Find the values of $f^{2018}(n)$

My solution is similar to Brian's, but not identical.

If the three digits of $n$ are identical, then $f(n)=0$ and $f^k(n)=0$ for all $k$ . From now on let's assume that $n$ has three digits $a,b,c$ , in any order, with $a\geq b\geq c$ and $a>c$ . Then $f(n)=100a+10b+c-(100c+10b+a)=99(a-c)$ . Thus $f(n)$ will be one of the nine numbers $99k$ for $k=1,...,9$ . Note that the middle digit of $99k$ is always $9$ , the first is $k-1$ and the last is $10-k$ . The largest digit is $a=9$ and the smallest digit $c$ is the minimum of $k-1$ and $10-k$ .

For $k=1,2,3,4,5$ , the smallest digit of $99k$ is $c=k-1$ , so $f(99k)=99(a-c)=99(10-k)$ . In particular, $f(5\times99)=f(495)=495$ . For $k=6,7,8,9$ , the smallest digit is $c=10-k$ , so $f(99k)=99(a-c)=99(k-1)$ .

It follows that we will reach the fixed point 495 after at most 6 iterations, and certainly after 2018. The longest chain of distinct numbers is generated when the "seed" maps to 99, as in $100 \rightarrow 99 \rightarrow 9\times 99=891 \rightarrow 8\times 99 =792 \rightarrow 7\times 99 = 693 \rightarrow 6\times 99=594 \rightarrow 5\times 99 = 495.$

Thus the answer is $0+495=\boxed{495}$

We show that $f^{2018}(n) = 0, 495$ . We will see that the number $2018$ is not too relevant; the answer is the same for any iteration of at least $6$ .

Given $n \in \mathbb{N}_{\leq 999}$ , let its digits be $a \geq b \geq c$ . If $a = b = c$ , then $n = \overline{aaa}$ so that $f(n) = \overline{aaa} - \overline{aaa} = 0$ . As $f(0) = 0$ , it follows that $f^{2018}(n) = 0$ . Otherwise, not all digits of $n$ are distinct, whereupon we apply the following useful result.

Lemma : Let $m$ be a number with digits $a \geq b > c$ . Then $f(m) = \overline{d9e}$ where $d = a - c - 1$ and $e = 9 - d$ . Furthermore, $d < 9$ and $e > 0$ .

Proof : By the definition of $f$ , note that $\begin{aligned}f(m) & = \overline{abc} - \overline{cba} \\ & = 100a + 10b + c - 100c - 10b - a \\ & = 100(a - c) + c - a \\ & = 100(a - c - 1) + 10(9) + (10 - (a - c)) && \text{since } c - a < 0 \\ & = \overline{d9e}, \end{aligned}$ where $d = a - c - 1$ and $e = 10 - (a - c) = 9 - d$ . Note that since $a \leq 9$ and $c \geq 0$ , $a - c \leq 9$ , and since $a > c$ by assumption, $a - c > 0$ . This implies $1 \leq a - c \leq 9$ , which gives $0 \leq d \leq 8$ and $1 \leq e \leq 9$ (this last statement also shows that $d$ and $e$ are indeed digits). $\blacksquare$

Corollary : If a number $m$ 's digits are $4$ , $5$ , and $9$ in any order, then $f(m) = 495$ . In particular, $495$ is a fixed point of $f$ .

Proof : By the lemma, $f(m) = \overline{d9e}$ where $d = 9 - 4 - 1 = 4$ and $e = 9 - d = 5$ . $\blacksquare$

Letting the digits of $n$ be denoted by $a \geq b > c$ , we have by the lemma that $f(n) = \overline{d9e}$ with $e = 9 - d$ . Note that the corollary classifies the behavior of $f$ when $d = 4$ ; in that case $f(n) = \overline{d9e} = 495$ , so since $f(495) = 495$ , $f^{2018}(n) = 495$ . We consider the remaining two cases.

• If $5 \leq d \leq 8$ , then $9 > d > e > 0$ , so by the lemma, $f(\overline{d9e}) = \overline{d'9e'}$ where $d' = 9 - e - 1 = d - 1$ and $e' = 9 - d'$ . In particular, the new hundreds digit has decreased by exactly one. Thus, it takes $d - 4$ applications of $f$ to $\overline{d9e}$ before the hundreds digit becomes $4$ , at which point we will have reached the fixed point at $495$ . In other words, $f^{d - 4}(\overline{d9e}) = 495$ . Thus, since $f(n) = \overline{d9e}$ , $f^{d - 3}(n) = 495$ . By the corollary, we have $f^{2018}(n) = 495$ .
• If $0 \leq d \leq 3$ , then $6 \leq e \leq 9$ . This implies $9 \geq e > d + 2 > d$ , so by the lemma, $f(\overline{d9e}) = d'9e'$ where $d' = 9 - d - 1 = e - 1$ and $e' = 9 - d' = d + 1$ . In particular, since $e > d + 2$ , we have $d' > e'$ , which puts the number $\overline{d'9e'}$ precisely in the first case, since now $5 \leq d' \leq 8$ . By the same argument in the first case, it takes $d' - 4$ applications of $f$ to $\overline{d'9e'}$ to reach $495$ . Since $f^{d' - 4}(\overline{d'9e'}) = 495$ and $f^2(n) = \overline{d'9e'}$ , we have $f^{d' - 2}(n) = 495$ . By the corollary, we have $f^{2018}(n) = 495$ .

Therefore, when it is not the case that $a = b = c$ , $f^{2018}(n) = 495$ . This makes the answer $0 + 495 = \boxed{495}$ .

As an additional fact: the proof also shows that for any $n \in \mathbb{N}_{\leq 999}$ , either $f(n) = 0$ or $f^6(n) = 495$ .