28th Hellenic Mathematical Olympiad

We consider the decimal representation of the integers $$\begin{array}{rl}x-y& =1000\alpha +100\beta +10\gamma +\delta -1000\delta -100\gamma -10\beta -\alpha \\ & =999(\alpha -\delta )+90(\beta -\gamma )=9(111(\alpha -\delta )+10(\beta -\gamma )).\end{array}$$ Therefore it is enough to find the greatest and the lowest values of the expression: $$A=111(\alpha -\delta )+10(\beta -\gamma ).$$ Since $\alpha ,\beta ,\gamma ,\delta$ are digits pairwise different and no zero, we must have $\alpha >\delta$. The expression $A$ gets its maximal value when the integers $\alpha -\delta$ and $\beta -\gamma$ take their maximal value and moreover $\alpha -\delta >\beta -\gamma$. That is, $\alpha -\delta$ becomes maximal when $\alpha =9$ and $\delta =1$. The difference $\beta -\gamma$ becomes maximal when $\beta =8$ and $\gamma =2$. Hence $x=9821$, $y=1289$ and $x-y=9821-1289=8532$. The expression $A$ gets its minimal value when the integers $\alpha -\delta$ and $\beta -\gamma$ take their minimal values. The minimal value of the difference $\alpha -\delta$ is $1$ and so the pair $(\alpha ,\delta )$ has as possible values: $$(9,8),(8,7),(7,6),(6,5),(5,4),(4,3),(3,2)\text{ and }(2,1).$$ For all the above possible values of the pair $(\alpha ,\delta )$, the value of $A$ is: $$A=111+10(\beta -\gamma ).$$ The lowest value of the difference $\beta -\gamma$ is $-8$, for $\beta =1$ and $\gamma =9$. Taking in mind that the digits must be different we find the following table of possible values:

$x$	$y$	$x-y$
3192	2913	279
4193	3914	279
5194	4915	279
6195	5916	279
7196	6917	279
8197	7918	279