Japan Mathematical Olympiad

[18]
First, let us show that $g$ cannot exceed $18$ for any $n$. Denote by $a, b, c$ the $3$ digits of $n$, where we assume $a < b < c$. Both $100c+10b+a$ and $100c+10a+b$ are numbers obtained by permuting the digits of $n$. Hence $g$ is a divisor of $(100c+10b+a) - (100c+10a+b) = 9(b-a)$. Similarly, we get that $g$ is a divisor of $9(c-b)$ and of $9(c-a)$. If we set $x = b-a$, $y = c-b$, $z = c-a$, then $x, y, z$ are positive integers not exceeding $8$, and satisfy $x+y=z$. If we denote by $g'$ the greatest common divisor of $x, y, z$ then $g$ is a divisor of $9g'$.

(1) If $g' \ge 5$, there exists at most $1$ number less than or equal to $8$ divisible by $g'$, and we get a contradiction to the fact that $x + y = z$. So, $g' \ge 5$ is impossible.

(2) If $g' = 4$, then $4$ and $8$ are the only positive integers not bigger than $8$ and divisible by $4$, so we must have $(x, y, z) = (4, 4, 8)$. We then have $(a, b, c) = (1, 5, 9)$ and $g = 3$.

(3) If $g' = 3$, then $3$ and $6$ are the only positive integers not bigger than $8$ and divisible by $3$, so we must have $(x, y, z) = (3, 3, 6)$. Then $(a, b, c)$ must be $(1, 4, 7)$ or $(2, 5, 8)$ or $(3, 6, 9)$ and the value of $g$ is $3$, $3$, $9$, respectively.

(4) If $g' \le 2$, then since $g$ divides $9g'$, we must have $g \le 9g' \le 18$.

Thus we have shown that $g \le 18$. On the other hand if $n = 468$, then $g = 18$ and this shows that $18$ is the maximum possible value for $g$.