The 16th Japanese Mathematical Olympiad - The First Round

We can write $x=10a+b$, $y=10b+c$, $z=10c+a$ with positive integers $a,b,c$ less than $10$. Let $d$ be the greatest common divisor of $x$, $y$ and $z$. $11$ cannot divide $d$, since otherwise it contradicts the fact that $x$, $y$ and $z$ are distinct.

$x+y+z=(10a+b)+(10b+c)+(10c+a)=11(a+b+c)$ is divisible by $d$. Since $d$ is not divisible by $11$, $a+b+c$ is divisible by $d$. Thus $d\le a+b+c\le 27$.

Also, $100x-10y+z=1000a+100b-100b-10c+10c+a=1001a$ is divisible by $d$. Similarly, $1001b$ and $1001c$ are divisible by $d$. So $1001k$ and hence $91k$ is divisible by $d$, where $k$ is the greatest common divisor of $a$, $b$ and $c$.

Since $d\le 27$ and $k\le 4$ (otherwise $x=y=z$), $d$ is one of $1,2,3,4,7,13,14,21$ or $26$.

But, $d$ cannot be $21$, because a 2-digit multiple of $21$ must be $21,42,63$ or $84$, and we can't find $x$, $y$, $z$ among them with the required conditions. Similarly, $d$ cannot be $26$.

On the other hand, since $(x,y,z)=(32,21,13),(64,42,26),(96,63,39),(88,84,48),(42,21,14),(65,52,26)$ and $(84,42,28)$ give examples for $d=1,2,3,4,7,13$ and $14$, there are exactly $7$ such $d$'s.