S(S(N))$ can be zero, we must have $N ≤ 97$ . Thus $S(N) ≤ 17$ (using $N = 89$ ) and $S(S(N)) ≤ 9$ . This gives a crude lower bound for $N$ of $99-(17 + 9) = 73$ . If $S(N +1)$ has the same number of digits as $S(N)$ , then $T(N +1) = T(N)+3$ since $S(N + 1)$ will be one more than $S(N)$ and $S(S(N + 1))$ will be one more than $S(N)$ . Moreover, as long as $S(N + k)$ continues to have the same number of digits as $S(N)$ , we have $T(N + k)= T(N) + 3k$ . The first integer $M$ greater than $73$ such that $S(M)$ has only one digit is $M = 80$ . Thus the formula will work from $73$ up to $79$ , so for $1 ≤ k ≤ 6$ . We have $T(73) = 73 + S(73) + S(S(73)) = 73 + 10 + 1 = 84$ , and $99 - 84 = 15 = 3 · 5$ . Thus $99 = T(73) + 3 · 5 = T(78)$ . Therefore $N = 78$ is the smallest integer with $T(N) = 99$ .

I know it's a strange solution but... It's just how I solved it:

Case 1:

$S(N)$ is a one-digit integer. Then, we have that $N = 99 - 2S(N)$ (Because in this assumption we know that as $S(N)$ has one digit, $S(S(N)) = S(N)$ . If we check for $S(N)$ in the range $1 \leq S(N) \leq 9$ , we will get eventually the solution $N = 81$ .

Case 2:

$S(N)$ is a two-digit integer. In this case, we note various things. First, $N < 100$ in order to comply the equation given. Second, as $N < 100$ , the highest value $S(N)$ could take is $18$ , and this tells us that $S(S(N))$ will be a one digit number. Then, let's express our $N$ as $\overline{ab}$ , and let's express $S(S(N))$ as some one-digit integer $c$ . We know that $\overline{ab} = 10a + b$ . Also, $S(N)$ will simply be $a + b$ , and $S(S(N)) = c$ . Our equation is:

$11a + 2b + c = 99$ .

With some mental testing we can see that this equation requires $N$ to be at least bigger than $70$ . This is why I set $a = 7$ . The equation turns to be:

$2b + c = 22$ .

Here it's much easier, as we just have to test $7 \leq b \leq 9$ (otherwise, we would make $c$ be a two-digit integer). We get the following three triplets $(N, S(N), S(S(N)))$ :

$(77, 14, 5), (78, 15, 6), (79, 16, 7)$ .

Recall that after all $S(S(N)) = c$ . So here the only solution that isn't contradicting this equation (and that satisfies the original equation of the problem) is the second triplet. Hence, the answer is

$S(N) = \boxed{15}$ .

Obviously I'm open to any comment! :)