HMMT November, 2015

Solution:
Let $n = \overline{abc}$, and assume without loss of generality that $a \geq b \geq c$. We have $k \mid 100a + 10b + c$ and $k \mid 100a + 10c + b$, so $k \mid 9(b - c)$. Analogously, $k \mid 9(a - c)$ and $k \mid 9(a - b)$. Note that if $9 \mid n$, then $9$ also divides any permutation of $n$'s digits, so $9 \mid f(n)$ as well; ergo, $f(n) \geq 9$, implying that $k \geq 9$. If $k$ is not a multiple of $3$, then we have $k \mid c - a \Longrightarrow k \leq c - a < 9$, contradiction, so $3 \mid k$.

Let $x = \min(a - b, b - c, a - c)$. If $x = 1$, then we have $k \mid 9$, implying $k = 9$—irrelevant to our investigation. So we can assume $x \geq 2$. Note also that $x \leq 4$, as $2x \leq (a - b) + (b - c) = a - c \leq 9 - 1$, and if $x = 4$ we have $n = 951 \Longrightarrow f(n) = 3$. If $x = 3$, then since $3 \mid k \mid 100a + 10b + c \Longrightarrow 3 \mid a + b + c$, we have $a \equiv b \equiv c \pmod{3}$ (e.g., if $b - c = 3$, then $b \equiv c \pmod{3}$, so $a \equiv b \equiv c \pmod{3}$—the other cases are analogous). This gives us the possibilities $n = 147, 258, 369$, which give $f(n) = 3, 3, 9$ respectively.

Hence we can conclude that $x = 2$; therefore $k \mid 18$. We know also that $k \geq 9$, so either $k = 9$ or $k = 18$. If $k = 18$, then all the digits of $n$ must be even, and $n$ must be a multiple of $9$; it is clear that these are sufficient criteria. As $n$'s digits are all even, the sum of them is also even, and hence their sum is $18$. Since $a \geq b \geq c$, we have $a + b + c = 18 \leq 3a \Longrightarrow a \geq 6$, but if $a = 6$ then $a = b = c = 6$, contradicting the problem statement. Thus $a = 8$, and this gives us the solutions $n = 882, 864$ along with their permutations.

It remains to calculate the sum of the permutations of these solutions. In the $n = 882$ case, each digit is either $8, 8$, or $2$ (one time each), and in the $n = 864$ case, each digit is either $8, 6$, or $4$ (twice each). Hence the desired sum is $111(8 + 8 + 2) + 111(8 \cdot 2 + 6 \cdot 2 + 4 \cdot 2) = 111(54) = 5994$.